How do I match any character across multiple lines in a regular expression?

264

For example, this regex

(.*)<FooBar>

will match:

abcde<FooBar>

But how do I get it to match across multiple lines?

abcde

fghij<FooBar>

regex multiline

share|improve this question

edited Apr 19 '15 at 20:59

Bobulous

9,81642850

asked Oct 1 '08 at 18:48

andyukandyuk

23.5k154451

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  To clarify; I was originally using Eclipse to do a find and replace in multiple files. What I have discovered by the answers below is that my problem was the tool and not regex pattern.
  
  – andyuk
  Oct 2 '08 at 15:45
  

  
  
  
  


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  Your flag "eclipse" should be removed then because one looking for an eclipse solution will find this question (like I did) and then find a non-eclipse solution as accepted one.
  
  – acme
  Jun 13 '12 at 12:09
  

  
  
  
  


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  Now I'm finding this in the search engine because eclipse was mentioned. Oh the horror.
  
  – Brian Olsen
  Mar 20 '18 at 19:29
  

  
  
  
  

add a comment | 

264

For example, this regex

(.*)<FooBar>

will match:

abcde<FooBar>

But how do I get it to match across multiple lines?

abcde

fghij<FooBar>

regex multiline

share|improve this question

edited Apr 19 '15 at 20:59

Bobulous

9,81642850

asked Oct 1 '08 at 18:48

andyukandyuk

23.5k154451

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  To clarify; I was originally using Eclipse to do a find and replace in multiple files. What I have discovered by the answers below is that my problem was the tool and not regex pattern.
  
  – andyuk
  Oct 2 '08 at 15:45
  

  
  
  
  


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  2
  

  
  
  
  
  
  

  
  
  Your flag "eclipse" should be removed then because one looking for an eclipse solution will find this question (like I did) and then find a non-eclipse solution as accepted one.
  
  – acme
  Jun 13 '12 at 12:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Now I'm finding this in the search engine because eclipse was mentioned. Oh the horror.
  
  – Brian Olsen
  Mar 20 '18 at 19:29
  

  
  
  
  

add a comment | 

264

264

264

62

For example, this regex

(.*)<FooBar>

will match:

abcde<FooBar>

But how do I get it to match across multiple lines?

abcde

fghij<FooBar>

regex multiline

share|improve this question

edited Apr 19 '15 at 20:59

Bobulous

9,81642850

asked Oct 1 '08 at 18:48

andyukandyuk

23.5k154451

For example, this regex

(.*)<FooBar>

will match:

abcde<FooBar>

But how do I get it to match across multiple lines?

abcde

fghij<FooBar>

regex multiline

regex multiline

share|improve this question

edited Apr 19 '15 at 20:59

Bobulous

9,81642850

asked Oct 1 '08 at 18:48

andyukandyuk

23.5k154451

share|improve this question

edited Apr 19 '15 at 20:59

Bobulous

9,81642850

asked Oct 1 '08 at 18:48

andyukandyuk

23.5k154451

share|improve this question

share|improve this question

edited Apr 19 '15 at 20:59

Bobulous

9,81642850

edited Apr 19 '15 at 20:59

Bobulous

9,81642850

edited Apr 19 '15 at 20:59

Bobulous

9,81642850

9,81642850

asked Oct 1 '08 at 18:48

andyukandyuk

23.5k154451

asked Oct 1 '08 at 18:48

andyukandyuk

23.5k154451

asked Oct 1 '08 at 18:48

andyukandyuk

23.5k154451

23.5k154451

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  To clarify; I was originally using Eclipse to do a find and replace in multiple files. What I have discovered by the answers below is that my problem was the tool and not regex pattern.
  
  – andyuk
  Oct 2 '08 at 15:45
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  Your flag "eclipse" should be removed then because one looking for an eclipse solution will find this question (like I did) and then find a non-eclipse solution as accepted one.
  
  – acme
  Jun 13 '12 at 12:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Now I'm finding this in the search engine because eclipse was mentioned. Oh the horror.
  
  – Brian Olsen
  Mar 20 '18 at 19:29
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  To clarify; I was originally using Eclipse to do a find and replace in multiple files. What I have discovered by the answers below is that my problem was the tool and not regex pattern.
  
  – andyuk
  Oct 2 '08 at 15:45
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  Your flag "eclipse" should be removed then because one looking for an eclipse solution will find this question (like I did) and then find a non-eclipse solution as accepted one.
  
  – acme
  Jun 13 '12 at 12:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Now I'm finding this in the search engine because eclipse was mentioned. Oh the horror.
  
  – Brian Olsen
  Mar 20 '18 at 19:29
  

  
  
  
  

1

1

To clarify; I was originally using Eclipse to do a find and replace in multiple files. What I have discovered by the answers below is that my problem was the tool and not regex pattern.

– andyuk
Oct 2 '08 at 15:45

To clarify; I was originally using Eclipse to do a find and replace in multiple files. What I have discovered by the answers below is that my problem was the tool and not regex pattern.

– andyuk
Oct 2 '08 at 15:45

2

2

Your flag "eclipse" should be removed then because one looking for an eclipse solution will find this question (like I did) and then find a non-eclipse solution as accepted one.

– acme
Jun 13 '12 at 12:09

Your flag "eclipse" should be removed then because one looking for an eclipse solution will find this question (like I did) and then find a non-eclipse solution as accepted one.

– acme
Jun 13 '12 at 12:09

Now I'm finding this in the search engine because eclipse was mentioned. Oh the horror.

– Brian Olsen
Mar 20 '18 at 19:29

Now I'm finding this in the search engine because eclipse was mentioned. Oh the horror.

– Brian Olsen
Mar 20 '18 at 19:29

add a comment | 

21 Answers
21

active

oldest

votes

193

It depends on the language, but there should be a modifier that you can add to the regex pattern. In PHP it is:

/(.*)<FooBar>/s

The s at the end causes the dot to match all characters including newlines.

share|improve this answer

answered Oct 1 '08 at 18:52

Jeremy RutenJeremy Ruten

125k34157184

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  @Grace: use n to match a newline
  
  – Jeremy Ruten
  Apr 11 '11 at 21:05
  

  
  
  
  


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  rn works perfectly
  
  – Grace
  Apr 12 '11 at 8:08
  

  
  
  
  


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  The s flag is (now?) invalid, at least in Chrome/V8. Instead use /([sS]*)<FooBar>/ character class (match space and non-space] instead of the period matcher. See other answers for more info.
  
  – Allen
  May 9 '13 at 15:37
  
  
  

  
  
  
  


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  @Allen - JavaScript doesn't support the s modifier. Instead, do [^]* for the same effect.
  
  – Derek 朕會功夫
  Jul 12 '15 at 22:26
  

  
  
  
  


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  In Ruby, use the m modifier
  
  – Ryan Buckley
  Jul 15 '15 at 22:57
  

  
  
  
  

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show 6 more comments

264

Try this:

((.|n)*)<FooBar>

It basically says "any character or a newline" repeated zero or more times.

share|improve this answer

answered Oct 1 '08 at 18:52

leviklevik

72.9k236689

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  This is dependent on the language and/or tool you are using. Please let us know what you are using, eg Perl, PHP, CF, C#, sed, awk, etc.
  
  – Ben Doom
  Oct 1 '08 at 18:57
  

  
  
  
  


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  Depending on your line endings you might need ((.|n|r)*)<FooBar>
  
  – Potherca
  Mar 9 '12 at 17:27
  

  
  
  
  


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  He said he is using Eclipse. This is correct solution in my opinion. I have same problem and this solved it.
  
  – Danubian Sailor
  Apr 18 '12 at 8:14
  

  
  
  
  


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  Right - the question is about eclipse and so are the tags. But the accepted solution is a PHP solution. Yours should be the accepted solution...
  
  – acme
  Jun 13 '12 at 12:04
  

  
  
  
  


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  This is the worst regex for matching multiple line input. Please never use it unless you are using ElasticSearch. Use [sS]* or (?s).*.
  
  – Wiktor Stribiżew
  Jul 18 '16 at 11:05
  
  
  

  
  
  
  

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show 2 more comments

61

If you're using Eclipse search, you can enable the "DOTALL" option to make '.' match any character including line delimiters: just add "(?s)" at the beginning of your search string. Example:

(?s).*<FooBar>

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answered Nov 25 '11 at 13:16

Paulo MersonPaulo Merson

5,40143236

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  This is not eclipse-specific, should work anywhere.
  
  – Steven Soroka
  Oct 8 '13 at 16:50
  

  
  
  
  


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  Not anywhere, only in regex flavors supporting inline modifiers, and certainly not in Ruby where (?s) => (?m)
  
  – Wiktor Stribiżew
  Jul 18 '16 at 11:06
  

  
  
  
  


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  Anything for bash?
  
  – Pasupathi Rajamanickam
  Dec 19 '18 at 2:12
  

  
  
  
  

add a comment | 

46

The question is, can . pattern match any character? The answer varies from engine to engine. The main difference is whether the pattern is used by a POSIX or non-POSIX regex library.

Special note about lua-patterns: they are not considered regular expressions, but . matches any char there, same as POSIX based engines.

Another note on matlab and octave: the . matches any char by default (demo): str = "abcden fghij<Foobar>"; expression = '(.*)<Foobar>*'; [tokens,matches] = regexp(str,expression,'tokens','match'); (tokens contain a abcden fghij item).

Also, in all of boost's regex grammars the dot matches line breaks by default. Boost's ECMAScript grammar allows you to turn this off with regex_constants::no_mod_m (source).

As for oracle (it is POSIX based), use n option (demo): select regexp_substr('abcde' || chr(10) ||' fghij<Foobar>', '(.*)<Foobar>', 1, 1, 'n', 1) as results from dual

POSIX-based engines:

A mere . already matches line breaks, no need to use any modifiers, see bash (demo).

The tcl (demo), postgresql (demo), r (TRE, base R default engine with no perl=TRUE, for base R with perl=TRUE or for stringr/stringi patterns, use the (?s) inline modifier) (demo) also treat . the same way.

However, most POSIX based tools process input line by line. Hence, . does not match the line breaks just because they are not in scope. Here are some examples how to override this:

• 
  sed - There are multiple workarounds, the most precise but not very safe is sed 'H;1h;$!d;x; s/(.*)><Foobar>/1/' (H;1h;$!d;x; slurps the file into memory). If whole lines must be included, sed '/start_pattern/,/end_pattern/d' file (removing from start will end with matched lines included) or sed '/start_pattern/,/end_pattern/{{//!d;};}' file (with matching lines excluded) can be considered.


• 
  perl - perl -0pe 's/(.*)<FooBar>/$1/gs' <<< "$str" (-0 slurps the whole file into memory, -p prints the file after applying the script given by -e). Note that using -000pe will slurp the file and activate 'paragraph mode' where Perl uses consecutive newlines (nn) as the record separator.


• 
  gnu-grep - grep -Poz '(?si)abcK.*?(?=<Foobar>)' file. Here, z enables file slurping, (?s) enables the DOTALL mode for the . pattern, (?i) enables case insensitive mode, K omits the text matched so far, *? is a lazy quantifier, (?=<Foobar>) matches the location before <Foobar>.


• 
  pcregrep - pcregrep -Mi "(?si)abcK.*?(?=<Foobar>)" file (M enables file slurping here). Note pcregrep is a good solution for Mac OS grep users.

See demos.

Non-POSIX-based engines:

• 
  php - Use s modifier PCRE_DOTALL modifier: preg_match('~(.*)<Foobar>~s', $s, $m) (demo)


• 
  c# - Use RegexOptions.Singleline flag (demo):
  - var result = Regex.Match(s, @"(.*)<Foobar>", RegexOptions.Singleline).Groups[1].Value;
  - var result = Regex.Match(s, @"(?s)(.*)<Foobar>").Groups[1].Value;
  


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  powershell - Use (?s) inline option: $s = "abcde`nfghij<FooBar>"; $s -match "(?s)(.*)<Foobar>"; $matches[1]
  


• 
  perl - Use s modifier (or (?s) inline version at the start) (demo): /(.*)<FooBar>/s
  


• 
  python - Use re.DOTALL (or re.S) flags or (?s) inline modifier (demo): m = re.search(r"(.*)<FooBar>", s, flags=re.S) (and then if m:, print(m.group(1)))


• 
  java - Use Pattern.DOTALL modifier (or inline (?s) flag) (demo): Pattern.compile("(.*)<FooBar>", Pattern.DOTALL)
  


• 
  groovy - Use (?s) in-pattern modifier (demo): regex = /(?s)(.*)<FooBar>/
  


• 
  scala - Use (?s) modifier (demo): "(?s)(.*)<Foobar>".r.findAllIn("abcden fghij<Foobar>").matchData foreach { m => println(m.group(1)) }
  


• 
  javascript - Use [^] or workarounds [dD] / [wW] / [sS] (demo): s.match(/([sS]*)<FooBar>/)[1]
  


• 
  c++ (std::regex) Use [sS] or the JS workarounds (demo): regex rex(R"(([sS]*)<FooBar>)");
  


• 
  vba - Use the same approach as in JavaScript, ([sS]*)<Foobar>.


• 
  ruby - Use /m MULTILINE modifier (demo): s[/(.*)<Foobar>/m, 1]
  


• 
  go - Use the inline modifier (?s) at the start (demo): re: = regexp.MustCompile(`(?s)(.*)<FooBar>`)
  


• 
  swift - Use dotMatchesLineSeparators or (easier) pass the (?s) inline modifier to the pattern: let rx = "(?s)(.*)<Foobar>"
  


• 
  objective-c - Same as Swift, (?s) works the easiest, but here is how the option can be used: NSRegularExpression* regex = [NSRegularExpression regularExpressionWithPattern:pattern
options:NSRegularExpressionDotMatchesLineSeparators error:&regexError];
  


• 
  re2, google-apps-script - Use (?s) modifier (demo): "(?s)(.*)<Foobar>" (in Google Spreadsheets, =REGEXEXTRACT(A2,"(?s)(.*)<Foobar>"))

NOTES ON (?s):

In most non-POSIX engines, (?s) inline modifier (or embedded flag option) can be used to enforce . to match line breaks.

If placed at the start of the pattern, (?s) changes the bahavior of all . in the pattern. If the (?s) is placed somewhere after the beginning, only those . will be affected that are located to the right of it unless this is a pattern passed to Python re. In Python re, regardless of the (?s) location, the whole pattern . are affected. The (?s) effect is stopped using (?-s). A modified group can be used to only affect a specified range of a regex pattern (e.g. Delim1(?s:.*?)nDelim2.* will make the first .*? match across newlines and the second .* will only match the rest of the line).

POSIX note:

In non-regex engines, to match any char, [sS] / [dD] / [wW] constructs can be used.

In POSIX, [sS] is not matching any char (as in JavaScript or any non-POSIX engine) because regex escape sequences are not supported inside bracket expressions. [sS] is parsed as bracket expressions that match a single char, or s or S.

share|improve this answer

edited Dec 20 '18 at 9:49

answered Aug 31 '17 at 12:47

Wiktor StribiżewWiktor Stribiżew

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  You should link to this excellent overview from your profile page or something (+1).
  
  – Jan
  Oct 15 '17 at 20:15
  

  
  
  
  


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  You may want to add this to the boost item: In the regex_constants namespace, flag_type_'s : perl = ECMAScript = JavaScript = JScript = ::boost::regbase::normal = 0 which defaults to Perl. Programmers will set a base flag definition #define MOD regex_constants::perl | boost::regex::no_mod_s | boost::regex::no_mod_m for thier regex flags to reflect that. And the arbitor is always the inline modifiers. Where (?-sm)(?s).* resets.
  
  – sln
  Apr 26 '18 at 21:30
  
  
  

  
  
  
  


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  Can you also add for bash please?
  
  – Pasupathi Rajamanickam
  Dec 19 '18 at 2:12
  

  
  
  
  


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  @PasupathiRajamanickam Bash uses a POSIX regex engine, the . matches any char there (including line breaks). See this online Bash demo.
  
  – Wiktor Stribiżew
  Dec 19 '18 at 7:33
  
  
  

  
  
  
  

add a comment | 

31

In JavaScript, use /[Ss]*<Foobar>/. Source

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edited Feb 23 '17 at 1:15

Nathan Arthur

3,00032447

answered Jul 30 '11 at 13:03

Abbas ShahzadehAbbas Shahzadeh

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  From that link: "JavaScript and VBScript do not have an option to make the dot match line break characters. In those languages, you can use a character class such as [sS] to match any character." Instead of the . use [sS] (match spaces and non-spaces) instead.
  
  – Allen
  May 9 '13 at 15:34
  

  
  
  
  

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27

([sS]*)<FooBar>

The dot matches all except newlines (rn). So use sS, which will match ALL characters.

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answered Jul 19 '12 at 17:59

samwizesamwize

14.7k882140

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  This solve the problem if you are using the Objective-C [text rangeOfString:regEx options:NSRegularExpressionSearch]. Thanks!
  
  – J. Costa
  Aug 24 '12 at 22:29
  
  
  

  
  
  
  


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  This works in intelliJ's find&replace regex, thanks.
  
  – barclay
  Sep 16 '15 at 22:14
  

  
  
  
  


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  This works. But it needs to be the first occurrence of <FooBar>
  
  – Ozkan
  Sep 26 '17 at 14:16
  

  
  
  
  

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18

In Ruby ruby you can use the 'm' option (multiline):

/YOUR_REGEXP/m

See the Regexp documentation on ruby-doc.org for more information.

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edited Aug 31 '17 at 7:54

Wiktor Stribiżew

320k16140222

answered Aug 3 '12 at 7:52

vibaihervibaiher

38228

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10

we can also use

(.*?n)*?

to match everything including newline without greedy

This will make the new line optional

(.*?|n)*?

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answered Aug 6 '18 at 7:48

RAN_0915RAN_0915

861216

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8

"." normally doesn't match line-breaks. Most regex engines allows you to add the S-flag (also called DOTALL and SINGLELINE) to make "." also match newlines.
If that fails, you could do something like [Ss].

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edited Apr 25 '09 at 21:09

answered Oct 1 '08 at 18:52

Markus JarderotMarkus Jarderot

66.8k12112120

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7

For Eclipse worked following expression:

Foo

jadajada Bar"

Regular-Expression:

Foo[Ss]{1,10}.*Bar*

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edited Jan 3 '13 at 11:52

devOp

2,91411232

answered Jan 3 '13 at 11:32

GordonGordon

7111

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5

/(.*)<FooBar>/s

the s causes Dot (.) to match carriage returns

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answered Oct 1 '08 at 18:54

BillBill

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  Seems like this is invalid (Chrome): text.match(/a/s) SyntaxError: Invalid flags supplied to RegExp constructor 's'
  
  – Allen
  May 9 '13 at 15:31
  
  
  

  
  
  
  


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  Because it is unsupported in JavaScript RegEx engines. The s flags exists in PCRE, the most complete engine (available in Perl and PHP). PCRE has 10 flags (and a lot of other features) while JavaScript has only 3 flags (gmi).
  
  – Morgan Touverey Quilling
  Apr 20 '16 at 18:51
  

  
  
  
  

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4

In java based regular expression you can use [sS]

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edited May 25 '18 at 19:01

revo

33.3k135085

answered Jun 3 '13 at 6:22

KamahireKamahire

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  Shouldn't those be backslashes?
  
  – Paul Draper
  Oct 19 '13 at 6:48
  

  
  
  
  


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  They go at the end of the Regular Expression, not within in. Example: /blah/s
  
  – RandomInsano
  Dec 21 '13 at 20:12
  

  
  
  
  


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  I guess you mean JavaScript, not Java? Since you can just add the s flag to the pattern in Java and JavaScript doesn't have the s flag.
  
  – 3limin4t0r
  Sep 25 '18 at 17:47
  
  
  

  
  
  
  

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3

Note that (.|n)* can be less efficient than (for example) [sS]* (if your language's regexes support such escapes) and than finding how to specify the modifier that makes . also match newlines. Or you can go with POSIXy alternatives like [[:space:][:^space:]]*.

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answered Oct 2 '08 at 3:31

tyetye

1,072911

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3

Use RegexOptions.Singleline, it changes the meaning of . to include newlines

Regex.Replace(content, searchText, replaceText, RegexOptions.Singleline);

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answered Apr 13 '10 at 0:42

shmallshmall

311

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2

Solution:

Use pattern modifier sU will get the desired matching in PHP.

example:

preg_match('/(.*)/sU',$content,$match);

Source:

http://dreamluverz.com/developers-tools/regex-match-all-including-new-line
http://php.net/manual/en/reference.pcre.pattern.modifiers.php

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answered Apr 4 '12 at 11:00

Sian Lerk LauSian Lerk Lau

1088

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1

In the context of use within languages, regular expressions act on strings, not lines. So you should be able to use the regex normally, assuming that the input string has multiple lines.

In this case, the given regex will match the entire string, since "<FooBar>" is present. Depending on the specifics of the regex implementation, the $1 value (obtained from the "(.*)") will either be "fghij" or "abcdenfghij". As others have said, some implementations allow you to control whether the "." will match the newline, giving you the choice.

Line-based regular expression use is usually for command line things like egrep.

share|improve this answer

edited Oct 1 '08 at 18:54

answered Oct 1 '08 at 18:49

nsayernsayer

13.5k22645

add a comment | 

1

I had the same problem and solved it in probably not the best way but it works. I replaced all line breaks before I did my real match:

mystring= Regex.Replace(mystring, "rn", "")

I am manipulating HTML so line breaks don't really matter to me in this case.

I tried all of the suggestions above with no luck, I am using .Net 3.5 FYI

share|improve this answer

answered Mar 26 '09 at 14:57

SleeSlee

11.4k41129232

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I am using .NET too and (s|S) seems to do the trick for me!
  
  – Vamshi Krishna
  May 18 '18 at 7:26
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @VamshiKrishna In .NET, use (?s) to make . match any chars. Do not use (s|S) that will slow down performance.
  
  – Wiktor Stribiżew
  Sep 14 '18 at 20:35
  

  
  
  
  

add a comment | 

0

generally . doesn't match newlines, so try ((.|n)*)<foobar>

share|improve this answer

answered Oct 1 '08 at 18:52

tloachtloach

7,53112943

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  No, don't do that. If you need to match anything including line separators, use the DOTALL (a.k.a. /s or SingleLine) modifier. Not only does the (.|n) hack make the regex less efficient, it's not even correct. At the very least, it should match r (carriage return) as well as n (linefeed). There are other line separator characters, too, albeit rarely used. But if you use the DOTALL flag, you don't have to worry about them.
  
  – Alan Moore
  Apr 26 '09 at 3:17
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  R is the platform-independent match for newlines in Eclipse.
  
  – opyate
  Nov 30 '09 at 11:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @opyate You should post this as an answer as this little gem is incredibly useful.
  
  – jeckhart
  Oct 15 '12 at 21:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  You could try this instead. It won't match the inner brackets and also consider the optionalr.: ((?:.|r?n)*)<foobar>
  
  – ssc-hrep3
  Nov 29 '16 at 9:52
  

  
  
  
  

add a comment | 

0

I wanted to match a particular if block in java

   ...

   ...

   if(isTrue){

       doAction();



   }

...

...

}

If I use the regExp

if (isTrue(.|n)*}

it included the closing brace for the method block so I used

if (!isTrue([^}.]|n)*}

to exclude the closing brace from the wildcard match.

share|improve this answer

answered Jan 18 '11 at 9:31

SpangenSpangen

1,89031824

add a comment | 

0

Often we have to modify a substring with a few keywords spread across lines preceding the substring. Consider an xml element:

<TASK>

  <UID>21</UID>

  <Name>Architectural design</Name>

  <PercentComplete>81</PercentComplete>

</TASK>

Suppose we want to modify the 81, to some other value, say 40. First identify .UID.21..UID., then skip all characters including n till .PercentCompleted.. The regular expression pattern and the replace specification are:

String hw = new String("<TASK>n  <UID>21</UID>n  <Name>Architectural design</Name>n  <PercentComplete>81</PercentComplete>n</TASK>");

String pattern = new String ("(<UID>21</UID>)((.|n)*?)(<PercentComplete>)(\d+)(</PercentComplete>)");

String replaceSpec = new String ("$1$2$440$6");

//note that the group (<PercentComplete>) is $4 and the group ((.|n)*?) is $2.



String  iw = hw.replaceFirst(pattern, replaceSpec);

System.out.println(iw);



<TASK>

  <UID>21</UID>

  <Name>Architectural design</Name>

  <PercentComplete>40</PercentComplete>

</TASK>

The subgroup (.|n) is probably the missing group $3. If we make it non-capturing by (?:.|n) then the $3 is (<PercentComplete>). So the pattern and replaceSpec can also be:

pattern = new String("(<UID>21</UID>)((?:.|n)*?)(<PercentComplete>)(\d+)(</PercentComplete>)");

replaceSpec = new String("$1$2$340$5")

and the replacement works correctly as before.

share|improve this answer

edited Oct 26 '12 at 8:49

deadly

1,1431223

answered Apr 21 '12 at 20:05

user1348737user1348737

92

add a comment | 

0

In Javascript you can use [^]* to search for zero to infinite characters, including line breaks.

$("#find_and_replace").click(function() {

  var text = $("#textarea").val();

  search_term = new RegExp("[^]*<Foobar>", "gi");;

  replace_term = "Replacement term";

  var new_text = text.replace(search_term, replace_term);

  $("#textarea").val(new_text);

});

<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>

<button id="find_and_replace">Find and replace</button>

<br>

<textarea ID="textarea">abcde

fghij&lt;Foobar&gt;</textarea>

share|improve this answer

answered yesterday

Paul JonesPaul Jones

329311

add a comment | 

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193

It depends on the language, but there should be a modifier that you can add to the regex pattern. In PHP it is:

/(.*)<FooBar>/s

The s at the end causes the dot to match all characters including newlines.

share|improve this answer

answered Oct 1 '08 at 18:52

Jeremy RutenJeremy Ruten

125k34157184

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  @Grace: use n to match a newline
  
  – Jeremy Ruten
  Apr 11 '11 at 21:05
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  
  rn works perfectly
  
  – Grace
  Apr 12 '11 at 8:08
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  The s flag is (now?) invalid, at least in Chrome/V8. Instead use /([sS]*)<FooBar>/ character class (match space and non-space] instead of the period matcher. See other answers for more info.
  
  – Allen
  May 9 '13 at 15:37
  
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  @Allen - JavaScript doesn't support the s modifier. Instead, do [^]* for the same effect.
  
  – Derek 朕會功夫
  Jul 12 '15 at 22:26
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  In Ruby, use the m modifier
  
  – Ryan Buckley
  Jul 15 '15 at 22:57
  

  
  
  
  

 | 
show 6 more comments

193

It depends on the language, but there should be a modifier that you can add to the regex pattern. In PHP it is:

/(.*)<FooBar>/s

The s at the end causes the dot to match all characters including newlines.

share|improve this answer

answered Oct 1 '08 at 18:52

Jeremy RutenJeremy Ruten

125k34157184

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  @Grace: use n to match a newline
  
  – Jeremy Ruten
  Apr 11 '11 at 21:05
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  
  rn works perfectly
  
  – Grace
  Apr 12 '11 at 8:08
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  The s flag is (now?) invalid, at least in Chrome/V8. Instead use /([sS]*)<FooBar>/ character class (match space and non-space] instead of the period matcher. See other answers for more info.
  
  – Allen
  May 9 '13 at 15:37
  
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  @Allen - JavaScript doesn't support the s modifier. Instead, do [^]* for the same effect.
  
  – Derek 朕會功夫
  Jul 12 '15 at 22:26
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  In Ruby, use the m modifier
  
  – Ryan Buckley
  Jul 15 '15 at 22:57
  

  
  
  
  

 | 
show 6 more comments

193

193

193

It depends on the language, but there should be a modifier that you can add to the regex pattern. In PHP it is:

/(.*)<FooBar>/s

The s at the end causes the dot to match all characters including newlines.

share|improve this answer

answered Oct 1 '08 at 18:52

Jeremy RutenJeremy Ruten

125k34157184

It depends on the language, but there should be a modifier that you can add to the regex pattern. In PHP it is:

/(.*)<FooBar>/s

The s at the end causes the dot to match all characters including newlines.

share|improve this answer

answered Oct 1 '08 at 18:52

Jeremy RutenJeremy Ruten

125k34157184

share|improve this answer

share|improve this answer

answered Oct 1 '08 at 18:52

Jeremy RutenJeremy Ruten

125k34157184

answered Oct 1 '08 at 18:52

Jeremy RutenJeremy Ruten

125k34157184

answered Oct 1 '08 at 18:52

Jeremy RutenJeremy Ruten

125k34157184

125k34157184

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  @Grace: use n to match a newline
  
  – Jeremy Ruten
  Apr 11 '11 at 21:05
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  
  rn works perfectly
  
  – Grace
  Apr 12 '11 at 8:08
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  The s flag is (now?) invalid, at least in Chrome/V8. Instead use /([sS]*)<FooBar>/ character class (match space and non-space] instead of the period matcher. See other answers for more info.
  
  – Allen
  May 9 '13 at 15:37
  
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  @Allen - JavaScript doesn't support the s modifier. Instead, do [^]* for the same effect.
  
  – Derek 朕會功夫
  Jul 12 '15 at 22:26
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  In Ruby, use the m modifier
  
  – Ryan Buckley
  Jul 15 '15 at 22:57
  

  
  
  
  

 | 
show 6 more comments

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  @Grace: use n to match a newline
  
  – Jeremy Ruten
  Apr 11 '11 at 21:05
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  
  rn works perfectly
  
  – Grace
  Apr 12 '11 at 8:08
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  The s flag is (now?) invalid, at least in Chrome/V8. Instead use /([sS]*)<FooBar>/ character class (match space and non-space] instead of the period matcher. See other answers for more info.
  
  – Allen
  May 9 '13 at 15:37
  
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  @Allen - JavaScript doesn't support the s modifier. Instead, do [^]* for the same effect.
  
  – Derek 朕會功夫
  Jul 12 '15 at 22:26
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  In Ruby, use the m modifier
  
  – Ryan Buckley
  Jul 15 '15 at 22:57
  

  
  
  
  

3

3

@Grace: use n to match a newline

– Jeremy Ruten
Apr 11 '11 at 21:05

@Grace: use n to match a newline

– Jeremy Ruten
Apr 11 '11 at 21:05

5

5

rn works perfectly

– Grace
Apr 12 '11 at 8:08

rn works perfectly

– Grace
Apr 12 '11 at 8:08

2

2

The s flag is (now?) invalid, at least in Chrome/V8. Instead use /([sS]*)<FooBar>/ character class (match space and non-space] instead of the period matcher. See other answers for more info.

– Allen
May 9 '13 at 15:37

The s flag is (now?) invalid, at least in Chrome/V8. Instead use /([sS]*)<FooBar>/ character class (match space and non-space] instead of the period matcher. See other answers for more info.

– Allen
May 9 '13 at 15:37

2

2

@Allen - JavaScript doesn't support the s modifier. Instead, do [^]* for the same effect.

– Derek 朕會功夫
Jul 12 '15 at 22:26

@Allen - JavaScript doesn't support the s modifier. Instead, do [^]* for the same effect.

– Derek 朕會功夫
Jul 12 '15 at 22:26

1

1

In Ruby, use the m modifier

– Ryan Buckley
Jul 15 '15 at 22:57

In Ruby, use the m modifier

– Ryan Buckley
Jul 15 '15 at 22:57

 | 
show 6 more comments

264

Try this:

((.|n)*)<FooBar>

It basically says "any character or a newline" repeated zero or more times.

share|improve this answer

answered Oct 1 '08 at 18:52

leviklevik

72.9k236689

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  This is dependent on the language and/or tool you are using. Please let us know what you are using, eg Perl, PHP, CF, C#, sed, awk, etc.
  
  – Ben Doom
  Oct 1 '08 at 18:57
  

  
  
  
  


• 
  
  
  

  
  31
  

  
  
  
  
  
  

  
  
  Depending on your line endings you might need ((.|n|r)*)<FooBar>
  
  – Potherca
  Mar 9 '12 at 17:27
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  He said he is using Eclipse. This is correct solution in my opinion. I have same problem and this solved it.
  
  – Danubian Sailor
  Apr 18 '12 at 8:14
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  
  Right - the question is about eclipse and so are the tags. But the accepted solution is a PHP solution. Yours should be the accepted solution...
  
  – acme
  Jun 13 '12 at 12:04
  

  
  
  
  


• 
  
  
  

  
  6
  

  
  
  
  
  
  

  
  
  This is the worst regex for matching multiple line input. Please never use it unless you are using ElasticSearch. Use [sS]* or (?s).*.
  
  – Wiktor Stribiżew
  Jul 18 '16 at 11:05
  
  
  

  
  
  
  

 | 
show 2 more comments

264

Try this:

((.|n)*)<FooBar>

It basically says "any character or a newline" repeated zero or more times.

share|improve this answer

answered Oct 1 '08 at 18:52

leviklevik

72.9k236689

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  This is dependent on the language and/or tool you are using. Please let us know what you are using, eg Perl, PHP, CF, C#, sed, awk, etc.
  
  – Ben Doom
  Oct 1 '08 at 18:57
  

  
  
  
  


• 
  
  
  

  
  31
  

  
  
  
  
  
  

  
  
  Depending on your line endings you might need ((.|n|r)*)<FooBar>
  
  – Potherca
  Mar 9 '12 at 17:27
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  He said he is using Eclipse. This is correct solution in my opinion. I have same problem and this solved it.
  
  – Danubian Sailor
  Apr 18 '12 at 8:14
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  
  Right - the question is about eclipse and so are the tags. But the accepted solution is a PHP solution. Yours should be the accepted solution...
  
  – acme
  Jun 13 '12 at 12:04
  

  
  
  
  


• 
  
  
  

  
  6
  

  
  
  
  
  
  

  
  
  This is the worst regex for matching multiple line input. Please never use it unless you are using ElasticSearch. Use [sS]* or (?s).*.
  
  – Wiktor Stribiżew
  Jul 18 '16 at 11:05
  
  
  

  
  
  
  

 | 
show 2 more comments

264

264

264

Try this:

((.|n)*)<FooBar>

It basically says "any character or a newline" repeated zero or more times.

share|improve this answer

answered Oct 1 '08 at 18:52

leviklevik

72.9k236689

Try this:

((.|n)*)<FooBar>

It basically says "any character or a newline" repeated zero or more times.

share|improve this answer

answered Oct 1 '08 at 18:52

leviklevik

72.9k236689

share|improve this answer

share|improve this answer

answered Oct 1 '08 at 18:52

leviklevik

72.9k236689

answered Oct 1 '08 at 18:52

leviklevik

72.9k236689

answered Oct 1 '08 at 18:52

leviklevik

72.9k236689

72.9k236689

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  This is dependent on the language and/or tool you are using. Please let us know what you are using, eg Perl, PHP, CF, C#, sed, awk, etc.
  
  – Ben Doom
  Oct 1 '08 at 18:57
  

  
  
  
  


• 
  
  
  

  
  31
  

  
  
  
  
  
  

  
  
  Depending on your line endings you might need ((.|n|r)*)<FooBar>
  
  – Potherca
  Mar 9 '12 at 17:27
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  He said he is using Eclipse. This is correct solution in my opinion. I have same problem and this solved it.
  
  – Danubian Sailor
  Apr 18 '12 at 8:14
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  
  Right - the question is about eclipse and so are the tags. But the accepted solution is a PHP solution. Yours should be the accepted solution...
  
  – acme
  Jun 13 '12 at 12:04
  

  
  
  
  


• 
  
  
  

  
  6
  

  
  
  
  
  
  

  
  
  This is the worst regex for matching multiple line input. Please never use it unless you are using ElasticSearch. Use [sS]* or (?s).*.
  
  – Wiktor Stribiżew
  Jul 18 '16 at 11:05
  
  
  

  
  
  
  

 | 
show 2 more comments

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  This is dependent on the language and/or tool you are using. Please let us know what you are using, eg Perl, PHP, CF, C#, sed, awk, etc.
  
  – Ben Doom
  Oct 1 '08 at 18:57
  

  
  
  
  


• 
  
  
  

  
  31
  

  
  
  
  
  
  

  
  
  Depending on your line endings you might need ((.|n|r)*)<FooBar>
  
  – Potherca
  Mar 9 '12 at 17:27
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  He said he is using Eclipse. This is correct solution in my opinion. I have same problem and this solved it.
  
  – Danubian Sailor
  Apr 18 '12 at 8:14
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  
  Right - the question is about eclipse and so are the tags. But the accepted solution is a PHP solution. Yours should be the accepted solution...
  
  – acme
  Jun 13 '12 at 12:04
  

  
  
  
  


• 
  
  
  

  
  6
  

  
  
  
  
  
  

  
  
  This is the worst regex for matching multiple line input. Please never use it unless you are using ElasticSearch. Use [sS]* or (?s).*.
  
  – Wiktor Stribiżew
  Jul 18 '16 at 11:05
  
  
  

  
  
  
  

3

3

This is dependent on the language and/or tool you are using. Please let us know what you are using, eg Perl, PHP, CF, C#, sed, awk, etc.

– Ben Doom
Oct 1 '08 at 18:57

This is dependent on the language and/or tool you are using. Please let us know what you are using, eg Perl, PHP, CF, C#, sed, awk, etc.

– Ben Doom
Oct 1 '08 at 18:57

31

31

Depending on your line endings you might need ((.|n|r)*)<FooBar>

– Potherca
Mar 9 '12 at 17:27

Depending on your line endings you might need ((.|n|r)*)<FooBar>

– Potherca
Mar 9 '12 at 17:27

3

3

He said he is using Eclipse. This is correct solution in my opinion. I have same problem and this solved it.

– Danubian Sailor
Apr 18 '12 at 8:14

He said he is using Eclipse. This is correct solution in my opinion. I have same problem and this solved it.

– Danubian Sailor
Apr 18 '12 at 8:14

4

4

Right - the question is about eclipse and so are the tags. But the accepted solution is a PHP solution. Yours should be the accepted solution...

– acme
Jun 13 '12 at 12:04

Right - the question is about eclipse and so are the tags. But the accepted solution is a PHP solution. Yours should be the accepted solution...

– acme
Jun 13 '12 at 12:04

6

6

This is the worst regex for matching multiple line input. Please never use it unless you are using ElasticSearch. Use [sS]* or (?s).*.

– Wiktor Stribiżew
Jul 18 '16 at 11:05

This is the worst regex for matching multiple line input. Please never use it unless you are using ElasticSearch. Use [sS]* or (?s).*.

– Wiktor Stribiżew
Jul 18 '16 at 11:05

 | 
show 2 more comments

61

If you're using Eclipse search, you can enable the "DOTALL" option to make '.' match any character including line delimiters: just add "(?s)" at the beginning of your search string. Example:

(?s).*<FooBar>

share|improve this answer

answered Nov 25 '11 at 13:16

Paulo MersonPaulo Merson

5,40143236

• 
  
  
  

  
  8
  

  
  
  
  
  
  

  
  
  This is not eclipse-specific, should work anywhere.
  
  – Steven Soroka
  Oct 8 '13 at 16:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Not anywhere, only in regex flavors supporting inline modifiers, and certainly not in Ruby where (?s) => (?m)
  
  – Wiktor Stribiżew
  Jul 18 '16 at 11:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Anything for bash?
  
  – Pasupathi Rajamanickam
  Dec 19 '18 at 2:12
  

  
  
  
  

add a comment | 

61

If you're using Eclipse search, you can enable the "DOTALL" option to make '.' match any character including line delimiters: just add "(?s)" at the beginning of your search string. Example:

(?s).*<FooBar>

share|improve this answer

answered Nov 25 '11 at 13:16

Paulo MersonPaulo Merson

5,40143236

• 
  
  
  

  
  8
  

  
  
  
  
  
  

  
  
  This is not eclipse-specific, should work anywhere.
  
  – Steven Soroka
  Oct 8 '13 at 16:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Not anywhere, only in regex flavors supporting inline modifiers, and certainly not in Ruby where (?s) => (?m)
  
  – Wiktor Stribiżew
  Jul 18 '16 at 11:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Anything for bash?
  
  – Pasupathi Rajamanickam
  Dec 19 '18 at 2:12
  

  
  
  
  

add a comment | 

61

61

61

If you're using Eclipse search, you can enable the "DOTALL" option to make '.' match any character including line delimiters: just add "(?s)" at the beginning of your search string. Example:

(?s).*<FooBar>

share|improve this answer

answered Nov 25 '11 at 13:16

Paulo MersonPaulo Merson

5,40143236

If you're using Eclipse search, you can enable the "DOTALL" option to make '.' match any character including line delimiters: just add "(?s)" at the beginning of your search string. Example:

(?s).*<FooBar>

share|improve this answer

answered Nov 25 '11 at 13:16

Paulo MersonPaulo Merson

5,40143236

share|improve this answer

share|improve this answer

answered Nov 25 '11 at 13:16

Paulo MersonPaulo Merson

5,40143236

answered Nov 25 '11 at 13:16

Paulo MersonPaulo Merson

5,40143236

answered Nov 25 '11 at 13:16

Paulo MersonPaulo Merson

5,40143236

5,40143236

• 
  
  
  

  
  8
  

  
  
  
  
  
  

  
  
  This is not eclipse-specific, should work anywhere.
  
  – Steven Soroka
  Oct 8 '13 at 16:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Not anywhere, only in regex flavors supporting inline modifiers, and certainly not in Ruby where (?s) => (?m)
  
  – Wiktor Stribiżew
  Jul 18 '16 at 11:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Anything for bash?
  
  – Pasupathi Rajamanickam
  Dec 19 '18 at 2:12
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  8
  

  
  
  
  
  
  

  
  
  This is not eclipse-specific, should work anywhere.
  
  – Steven Soroka
  Oct 8 '13 at 16:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Not anywhere, only in regex flavors supporting inline modifiers, and certainly not in Ruby where (?s) => (?m)
  
  – Wiktor Stribiżew
  Jul 18 '16 at 11:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Anything for bash?
  
  – Pasupathi Rajamanickam
  Dec 19 '18 at 2:12
  

  
  
  
  

8

8

This is not eclipse-specific, should work anywhere.

– Steven Soroka
Oct 8 '13 at 16:50

This is not eclipse-specific, should work anywhere.

– Steven Soroka
Oct 8 '13 at 16:50

Not anywhere, only in regex flavors supporting inline modifiers, and certainly not in Ruby where (?s) => (?m)

– Wiktor Stribiżew
Jul 18 '16 at 11:06

Not anywhere, only in regex flavors supporting inline modifiers, and certainly not in Ruby where (?s) => (?m)

– Wiktor Stribiżew
Jul 18 '16 at 11:06

Anything for bash?

– Pasupathi Rajamanickam
Dec 19 '18 at 2:12

Anything for bash?

– Pasupathi Rajamanickam
Dec 19 '18 at 2:12

add a comment | 

46

The question is, can . pattern match any character? The answer varies from engine to engine. The main difference is whether the pattern is used by a POSIX or non-POSIX regex library.

Special note about lua-patterns: they are not considered regular expressions, but . matches any char there, same as POSIX based engines.

Another note on matlab and octave: the . matches any char by default (demo): str = "abcden fghij<Foobar>"; expression = '(.*)<Foobar>*'; [tokens,matches] = regexp(str,expression,'tokens','match'); (tokens contain a abcden fghij item).

Also, in all of boost's regex grammars the dot matches line breaks by default. Boost's ECMAScript grammar allows you to turn this off with regex_constants::no_mod_m (source).

As for oracle (it is POSIX based), use n option (demo): select regexp_substr('abcde' || chr(10) ||' fghij<Foobar>', '(.*)<Foobar>', 1, 1, 'n', 1) as results from dual

POSIX-based engines:

A mere . already matches line breaks, no need to use any modifiers, see bash (demo).

The tcl (demo), postgresql (demo), r (TRE, base R default engine with no perl=TRUE, for base R with perl=TRUE or for stringr/stringi patterns, use the (?s) inline modifier) (demo) also treat . the same way.

However, most POSIX based tools process input line by line. Hence, . does not match the line breaks just because they are not in scope. Here are some examples how to override this:

• 
  sed - There are multiple workarounds, the most precise but not very safe is sed 'H;1h;$!d;x; s/(.*)><Foobar>/1/' (H;1h;$!d;x; slurps the file into memory). If whole lines must be included, sed '/start_pattern/,/end_pattern/d' file (removing from start will end with matched lines included) or sed '/start_pattern/,/end_pattern/{{//!d;};}' file (with matching lines excluded) can be considered.


• 
  perl - perl -0pe 's/(.*)<FooBar>/$1/gs' <<< "$str" (-0 slurps the whole file into memory, -p prints the file after applying the script given by -e). Note that using -000pe will slurp the file and activate 'paragraph mode' where Perl uses consecutive newlines (nn) as the record separator.


• 
  gnu-grep - grep -Poz '(?si)abcK.*?(?=<Foobar>)' file. Here, z enables file slurping, (?s) enables the DOTALL mode for the . pattern, (?i) enables case insensitive mode, K omits the text matched so far, *? is a lazy quantifier, (?=<Foobar>) matches the location before <Foobar>.


• 
  pcregrep - pcregrep -Mi "(?si)abcK.*?(?=<Foobar>)" file (M enables file slurping here). Note pcregrep is a good solution for Mac OS grep users.

See demos.

Non-POSIX-based engines:

• 
  php - Use s modifier PCRE_DOTALL modifier: preg_match('~(.*)<Foobar>~s', $s, $m) (demo)


• 
  c# - Use RegexOptions.Singleline flag (demo):
  - var result = Regex.Match(s, @"(.*)<Foobar>", RegexOptions.Singleline).Groups[1].Value;
  - var result = Regex.Match(s, @"(?s)(.*)<Foobar>").Groups[1].Value;
  


• 
  powershell - Use (?s) inline option: $s = "abcde`nfghij<FooBar>"; $s -match "(?s)(.*)<Foobar>"; $matches[1]
  


• 
  perl - Use s modifier (or (?s) inline version at the start) (demo): /(.*)<FooBar>/s
  


• 
  python - Use re.DOTALL (or re.S) flags or (?s) inline modifier (demo): m = re.search(r"(.*)<FooBar>", s, flags=re.S) (and then if m:, print(m.group(1)))


• 
  java - Use Pattern.DOTALL modifier (or inline (?s) flag) (demo): Pattern.compile("(.*)<FooBar>", Pattern.DOTALL)
  


• 
  groovy - Use (?s) in-pattern modifier (demo): regex = /(?s)(.*)<FooBar>/
  


• 
  scala - Use (?s) modifier (demo): "(?s)(.*)<Foobar>".r.findAllIn("abcden fghij<Foobar>").matchData foreach { m => println(m.group(1)) }
  


• 
  javascript - Use [^] or workarounds [dD] / [wW] / [sS] (demo): s.match(/([sS]*)<FooBar>/)[1]
  


• 
  c++ (std::regex) Use [sS] or the JS workarounds (demo): regex rex(R"(([sS]*)<FooBar>)");
  


• 
  vba - Use the same approach as in JavaScript, ([sS]*)<Foobar>.


• 
  ruby - Use /m MULTILINE modifier (demo): s[/(.*)<Foobar>/m, 1]
  


• 
  go - Use the inline modifier (?s) at the start (demo): re: = regexp.MustCompile(`(?s)(.*)<FooBar>`)
  


• 
  swift - Use dotMatchesLineSeparators or (easier) pass the (?s) inline modifier to the pattern: let rx = "(?s)(.*)<Foobar>"
  


• 
  objective-c - Same as Swift, (?s) works the easiest, but here is how the option can be used: NSRegularExpression* regex = [NSRegularExpression regularExpressionWithPattern:pattern
options:NSRegularExpressionDotMatchesLineSeparators error:&regexError];
  


• 
  re2, google-apps-script - Use (?s) modifier (demo): "(?s)(.*)<Foobar>" (in Google Spreadsheets, =REGEXEXTRACT(A2,"(?s)(.*)<Foobar>"))

NOTES ON (?s):

In most non-POSIX engines, (?s) inline modifier (or embedded flag option) can be used to enforce . to match line breaks.

If placed at the start of the pattern, (?s) changes the bahavior of all . in the pattern. If the (?s) is placed somewhere after the beginning, only those . will be affected that are located to the right of it unless this is a pattern passed to Python re. In Python re, regardless of the (?s) location, the whole pattern . are affected. The (?s) effect is stopped using (?-s). A modified group can be used to only affect a specified range of a regex pattern (e.g. Delim1(?s:.*?)nDelim2.* will make the first .*? match across newlines and the second .* will only match the rest of the line).

POSIX note:

In non-regex engines, to match any char, [sS] / [dD] / [wW] constructs can be used.

In POSIX, [sS] is not matching any char (as in JavaScript or any non-POSIX engine) because regex escape sequences are not supported inside bracket expressions. [sS] is parsed as bracket expressions that match a single char, or s or S.

share|improve this answer

edited Dec 20 '18 at 9:49

answered Aug 31 '17 at 12:47

Wiktor StribiżewWiktor Stribiżew

320k16140222

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  
  You should link to this excellent overview from your profile page or something (+1).
  
  – Jan
  Oct 15 '17 at 20:15
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  You may want to add this to the boost item: In the regex_constants namespace, flag_type_'s : perl = ECMAScript = JavaScript = JScript = ::boost::regbase::normal = 0 which defaults to Perl. Programmers will set a base flag definition #define MOD regex_constants::perl | boost::regex::no_mod_s | boost::regex::no_mod_m for thier regex flags to reflect that. And the arbitor is always the inline modifiers. Where (?-sm)(?s).* resets.
  
  – sln
  Apr 26 '18 at 21:30
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Can you also add for bash please?
  
  – Pasupathi Rajamanickam
  Dec 19 '18 at 2:12
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  @PasupathiRajamanickam Bash uses a POSIX regex engine, the . matches any char there (including line breaks). See this online Bash demo.
  
  – Wiktor Stribiżew
  Dec 19 '18 at 7:33
  
  
  

  
  
  
  

add a comment | 

46

The question is, can . pattern match any character? The answer varies from engine to engine. The main difference is whether the pattern is used by a POSIX or non-POSIX regex library.

Special note about lua-patterns: they are not considered regular expressions, but . matches any char there, same as POSIX based engines.

Another note on matlab and octave: the . matches any char by default (demo): str = "abcden fghij<Foobar>"; expression = '(.*)<Foobar>*'; [tokens,matches] = regexp(str,expression,'tokens','match'); (tokens contain a abcden fghij item).

Also, in all of boost's regex grammars the dot matches line breaks by default. Boost's ECMAScript grammar allows you to turn this off with regex_constants::no_mod_m (source).

As for oracle (it is POSIX based), use n option (demo): select regexp_substr('abcde' || chr(10) ||' fghij<Foobar>', '(.*)<Foobar>', 1, 1, 'n', 1) as results from dual

POSIX-based engines:

A mere . already matches line breaks, no need to use any modifiers, see bash (demo).

The tcl (demo), postgresql (demo), r (TRE, base R default engine with no perl=TRUE, for base R with perl=TRUE or for stringr/stringi patterns, use the (?s) inline modifier) (demo) also treat . the same way.

However, most POSIX based tools process input line by line. Hence, . does not match the line breaks just because they are not in scope. Here are some examples how to override this:

• 
  sed - There are multiple workarounds, the most precise but not very safe is sed 'H;1h;$!d;x; s/(.*)><Foobar>/1/' (H;1h;$!d;x; slurps the file into memory). If whole lines must be included, sed '/start_pattern/,/end_pattern/d' file (removing from start will end with matched lines included) or sed '/start_pattern/,/end_pattern/{{//!d;};}' file (with matching lines excluded) can be considered.


• 
  perl - perl -0pe 's/(.*)<FooBar>/$1/gs' <<< "$str" (-0 slurps the whole file into memory, -p prints the file after applying the script given by -e). Note that using -000pe will slurp the file and activate 'paragraph mode' where Perl uses consecutive newlines (nn) as the record separator.


• 
  gnu-grep - grep -Poz '(?si)abcK.*?(?=<Foobar>)' file. Here, z enables file slurping, (?s) enables the DOTALL mode for the . pattern, (?i) enables case insensitive mode, K omits the text matched so far, *? is a lazy quantifier, (?=<Foobar>) matches the location before <Foobar>.


• 
  pcregrep - pcregrep -Mi "(?si)abcK.*?(?=<Foobar>)" file (M enables file slurping here). Note pcregrep is a good solution for Mac OS grep users.

See demos.

Non-POSIX-based engines:

• 
  php - Use s modifier PCRE_DOTALL modifier: preg_match('~(.*)<Foobar>~s', $s, $m) (demo)


• 
  c# - Use RegexOptions.Singleline flag (demo):
  - var result = Regex.Match(s, @"(.*)<Foobar>", RegexOptions.Singleline).Groups[1].Value;
  - var result = Regex.Match(s, @"(?s)(.*)<Foobar>").Groups[1].Value;
  


• 
  powershell - Use (?s) inline option: $s = "abcde`nfghij<FooBar>"; $s -match "(?s)(.*)<Foobar>"; $matches[1]
  


• 
  perl - Use s modifier (or (?s) inline version at the start) (demo): /(.*)<FooBar>/s
  


• 
  python - Use re.DOTALL (or re.S) flags or (?s) inline modifier (demo): m = re.search(r"(.*)<FooBar>", s, flags=re.S) (and then if m:, print(m.group(1)))


• 
  java - Use Pattern.DOTALL modifier (or inline (?s) flag) (demo): Pattern.compile("(.*)<FooBar>", Pattern.DOTALL)
  


• 
  groovy - Use (?s) in-pattern modifier (demo): regex = /(?s)(.*)<FooBar>/
  


• 
  scala - Use (?s) modifier (demo): "(?s)(.*)<Foobar>".r.findAllIn("abcden fghij<Foobar>").matchData foreach { m => println(m.group(1)) }
  


• 
  javascript - Use [^] or workarounds [dD] / [wW] / [sS] (demo): s.match(/([sS]*)<FooBar>/)[1]
  


• 
  c++ (std::regex) Use [sS] or the JS workarounds (demo): regex rex(R"(([sS]*)<FooBar>)");
  


• 
  vba - Use the same approach as in JavaScript, ([sS]*)<Foobar>.


• 
  ruby - Use /m MULTILINE modifier (demo): s[/(.*)<Foobar>/m, 1]
  


• 
  go - Use the inline modifier (?s) at the start (demo): re: = regexp.MustCompile(`(?s)(.*)<FooBar>`)
  


• 
  swift - Use dotMatchesLineSeparators or (easier) pass the (?s) inline modifier to the pattern: let rx = "(?s)(.*)<Foobar>"
  


• 
  objective-c - Same as Swift, (?s) works the easiest, but here is how the option can be used: NSRegularExpression* regex = [NSRegularExpression regularExpressionWithPattern:pattern
options:NSRegularExpressionDotMatchesLineSeparators error:&regexError];
  


• 
  re2, google-apps-script - Use (?s) modifier (demo): "(?s)(.*)<Foobar>" (in Google Spreadsheets, =REGEXEXTRACT(A2,"(?s)(.*)<Foobar>"))

NOTES ON (?s):

In most non-POSIX engines, (?s) inline modifier (or embedded flag option) can be used to enforce . to match line breaks.

If placed at the start of the pattern, (?s) changes the bahavior of all . in the pattern. If the (?s) is placed somewhere after the beginning, only those . will be affected that are located to the right of it unless this is a pattern passed to Python re. In Python re, regardless of the (?s) location, the whole pattern . are affected. The (?s) effect is stopped using (?-s). A modified group can be used to only affect a specified range of a regex pattern (e.g. Delim1(?s:.*?)nDelim2.* will make the first .*? match across newlines and the second .* will only match the rest of the line).

POSIX note:

In non-regex engines, to match any char, [sS] / [dD] / [wW] constructs can be used.

In POSIX, [sS] is not matching any char (as in JavaScript or any non-POSIX engine) because regex escape sequences are not supported inside bracket expressions. [sS] is parsed as bracket expressions that match a single char, or s or S.

share|improve this answer

edited Dec 20 '18 at 9:49

answered Aug 31 '17 at 12:47

Wiktor StribiżewWiktor Stribiżew

320k16140222

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  
  You should link to this excellent overview from your profile page or something (+1).
  
  – Jan
  Oct 15 '17 at 20:15
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  You may want to add this to the boost item: In the regex_constants namespace, flag_type_'s : perl = ECMAScript = JavaScript = JScript = ::boost::regbase::normal = 0 which defaults to Perl. Programmers will set a base flag definition #define MOD regex_constants::perl | boost::regex::no_mod_s | boost::regex::no_mod_m for thier regex flags to reflect that. And the arbitor is always the inline modifiers. Where (?-sm)(?s).* resets.
  
  – sln
  Apr 26 '18 at 21:30
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Can you also add for bash please?
  
  – Pasupathi Rajamanickam
  Dec 19 '18 at 2:12
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  @PasupathiRajamanickam Bash uses a POSIX regex engine, the . matches any char there (including line breaks). See this online Bash demo.
  
  – Wiktor Stribiżew
  Dec 19 '18 at 7:33
  
  
  

  
  
  
  

add a comment | 

46

46

46

The question is, can . pattern match any character? The answer varies from engine to engine. The main difference is whether the pattern is used by a POSIX or non-POSIX regex library.

Special note about lua-patterns: they are not considered regular expressions, but . matches any char there, same as POSIX based engines.

Another note on matlab and octave: the . matches any char by default (demo): str = "abcden fghij<Foobar>"; expression = '(.*)<Foobar>*'; [tokens,matches] = regexp(str,expression,'tokens','match'); (tokens contain a abcden fghij item).

Also, in all of boost's regex grammars the dot matches line breaks by default. Boost's ECMAScript grammar allows you to turn this off with regex_constants::no_mod_m (source).

As for oracle (it is POSIX based), use n option (demo): select regexp_substr('abcde' || chr(10) ||' fghij<Foobar>', '(.*)<Foobar>', 1, 1, 'n', 1) as results from dual

POSIX-based engines:

A mere . already matches line breaks, no need to use any modifiers, see bash (demo).

The tcl (demo), postgresql (demo), r (TRE, base R default engine with no perl=TRUE, for base R with perl=TRUE or for stringr/stringi patterns, use the (?s) inline modifier) (demo) also treat . the same way.

However, most POSIX based tools process input line by line. Hence, . does not match the line breaks just because they are not in scope. Here are some examples how to override this:

• 
  sed - There are multiple workarounds, the most precise but not very safe is sed 'H;1h;$!d;x; s/(.*)><Foobar>/1/' (H;1h;$!d;x; slurps the file into memory). If whole lines must be included, sed '/start_pattern/,/end_pattern/d' file (removing from start will end with matched lines included) or sed '/start_pattern/,/end_pattern/{{//!d;};}' file (with matching lines excluded) can be considered.


• 
  perl - perl -0pe 's/(.*)<FooBar>/$1/gs' <<< "$str" (-0 slurps the whole file into memory, -p prints the file after applying the script given by -e). Note that using -000pe will slurp the file and activate 'paragraph mode' where Perl uses consecutive newlines (nn) as the record separator.


• 
  gnu-grep - grep -Poz '(?si)abcK.*?(?=<Foobar>)' file. Here, z enables file slurping, (?s) enables the DOTALL mode for the . pattern, (?i) enables case insensitive mode, K omits the text matched so far, *? is a lazy quantifier, (?=<Foobar>) matches the location before <Foobar>.


• 
  pcregrep - pcregrep -Mi "(?si)abcK.*?(?=<Foobar>)" file (M enables file slurping here). Note pcregrep is a good solution for Mac OS grep users.

See demos.

Non-POSIX-based engines:

• 
  php - Use s modifier PCRE_DOTALL modifier: preg_match('~(.*)<Foobar>~s', $s, $m) (demo)


• 
  c# - Use RegexOptions.Singleline flag (demo):
  - var result = Regex.Match(s, @"(.*)<Foobar>", RegexOptions.Singleline).Groups[1].Value;
  - var result = Regex.Match(s, @"(?s)(.*)<Foobar>").Groups[1].Value;
  


• 
  powershell - Use (?s) inline option: $s = "abcde`nfghij<FooBar>"; $s -match "(?s)(.*)<Foobar>"; $matches[1]
  


• 
  perl - Use s modifier (or (?s) inline version at the start) (demo): /(.*)<FooBar>/s
  


• 
  python - Use re.DOTALL (or re.S) flags or (?s) inline modifier (demo): m = re.search(r"(.*)<FooBar>", s, flags=re.S) (and then if m:, print(m.group(1)))


• 
  java - Use Pattern.DOTALL modifier (or inline (?s) flag) (demo): Pattern.compile("(.*)<FooBar>", Pattern.DOTALL)
  


• 
  groovy - Use (?s) in-pattern modifier (demo): regex = /(?s)(.*)<FooBar>/
  


• 
  scala - Use (?s) modifier (demo): "(?s)(.*)<Foobar>".r.findAllIn("abcden fghij<Foobar>").matchData foreach { m => println(m.group(1)) }
  


• 
  javascript - Use [^] or workarounds [dD] / [wW] / [sS] (demo): s.match(/([sS]*)<FooBar>/)[1]
  


• 
  c++ (std::regex) Use [sS] or the JS workarounds (demo): regex rex(R"(([sS]*)<FooBar>)");
  


• 
  vba - Use the same approach as in JavaScript, ([sS]*)<Foobar>.


• 
  ruby - Use /m MULTILINE modifier (demo): s[/(.*)<Foobar>/m, 1]
  


• 
  go - Use the inline modifier (?s) at the start (demo): re: = regexp.MustCompile(`(?s)(.*)<FooBar>`)
  


• 
  swift - Use dotMatchesLineSeparators or (easier) pass the (?s) inline modifier to the pattern: let rx = "(?s)(.*)<Foobar>"
  


• 
  objective-c - Same as Swift, (?s) works the easiest, but here is how the option can be used: NSRegularExpression* regex = [NSRegularExpression regularExpressionWithPattern:pattern
options:NSRegularExpressionDotMatchesLineSeparators error:&regexError];
  


• 
  re2, google-apps-script - Use (?s) modifier (demo): "(?s)(.*)<Foobar>" (in Google Spreadsheets, =REGEXEXTRACT(A2,"(?s)(.*)<Foobar>"))

NOTES ON (?s):

In most non-POSIX engines, (?s) inline modifier (or embedded flag option) can be used to enforce . to match line breaks.

If placed at the start of the pattern, (?s) changes the bahavior of all . in the pattern. If the (?s) is placed somewhere after the beginning, only those . will be affected that are located to the right of it unless this is a pattern passed to Python re. In Python re, regardless of the (?s) location, the whole pattern . are affected. The (?s) effect is stopped using (?-s). A modified group can be used to only affect a specified range of a regex pattern (e.g. Delim1(?s:.*?)nDelim2.* will make the first .*? match across newlines and the second .* will only match the rest of the line).

POSIX note:

In non-regex engines, to match any char, [sS] / [dD] / [wW] constructs can be used.

In POSIX, [sS] is not matching any char (as in JavaScript or any non-POSIX engine) because regex escape sequences are not supported inside bracket expressions. [sS] is parsed as bracket expressions that match a single char, or s or S.

share|improve this answer

edited Dec 20 '18 at 9:49

answered Aug 31 '17 at 12:47

Wiktor StribiżewWiktor Stribiżew

320k16140222

The question is, can . pattern match any character? The answer varies from engine to engine. The main difference is whether the pattern is used by a POSIX or non-POSIX regex library.

Special note about lua-patterns: they are not considered regular expressions, but . matches any char there, same as POSIX based engines.

Another note on matlab and octave: the . matches any char by default (demo): str = "abcden fghij<Foobar>"; expression = '(.*)<Foobar>*'; [tokens,matches] = regexp(str,expression,'tokens','match'); (tokens contain a abcden fghij item).

Also, in all of boost's regex grammars the dot matches line breaks by default. Boost's ECMAScript grammar allows you to turn this off with regex_constants::no_mod_m (source).

As for oracle (it is POSIX based), use n option (demo): select regexp_substr('abcde' || chr(10) ||' fghij<Foobar>', '(.*)<Foobar>', 1, 1, 'n', 1) as results from dual

POSIX-based engines:

A mere . already matches line breaks, no need to use any modifiers, see bash (demo).

The tcl (demo), postgresql (demo), r (TRE, base R default engine with no perl=TRUE, for base R with perl=TRUE or for stringr/stringi patterns, use the (?s) inline modifier) (demo) also treat . the same way.

However, most POSIX based tools process input line by line. Hence, . does not match the line breaks just because they are not in scope. Here are some examples how to override this:

• 
  sed - There are multiple workarounds, the most precise but not very safe is sed 'H;1h;$!d;x; s/(.*)><Foobar>/1/' (H;1h;$!d;x; slurps the file into memory). If whole lines must be included, sed '/start_pattern/,/end_pattern/d' file (removing from start will end with matched lines included) or sed '/start_pattern/,/end_pattern/{{//!d;};}' file (with matching lines excluded) can be considered.


• 
  perl - perl -0pe 's/(.*)<FooBar>/$1/gs' <<< "$str" (-0 slurps the whole file into memory, -p prints the file after applying the script given by -e). Note that using -000pe will slurp the file and activate 'paragraph mode' where Perl uses consecutive newlines (nn) as the record separator.


• 
  gnu-grep - grep -Poz '(?si)abcK.*?(?=<Foobar>)' file. Here, z enables file slurping, (?s) enables the DOTALL mode for the . pattern, (?i) enables case insensitive mode, K omits the text matched so far, *? is a lazy quantifier, (?=<Foobar>) matches the location before <Foobar>.


• 
  pcregrep - pcregrep -Mi "(?si)abcK.*?(?=<Foobar>)" file (M enables file slurping here). Note pcregrep is a good solution for Mac OS grep users.

See demos.

Non-POSIX-based engines:

• 
  php - Use s modifier PCRE_DOTALL modifier: preg_match('~(.*)<Foobar>~s', $s, $m) (demo)


• 
  c# - Use RegexOptions.Singleline flag (demo):
  - var result = Regex.Match(s, @"(.*)<Foobar>", RegexOptions.Singleline).Groups[1].Value;
  - var result = Regex.Match(s, @"(?s)(.*)<Foobar>").Groups[1].Value;
  


• 
  powershell - Use (?s) inline option: $s = "abcde`nfghij<FooBar>"; $s -match "(?s)(.*)<Foobar>"; $matches[1]
  


• 
  perl - Use s modifier (or (?s) inline version at the start) (demo): /(.*)<FooBar>/s
  


• 
  python - Use re.DOTALL (or re.S) flags or (?s) inline modifier (demo): m = re.search(r"(.*)<FooBar>", s, flags=re.S) (and then if m:, print(m.group(1)))


• 
  java - Use Pattern.DOTALL modifier (or inline (?s) flag) (demo): Pattern.compile("(.*)<FooBar>", Pattern.DOTALL)
  


• 
  groovy - Use (?s) in-pattern modifier (demo): regex = /(?s)(.*)<FooBar>/
  


• 
  scala - Use (?s) modifier (demo): "(?s)(.*)<Foobar>".r.findAllIn("abcden fghij<Foobar>").matchData foreach { m => println(m.group(1)) }
  


• 
  javascript - Use [^] or workarounds [dD] / [wW] / [sS] (demo): s.match(/([sS]*)<FooBar>/)[1]
  


• 
  c++ (std::regex) Use [sS] or the JS workarounds (demo): regex rex(R"(([sS]*)<FooBar>)");
  


• 
  vba - Use the same approach as in JavaScript, ([sS]*)<Foobar>.


• 
  ruby - Use /m MULTILINE modifier (demo): s[/(.*)<Foobar>/m, 1]
  


• 
  go - Use the inline modifier (?s) at the start (demo): re: = regexp.MustCompile(`(?s)(.*)<FooBar>`)
  


• 
  swift - Use dotMatchesLineSeparators or (easier) pass the (?s) inline modifier to the pattern: let rx = "(?s)(.*)<Foobar>"
  


• 
  objective-c - Same as Swift, (?s) works the easiest, but here is how the option can be used: NSRegularExpression* regex = [NSRegularExpression regularExpressionWithPattern:pattern
options:NSRegularExpressionDotMatchesLineSeparators error:&regexError];
  


• 
  re2, google-apps-script - Use (?s) modifier (demo): "(?s)(.*)<Foobar>" (in Google Spreadsheets, =REGEXEXTRACT(A2,"(?s)(.*)<Foobar>"))

NOTES ON (?s):

In most non-POSIX engines, (?s) inline modifier (or embedded flag option) can be used to enforce . to match line breaks.

If placed at the start of the pattern, (?s) changes the bahavior of all . in the pattern. If the (?s) is placed somewhere after the beginning, only those . will be affected that are located to the right of it unless this is a pattern passed to Python re. In Python re, regardless of the (?s) location, the whole pattern . are affected. The (?s) effect is stopped using (?-s). A modified group can be used to only affect a specified range of a regex pattern (e.g. Delim1(?s:.*?)nDelim2.* will make the first .*? match across newlines and the second .* will only match the rest of the line).

POSIX note:

In non-regex engines, to match any char, [sS] / [dD] / [wW] constructs can be used.

In POSIX, [sS] is not matching any char (as in JavaScript or any non-POSIX engine) because regex escape sequences are not supported inside bracket expressions. [sS] is parsed as bracket expressions that match a single char, or s or S.

share|improve this answer

edited Dec 20 '18 at 9:49

answered Aug 31 '17 at 12:47

Wiktor StribiżewWiktor Stribiżew

320k16140222

share|improve this answer

share|improve this answer

edited Dec 20 '18 at 9:49

edited Dec 20 '18 at 9:49

edited Dec 20 '18 at 9:49

answered Aug 31 '17 at 12:47

Wiktor StribiżewWiktor Stribiżew

320k16140222

answered Aug 31 '17 at 12:47

Wiktor StribiżewWiktor Stribiżew

320k16140222

answered Aug 31 '17 at 12:47

Wiktor StribiżewWiktor Stribiżew

320k16140222

320k16140222

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  
  You should link to this excellent overview from your profile page or something (+1).
  
  – Jan
  Oct 15 '17 at 20:15
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  You may want to add this to the boost item: In the regex_constants namespace, flag_type_'s : perl = ECMAScript = JavaScript = JScript = ::boost::regbase::normal = 0 which defaults to Perl. Programmers will set a base flag definition #define MOD regex_constants::perl | boost::regex::no_mod_s | boost::regex::no_mod_m for thier regex flags to reflect that. And the arbitor is always the inline modifiers. Where (?-sm)(?s).* resets.
  
  – sln
  Apr 26 '18 at 21:30
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Can you also add for bash please?
  
  – Pasupathi Rajamanickam
  Dec 19 '18 at 2:12
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  @PasupathiRajamanickam Bash uses a POSIX regex engine, the . matches any char there (including line breaks). See this online Bash demo.
  
  – Wiktor Stribiżew
  Dec 19 '18 at 7:33
  
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  
  You should link to this excellent overview from your profile page or something (+1).
  
  – Jan
  Oct 15 '17 at 20:15
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  You may want to add this to the boost item: In the regex_constants namespace, flag_type_'s : perl = ECMAScript = JavaScript = JScript = ::boost::regbase::normal = 0 which defaults to Perl. Programmers will set a base flag definition #define MOD regex_constants::perl | boost::regex::no_mod_s | boost::regex::no_mod_m for thier regex flags to reflect that. And the arbitor is always the inline modifiers. Where (?-sm)(?s).* resets.
  
  – sln
  Apr 26 '18 at 21:30
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Can you also add for bash please?
  
  – Pasupathi Rajamanickam
  Dec 19 '18 at 2:12
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  @PasupathiRajamanickam Bash uses a POSIX regex engine, the . matches any char there (including line breaks). See this online Bash demo.
  
  – Wiktor Stribiżew
  Dec 19 '18 at 7:33
  
  
  

  
  
  
  

4

4

You should link to this excellent overview from your profile page or something (+1).

– Jan
Oct 15 '17 at 20:15

You should link to this excellent overview from your profile page or something (+1).

– Jan
Oct 15 '17 at 20:15

1

1

You may want to add this to the boost item: In the regex_constants namespace, flag_type_'s : perl = ECMAScript = JavaScript = JScript = ::boost::regbase::normal = 0 which defaults to Perl. Programmers will set a base flag definition #define MOD regex_constants::perl | boost::regex::no_mod_s | boost::regex::no_mod_m for thier regex flags to reflect that. And the arbitor is always the inline modifiers. Where (?-sm)(?s).* resets.

– sln
Apr 26 '18 at 21:30

You may want to add this to the boost item: In the regex_constants namespace, flag_type_'s : perl = ECMAScript = JavaScript = JScript = ::boost::regbase::normal = 0 which defaults to Perl. Programmers will set a base flag definition #define MOD regex_constants::perl | boost::regex::no_mod_s | boost::regex::no_mod_m for thier regex flags to reflect that. And the arbitor is always the inline modifiers. Where (?-sm)(?s).* resets.

– sln
Apr 26 '18 at 21:30

1

1

Can you also add for bash please?

– Pasupathi Rajamanickam
Dec 19 '18 at 2:12

Can you also add for bash please?

– Pasupathi Rajamanickam
Dec 19 '18 at 2:12

2

2

@PasupathiRajamanickam Bash uses a POSIX regex engine, the . matches any char there (including line breaks). See this online Bash demo.

– Wiktor Stribiżew
Dec 19 '18 at 7:33

@PasupathiRajamanickam Bash uses a POSIX regex engine, the . matches any char there (including line breaks). See this online Bash demo.

– Wiktor Stribiżew
Dec 19 '18 at 7:33

add a comment | 

31

In JavaScript, use /[Ss]*<Foobar>/. Source

share|improve this answer

edited Feb 23 '17 at 1:15

Nathan Arthur

3,00032447

answered Jul 30 '11 at 13:03

Abbas ShahzadehAbbas Shahzadeh

31132

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  From that link: "JavaScript and VBScript do not have an option to make the dot match line break characters. In those languages, you can use a character class such as [sS] to match any character." Instead of the . use [sS] (match spaces and non-spaces) instead.
  
  – Allen
  May 9 '13 at 15:34
  

  
  
  
  

add a comment | 

31

In JavaScript, use /[Ss]*<Foobar>/. Source

share|improve this answer

edited Feb 23 '17 at 1:15

Nathan Arthur

3,00032447

answered Jul 30 '11 at 13:03

Abbas ShahzadehAbbas Shahzadeh

31132

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  From that link: "JavaScript and VBScript do not have an option to make the dot match line break characters. In those languages, you can use a character class such as [sS] to match any character." Instead of the . use [sS] (match spaces and non-spaces) instead.
  
  – Allen
  May 9 '13 at 15:34
  

  
  
  
  

add a comment | 

31

31

31

In JavaScript, use /[Ss]*<Foobar>/. Source

share|improve this answer

edited Feb 23 '17 at 1:15

Nathan Arthur

3,00032447

answered Jul 30 '11 at 13:03

Abbas ShahzadehAbbas Shahzadeh

31132

In JavaScript, use /[Ss]*<Foobar>/. Source

share|improve this answer

edited Feb 23 '17 at 1:15

Nathan Arthur

3,00032447

answered Jul 30 '11 at 13:03

Abbas ShahzadehAbbas Shahzadeh

31132

share|improve this answer

share|improve this answer

edited Feb 23 '17 at 1:15

Nathan Arthur

3,00032447

edited Feb 23 '17 at 1:15

Nathan Arthur

3,00032447

edited Feb 23 '17 at 1:15

Nathan Arthur

3,00032447

3,00032447

answered Jul 30 '11 at 13:03

Abbas ShahzadehAbbas Shahzadeh

31132

answered Jul 30 '11 at 13:03

Abbas ShahzadehAbbas Shahzadeh

31132

answered Jul 30 '11 at 13:03

Abbas ShahzadehAbbas Shahzadeh

31132

31132

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  From that link: "JavaScript and VBScript do not have an option to make the dot match line break characters. In those languages, you can use a character class such as [sS] to match any character." Instead of the . use [sS] (match spaces and non-spaces) instead.
  
  – Allen
  May 9 '13 at 15:34
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  From that link: "JavaScript and VBScript do not have an option to make the dot match line break characters. In those languages, you can use a character class such as [sS] to match any character." Instead of the . use [sS] (match spaces and non-spaces) instead.
  
  – Allen
  May 9 '13 at 15:34
  

  
  
  
  

2

2

From that link: "JavaScript and VBScript do not have an option to make the dot match line break characters. In those languages, you can use a character class such as [sS] to match any character." Instead of the . use [sS] (match spaces and non-spaces) instead.

– Allen
May 9 '13 at 15:34

From that link: "JavaScript and VBScript do not have an option to make the dot match line break characters. In those languages, you can use a character class such as [sS] to match any character." Instead of the . use [sS] (match spaces and non-spaces) instead.

– Allen
May 9 '13 at 15:34

add a comment | 

27

([sS]*)<FooBar>

The dot matches all except newlines (rn). So use sS, which will match ALL characters.

share|improve this answer

answered Jul 19 '12 at 17:59

samwizesamwize

14.7k882140

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  This solve the problem if you are using the Objective-C [text rangeOfString:regEx options:NSRegularExpressionSearch]. Thanks!
  
  – J. Costa
  Aug 24 '12 at 22:29
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  This works in intelliJ's find&replace regex, thanks.
  
  – barclay
  Sep 16 '15 at 22:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  This works. But it needs to be the first occurrence of <FooBar>
  
  – Ozkan
  Sep 26 '17 at 14:16
  

  
  
  
  

add a comment | 

27

([sS]*)<FooBar>

The dot matches all except newlines (rn). So use sS, which will match ALL characters.

share|improve this answer

answered Jul 19 '12 at 17:59

samwizesamwize

14.7k882140

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  This solve the problem if you are using the Objective-C [text rangeOfString:regEx options:NSRegularExpressionSearch]. Thanks!
  
  – J. Costa
  Aug 24 '12 at 22:29
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  This works in intelliJ's find&replace regex, thanks.
  
  – barclay
  Sep 16 '15 at 22:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  This works. But it needs to be the first occurrence of <FooBar>
  
  – Ozkan
  Sep 26 '17 at 14:16
  

  
  
  
  

add a comment | 

27

27

27

([sS]*)<FooBar>

The dot matches all except newlines (rn). So use sS, which will match ALL characters.

share|improve this answer

answered Jul 19 '12 at 17:59

samwizesamwize

14.7k882140

([sS]*)<FooBar>

The dot matches all except newlines (rn). So use sS, which will match ALL characters.

share|improve this answer

answered Jul 19 '12 at 17:59

samwizesamwize

14.7k882140

share|improve this answer

share|improve this answer

answered Jul 19 '12 at 17:59

samwizesamwize

14.7k882140

answered Jul 19 '12 at 17:59

samwizesamwize

14.7k882140

answered Jul 19 '12 at 17:59

samwizesamwize

14.7k882140

14.7k882140

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  This solve the problem if you are using the Objective-C [text rangeOfString:regEx options:NSRegularExpressionSearch]. Thanks!
  
  – J. Costa
  Aug 24 '12 at 22:29
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  This works in intelliJ's find&replace regex, thanks.
  
  – barclay
  Sep 16 '15 at 22:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  This works. But it needs to be the first occurrence of <FooBar>
  
  – Ozkan
  Sep 26 '17 at 14:16
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  This solve the problem if you are using the Objective-C [text rangeOfString:regEx options:NSRegularExpressionSearch]. Thanks!
  
  – J. Costa
  Aug 24 '12 at 22:29
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  This works in intelliJ's find&replace regex, thanks.
  
  – barclay
  Sep 16 '15 at 22:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  This works. But it needs to be the first occurrence of <FooBar>
  
  – Ozkan
  Sep 26 '17 at 14:16
  

  
  
  
  

This solve the problem if you are using the Objective-C [text rangeOfString:regEx options:NSRegularExpressionSearch]. Thanks!

– J. Costa
Aug 24 '12 at 22:29

This solve the problem if you are using the Objective-C [text rangeOfString:regEx options:NSRegularExpressionSearch]. Thanks!

– J. Costa
Aug 24 '12 at 22:29

This works in intelliJ's find&replace regex, thanks.

– barclay
Sep 16 '15 at 22:14

This works in intelliJ's find&replace regex, thanks.

– barclay
Sep 16 '15 at 22:14

This works. But it needs to be the first occurrence of <FooBar>

– Ozkan
Sep 26 '17 at 14:16

This works. But it needs to be the first occurrence of <FooBar>

– Ozkan
Sep 26 '17 at 14:16

add a comment | 

18

In Ruby ruby you can use the 'm' option (multiline):

/YOUR_REGEXP/m

See the Regexp documentation on ruby-doc.org for more information.

share|improve this answer

edited Aug 31 '17 at 7:54

Wiktor Stribiżew

320k16140222

answered Aug 3 '12 at 7:52

vibaihervibaiher

38228

add a comment | 

18

In Ruby ruby you can use the 'm' option (multiline):

/YOUR_REGEXP/m

See the Regexp documentation on ruby-doc.org for more information.

share|improve this answer

edited Aug 31 '17 at 7:54

Wiktor Stribiżew

320k16140222

answered Aug 3 '12 at 7:52

vibaihervibaiher

38228

add a comment | 

18

18

18

In Ruby ruby you can use the 'm' option (multiline):

/YOUR_REGEXP/m

See the Regexp documentation on ruby-doc.org for more information.

share|improve this answer

edited Aug 31 '17 at 7:54

Wiktor Stribiżew

320k16140222

answered Aug 3 '12 at 7:52

vibaihervibaiher

38228

In Ruby ruby you can use the 'm' option (multiline):

/YOUR_REGEXP/m

See the Regexp documentation on ruby-doc.org for more information.

share|improve this answer

edited Aug 31 '17 at 7:54

Wiktor Stribiżew

320k16140222

answered Aug 3 '12 at 7:52

vibaihervibaiher

38228

share|improve this answer

share|improve this answer

edited Aug 31 '17 at 7:54

Wiktor Stribiżew

320k16140222

edited Aug 31 '17 at 7:54

Wiktor Stribiżew

320k16140222

edited Aug 31 '17 at 7:54

Wiktor Stribiżew

320k16140222

320k16140222

answered Aug 3 '12 at 7:52

vibaihervibaiher

38228

answered Aug 3 '12 at 7:52

vibaihervibaiher

38228

answered Aug 3 '12 at 7:52

vibaihervibaiher

38228

38228

add a comment | 

add a comment | 

10

we can also use

(.*?n)*?

to match everything including newline without greedy

This will make the new line optional

(.*?|n)*?

share|improve this answer

answered Aug 6 '18 at 7:48

RAN_0915RAN_0915

861216

add a comment | 

10

we can also use

(.*?n)*?

to match everything including newline without greedy

This will make the new line optional

(.*?|n)*?

share|improve this answer

answered Aug 6 '18 at 7:48

RAN_0915RAN_0915

861216

add a comment | 

10

10

10

we can also use

(.*?n)*?

to match everything including newline without greedy

This will make the new line optional

(.*?|n)*?

share|improve this answer

answered Aug 6 '18 at 7:48

RAN_0915RAN_0915

861216

we can also use

(.*?n)*?

to match everything including newline without greedy

This will make the new line optional

(.*?|n)*?

share|improve this answer

answered Aug 6 '18 at 7:48

RAN_0915RAN_0915

861216

share|improve this answer

share|improve this answer

answered Aug 6 '18 at 7:48

RAN_0915RAN_0915

861216

answered Aug 6 '18 at 7:48

RAN_0915RAN_0915

861216

answered Aug 6 '18 at 7:48

RAN_0915RAN_0915

861216

861216

add a comment | 

add a comment | 

8

"." normally doesn't match line-breaks. Most regex engines allows you to add the S-flag (also called DOTALL and SINGLELINE) to make "." also match newlines.
If that fails, you could do something like [Ss].

share|improve this answer

edited Apr 25 '09 at 21:09

answered Oct 1 '08 at 18:52

Markus JarderotMarkus Jarderot

66.8k12112120

add a comment | 

8

"." normally doesn't match line-breaks. Most regex engines allows you to add the S-flag (also called DOTALL and SINGLELINE) to make "." also match newlines.
If that fails, you could do something like [Ss].

share|improve this answer

edited Apr 25 '09 at 21:09

answered Oct 1 '08 at 18:52

Markus JarderotMarkus Jarderot

66.8k12112120

add a comment | 

8

8

8

"." normally doesn't match line-breaks. Most regex engines allows you to add the S-flag (also called DOTALL and SINGLELINE) to make "." also match newlines.
If that fails, you could do something like [Ss].

share|improve this answer

edited Apr 25 '09 at 21:09

answered Oct 1 '08 at 18:52

Markus JarderotMarkus Jarderot

66.8k12112120

"." normally doesn't match line-breaks. Most regex engines allows you to add the S-flag (also called DOTALL and SINGLELINE) to make "." also match newlines.
If that fails, you could do something like [Ss].

share|improve this answer

edited Apr 25 '09 at 21:09

answered Oct 1 '08 at 18:52

Markus JarderotMarkus Jarderot

66.8k12112120

share|improve this answer

share|improve this answer

edited Apr 25 '09 at 21:09

edited Apr 25 '09 at 21:09

edited Apr 25 '09 at 21:09

answered Oct 1 '08 at 18:52

Markus JarderotMarkus Jarderot

66.8k12112120

answered Oct 1 '08 at 18:52

Markus JarderotMarkus Jarderot

66.8k12112120

answered Oct 1 '08 at 18:52

Markus JarderotMarkus Jarderot

66.8k12112120

66.8k12112120

add a comment | 

add a comment | 

7

For Eclipse worked following expression:

Foo

jadajada Bar"

Regular-Expression:

Foo[Ss]{1,10}.*Bar*

share|improve this answer

edited Jan 3 '13 at 11:52

devOp

2,91411232

answered Jan 3 '13 at 11:32

GordonGordon

7111

add a comment | 

7

For Eclipse worked following expression:

Foo

jadajada Bar"

Regular-Expression:

Foo[Ss]{1,10}.*Bar*

share|improve this answer

edited Jan 3 '13 at 11:52

devOp

2,91411232

answered Jan 3 '13 at 11:32

GordonGordon

7111

add a comment | 

7

7

7

For Eclipse worked following expression:

Foo

jadajada Bar"

Regular-Expression:

Foo[Ss]{1,10}.*Bar*

share|improve this answer

edited Jan 3 '13 at 11:52

devOp

2,91411232

answered Jan 3 '13 at 11:32

GordonGordon

7111

For Eclipse worked following expression:

Foo

jadajada Bar"

Regular-Expression:

Foo[Ss]{1,10}.*Bar*

share|improve this answer

edited Jan 3 '13 at 11:52

devOp

2,91411232

answered Jan 3 '13 at 11:32

GordonGordon

7111

share|improve this answer

share|improve this answer

edited Jan 3 '13 at 11:52

devOp

2,91411232

edited Jan 3 '13 at 11:52

devOp

2,91411232

edited Jan 3 '13 at 11:52

devOp

2,91411232

2,91411232

answered Jan 3 '13 at 11:32

GordonGordon

7111

answered Jan 3 '13 at 11:32

GordonGordon

7111

answered Jan 3 '13 at 11:32

GordonGordon

7111

7111

add a comment | 

add a comment | 

5

/(.*)<FooBar>/s

the s causes Dot (.) to match carriage returns

share|improve this answer

answered Oct 1 '08 at 18:54

BillBill

2,32882429

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Seems like this is invalid (Chrome): text.match(/a/s) SyntaxError: Invalid flags supplied to RegExp constructor 's'
  
  – Allen
  May 9 '13 at 15:31
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Because it is unsupported in JavaScript RegEx engines. The s flags exists in PCRE, the most complete engine (available in Perl and PHP). PCRE has 10 flags (and a lot of other features) while JavaScript has only 3 flags (gmi).
  
  – Morgan Touverey Quilling
  Apr 20 '16 at 18:51
  

  
  
  
  

add a comment | 

5

/(.*)<FooBar>/s

the s causes Dot (.) to match carriage returns

share|improve this answer

answered Oct 1 '08 at 18:54

BillBill

2,32882429

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Seems like this is invalid (Chrome): text.match(/a/s) SyntaxError: Invalid flags supplied to RegExp constructor 's'
  
  – Allen
  May 9 '13 at 15:31
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Because it is unsupported in JavaScript RegEx engines. The s flags exists in PCRE, the most complete engine (available in Perl and PHP). PCRE has 10 flags (and a lot of other features) while JavaScript has only 3 flags (gmi).
  
  – Morgan Touverey Quilling
  Apr 20 '16 at 18:51
  

  
  
  
  

add a comment | 

5

5

5

/(.*)<FooBar>/s

the s causes Dot (.) to match carriage returns

share|improve this answer

answered Oct 1 '08 at 18:54

BillBill

2,32882429

/(.*)<FooBar>/s

the s causes Dot (.) to match carriage returns

share|improve this answer

answered Oct 1 '08 at 18:54

BillBill

2,32882429

share|improve this answer

share|improve this answer

answered Oct 1 '08 at 18:54

BillBill

2,32882429

answered Oct 1 '08 at 18:54

BillBill

2,32882429

answered Oct 1 '08 at 18:54

BillBill

2,32882429

2,32882429

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Seems like this is invalid (Chrome): text.match(/a/s) SyntaxError: Invalid flags supplied to RegExp constructor 's'
  
  – Allen
  May 9 '13 at 15:31
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Because it is unsupported in JavaScript RegEx engines. The s flags exists in PCRE, the most complete engine (available in Perl and PHP). PCRE has 10 flags (and a lot of other features) while JavaScript has only 3 flags (gmi).
  
  – Morgan Touverey Quilling
  Apr 20 '16 at 18:51
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Seems like this is invalid (Chrome): text.match(/a/s) SyntaxError: Invalid flags supplied to RegExp constructor 's'
  
  – Allen
  May 9 '13 at 15:31
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Because it is unsupported in JavaScript RegEx engines. The s flags exists in PCRE, the most complete engine (available in Perl and PHP). PCRE has 10 flags (and a lot of other features) while JavaScript has only 3 flags (gmi).
  
  – Morgan Touverey Quilling
  Apr 20 '16 at 18:51
  

  
  
  
  

Seems like this is invalid (Chrome): text.match(/a/s) SyntaxError: Invalid flags supplied to RegExp constructor 's'

– Allen
May 9 '13 at 15:31

Seems like this is invalid (Chrome): text.match(/a/s) SyntaxError: Invalid flags supplied to RegExp constructor 's'

– Allen
May 9 '13 at 15:31

Because it is unsupported in JavaScript RegEx engines. The s flags exists in PCRE, the most complete engine (available in Perl and PHP). PCRE has 10 flags (and a lot of other features) while JavaScript has only 3 flags (gmi).

– Morgan Touverey Quilling
Apr 20 '16 at 18:51

Because it is unsupported in JavaScript RegEx engines. The s flags exists in PCRE, the most complete engine (available in Perl and PHP). PCRE has 10 flags (and a lot of other features) while JavaScript has only 3 flags (gmi).

– Morgan Touverey Quilling
Apr 20 '16 at 18:51

add a comment | 

4

In java based regular expression you can use [sS]

share|improve this answer

edited May 25 '18 at 19:01

revo

33.3k135085

answered Jun 3 '13 at 6:22

KamahireKamahire

1,47721646

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Shouldn't those be backslashes?
  
  – Paul Draper
  Oct 19 '13 at 6:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  They go at the end of the Regular Expression, not within in. Example: /blah/s
  
  – RandomInsano
  Dec 21 '13 at 20:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I guess you mean JavaScript, not Java? Since you can just add the s flag to the pattern in Java and JavaScript doesn't have the s flag.
  
  – 3limin4t0r
  Sep 25 '18 at 17:47
  
  
  

  
  
  
  

add a comment | 

4

In java based regular expression you can use [sS]

share|improve this answer

edited May 25 '18 at 19:01

revo

33.3k135085

answered Jun 3 '13 at 6:22

KamahireKamahire

1,47721646

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Shouldn't those be backslashes?
  
  – Paul Draper
  Oct 19 '13 at 6:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  They go at the end of the Regular Expression, not within in. Example: /blah/s
  
  – RandomInsano
  Dec 21 '13 at 20:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I guess you mean JavaScript, not Java? Since you can just add the s flag to the pattern in Java and JavaScript doesn't have the s flag.
  
  – 3limin4t0r
  Sep 25 '18 at 17:47
  
  
  

  
  
  
  

add a comment | 

4

4

4

In java based regular expression you can use [sS]

share|improve this answer

edited May 25 '18 at 19:01

revo

33.3k135085

answered Jun 3 '13 at 6:22

KamahireKamahire

1,47721646

In java based regular expression you can use [sS]

share|improve this answer

edited May 25 '18 at 19:01

revo

33.3k135085

answered Jun 3 '13 at 6:22

KamahireKamahire

1,47721646

share|improve this answer

share|improve this answer

edited May 25 '18 at 19:01

revo

33.3k135085

edited May 25 '18 at 19:01

revo

33.3k135085

edited May 25 '18 at 19:01

revo

33.3k135085

33.3k135085

answered Jun 3 '13 at 6:22

KamahireKamahire

1,47721646

answered Jun 3 '13 at 6:22

KamahireKamahire

1,47721646

answered Jun 3 '13 at 6:22

KamahireKamahire

1,47721646

1,47721646

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Shouldn't those be backslashes?
  
  – Paul Draper
  Oct 19 '13 at 6:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  They go at the end of the Regular Expression, not within in. Example: /blah/s
  
  – RandomInsano
  Dec 21 '13 at 20:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I guess you mean JavaScript, not Java? Since you can just add the s flag to the pattern in Java and JavaScript doesn't have the s flag.
  
  – 3limin4t0r
  Sep 25 '18 at 17:47
  
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Shouldn't those be backslashes?
  
  – Paul Draper
  Oct 19 '13 at 6:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  They go at the end of the Regular Expression, not within in. Example: /blah/s
  
  – RandomInsano
  Dec 21 '13 at 20:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I guess you mean JavaScript, not Java? Since you can just add the s flag to the pattern in Java and JavaScript doesn't have the s flag.
  
  – 3limin4t0r
  Sep 25 '18 at 17:47
  
  
  

  
  
  
  

1

1

Shouldn't those be backslashes?

– Paul Draper
Oct 19 '13 at 6:48

Shouldn't those be backslashes?

– Paul Draper
Oct 19 '13 at 6:48

They go at the end of the Regular Expression, not within in. Example: /blah/s

– RandomInsano
Dec 21 '13 at 20:12

They go at the end of the Regular Expression, not within in. Example: /blah/s

– RandomInsano
Dec 21 '13 at 20:12

I guess you mean JavaScript, not Java? Since you can just add the s flag to the pattern in Java and JavaScript doesn't have the s flag.

– 3limin4t0r
Sep 25 '18 at 17:47

I guess you mean JavaScript, not Java? Since you can just add the s flag to the pattern in Java and JavaScript doesn't have the s flag.

– 3limin4t0r
Sep 25 '18 at 17:47

add a comment | 

3

Note that (.|n)* can be less efficient than (for example) [sS]* (if your language's regexes support such escapes) and than finding how to specify the modifier that makes . also match newlines. Or you can go with POSIXy alternatives like [[:space:][:^space:]]*.

share|improve this answer

answered Oct 2 '08 at 3:31

tyetye

1,072911

add a comment | 

3

Note that (.|n)* can be less efficient than (for example) [sS]* (if your language's regexes support such escapes) and than finding how to specify the modifier that makes . also match newlines. Or you can go with POSIXy alternatives like [[:space:][:^space:]]*.

share|improve this answer

answered Oct 2 '08 at 3:31

tyetye

1,072911

add a comment | 

3

3

3

Note that (.|n)* can be less efficient than (for example) [sS]* (if your language's regexes support such escapes) and than finding how to specify the modifier that makes . also match newlines. Or you can go with POSIXy alternatives like [[:space:][:^space:]]*.

share|improve this answer

answered Oct 2 '08 at 3:31

tyetye

1,072911

Note that (.|n)* can be less efficient than (for example) [sS]* (if your language's regexes support such escapes) and than finding how to specify the modifier that makes . also match newlines. Or you can go with POSIXy alternatives like [[:space:][:^space:]]*.

share|improve this answer

answered Oct 2 '08 at 3:31

tyetye

1,072911

share|improve this answer

share|improve this answer

answered Oct 2 '08 at 3:31

tyetye

1,072911

answered Oct 2 '08 at 3:31

tyetye

1,072911

answered Oct 2 '08 at 3:31

tyetye

1,072911

1,072911

add a comment | 

add a comment | 

3

Use RegexOptions.Singleline, it changes the meaning of . to include newlines

Regex.Replace(content, searchText, replaceText, RegexOptions.Singleline);

share|improve this answer

answered Apr 13 '10 at 0:42

shmallshmall

311

add a comment | 

3

Use RegexOptions.Singleline, it changes the meaning of . to include newlines

Regex.Replace(content, searchText, replaceText, RegexOptions.Singleline);

share|improve this answer

answered Apr 13 '10 at 0:42

shmallshmall

311

add a comment | 

3

3

3

Use RegexOptions.Singleline, it changes the meaning of . to include newlines

Regex.Replace(content, searchText, replaceText, RegexOptions.Singleline);

share|improve this answer

answered Apr 13 '10 at 0:42

shmallshmall

311

Use RegexOptions.Singleline, it changes the meaning of . to include newlines

Regex.Replace(content, searchText, replaceText, RegexOptions.Singleline);

share|improve this answer

answered Apr 13 '10 at 0:42

shmallshmall

311

share|improve this answer

share|improve this answer

answered Apr 13 '10 at 0:42

shmallshmall

311

answered Apr 13 '10 at 0:42

shmallshmall

311

answered Apr 13 '10 at 0:42

shmallshmall

311

311

add a comment | 

add a comment | 

2

Solution:

Use pattern modifier sU will get the desired matching in PHP.

example:

preg_match('/(.*)/sU',$content,$match);

Source:

http://dreamluverz.com/developers-tools/regex-match-all-including-new-line
http://php.net/manual/en/reference.pcre.pattern.modifiers.php

share|improve this answer

answered Apr 4 '12 at 11:00

Sian Lerk LauSian Lerk Lau

1088

add a comment | 

2

Solution:

Use pattern modifier sU will get the desired matching in PHP.

example:

preg_match('/(.*)/sU',$content,$match);

Source:

http://dreamluverz.com/developers-tools/regex-match-all-including-new-line
http://php.net/manual/en/reference.pcre.pattern.modifiers.php

share|improve this answer

answered Apr 4 '12 at 11:00

Sian Lerk LauSian Lerk Lau

1088

add a comment | 

2

2

2

Solution:

Use pattern modifier sU will get the desired matching in PHP.

example:

preg_match('/(.*)/sU',$content,$match);

Source:

http://dreamluverz.com/developers-tools/regex-match-all-including-new-line
http://php.net/manual/en/reference.pcre.pattern.modifiers.php

share|improve this answer

answered Apr 4 '12 at 11:00

Sian Lerk LauSian Lerk Lau

1088

Solution:

Use pattern modifier sU will get the desired matching in PHP.

example:

preg_match('/(.*)/sU',$content,$match);

Source:

http://dreamluverz.com/developers-tools/regex-match-all-including-new-line
http://php.net/manual/en/reference.pcre.pattern.modifiers.php

share|improve this answer

answered Apr 4 '12 at 11:00

Sian Lerk LauSian Lerk Lau

1088

share|improve this answer

share|improve this answer

answered Apr 4 '12 at 11:00

Sian Lerk LauSian Lerk Lau

1088

answered Apr 4 '12 at 11:00

Sian Lerk LauSian Lerk Lau

1088

answered Apr 4 '12 at 11:00

Sian Lerk LauSian Lerk Lau

1088

1088

add a comment | 

add a comment | 

1

In the context of use within languages, regular expressions act on strings, not lines. So you should be able to use the regex normally, assuming that the input string has multiple lines.

In this case, the given regex will match the entire string, since "<FooBar>" is present. Depending on the specifics of the regex implementation, the $1 value (obtained from the "(.*)") will either be "fghij" or "abcdenfghij". As others have said, some implementations allow you to control whether the "." will match the newline, giving you the choice.

Line-based regular expression use is usually for command line things like egrep.

share|improve this answer

edited Oct 1 '08 at 18:54

answered Oct 1 '08 at 18:49

nsayernsayer

13.5k22645

add a comment | 

1

In the context of use within languages, regular expressions act on strings, not lines. So you should be able to use the regex normally, assuming that the input string has multiple lines.

In this case, the given regex will match the entire string, since "<FooBar>" is present. Depending on the specifics of the regex implementation, the $1 value (obtained from the "(.*)") will either be "fghij" or "abcdenfghij". As others have said, some implementations allow you to control whether the "." will match the newline, giving you the choice.

Line-based regular expression use is usually for command line things like egrep.

share|improve this answer

edited Oct 1 '08 at 18:54

answered Oct 1 '08 at 18:49

nsayernsayer

13.5k22645

add a comment | 

1

1

1

In the context of use within languages, regular expressions act on strings, not lines. So you should be able to use the regex normally, assuming that the input string has multiple lines.

In this case, the given regex will match the entire string, since "<FooBar>" is present. Depending on the specifics of the regex implementation, the $1 value (obtained from the "(.*)") will either be "fghij" or "abcdenfghij". As others have said, some implementations allow you to control whether the "." will match the newline, giving you the choice.

Line-based regular expression use is usually for command line things like egrep.

share|improve this answer

edited Oct 1 '08 at 18:54

answered Oct 1 '08 at 18:49

nsayernsayer

13.5k22645

In the context of use within languages, regular expressions act on strings, not lines. So you should be able to use the regex normally, assuming that the input string has multiple lines.

In this case, the given regex will match the entire string, since "<FooBar>" is present. Depending on the specifics of the regex implementation, the $1 value (obtained from the "(.*)") will either be "fghij" or "abcdenfghij". As others have said, some implementations allow you to control whether the "." will match the newline, giving you the choice.

Line-based regular expression use is usually for command line things like egrep.

share|improve this answer

edited Oct 1 '08 at 18:54

answered Oct 1 '08 at 18:49

nsayernsayer

13.5k22645

share|improve this answer

share|improve this answer

edited Oct 1 '08 at 18:54

edited Oct 1 '08 at 18:54

edited Oct 1 '08 at 18:54

answered Oct 1 '08 at 18:49

nsayernsayer

13.5k22645

answered Oct 1 '08 at 18:49

nsayernsayer

13.5k22645

answered Oct 1 '08 at 18:49

nsayernsayer

13.5k22645

13.5k22645

add a comment | 

add a comment | 

1

I had the same problem and solved it in probably not the best way but it works. I replaced all line breaks before I did my real match:

mystring= Regex.Replace(mystring, "rn", "")

I am manipulating HTML so line breaks don't really matter to me in this case.

I tried all of the suggestions above with no luck, I am using .Net 3.5 FYI

share|improve this answer

answered Mar 26 '09 at 14:57

SleeSlee

11.4k41129232

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I am using .NET too and (s|S) seems to do the trick for me!
  
  – Vamshi Krishna
  May 18 '18 at 7:26
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @VamshiKrishna In .NET, use (?s) to make . match any chars. Do not use (s|S) that will slow down performance.
  
  – Wiktor Stribiżew
  Sep 14 '18 at 20:35
  

  
  
  
  

add a comment | 

1

I had the same problem and solved it in probably not the best way but it works. I replaced all line breaks before I did my real match:

mystring= Regex.Replace(mystring, "rn", "")

I am manipulating HTML so line breaks don't really matter to me in this case.

I tried all of the suggestions above with no luck, I am using .Net 3.5 FYI

share|improve this answer

answered Mar 26 '09 at 14:57

SleeSlee

11.4k41129232

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I am using .NET too and (s|S) seems to do the trick for me!
  
  – Vamshi Krishna
  May 18 '18 at 7:26
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @VamshiKrishna In .NET, use (?s) to make . match any chars. Do not use (s|S) that will slow down performance.
  
  – Wiktor Stribiżew
  Sep 14 '18 at 20:35
  

  
  
  
  

add a comment | 

1

1

1

I had the same problem and solved it in probably not the best way but it works. I replaced all line breaks before I did my real match:

mystring= Regex.Replace(mystring, "rn", "")

I am manipulating HTML so line breaks don't really matter to me in this case.

I tried all of the suggestions above with no luck, I am using .Net 3.5 FYI

share|improve this answer

answered Mar 26 '09 at 14:57

SleeSlee

11.4k41129232

I had the same problem and solved it in probably not the best way but it works. I replaced all line breaks before I did my real match:

mystring= Regex.Replace(mystring, "rn", "")

I am manipulating HTML so line breaks don't really matter to me in this case.

I tried all of the suggestions above with no luck, I am using .Net 3.5 FYI

share|improve this answer

answered Mar 26 '09 at 14:57

SleeSlee

11.4k41129232

share|improve this answer

share|improve this answer

answered Mar 26 '09 at 14:57

SleeSlee

11.4k41129232

answered Mar 26 '09 at 14:57

SleeSlee

11.4k41129232

answered Mar 26 '09 at 14:57

SleeSlee

11.4k41129232

11.4k41129232

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I am using .NET too and (s|S) seems to do the trick for me!
  
  – Vamshi Krishna
  May 18 '18 at 7:26
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @VamshiKrishna In .NET, use (?s) to make . match any chars. Do not use (s|S) that will slow down performance.
  
  – Wiktor Stribiżew
  Sep 14 '18 at 20:35
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I am using .NET too and (s|S) seems to do the trick for me!
  
  – Vamshi Krishna
  May 18 '18 at 7:26
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @VamshiKrishna In .NET, use (?s) to make . match any chars. Do not use (s|S) that will slow down performance.
  
  – Wiktor Stribiżew
  Sep 14 '18 at 20:35
  

  
  
  
  

I am using .NET too and (s|S) seems to do the trick for me!

– Vamshi Krishna
May 18 '18 at 7:26

I am using .NET too and (s|S) seems to do the trick for me!

– Vamshi Krishna
May 18 '18 at 7:26

@VamshiKrishna In .NET, use (?s) to make . match any chars. Do not use (s|S) that will slow down performance.

– Wiktor Stribiżew
Sep 14 '18 at 20:35

@VamshiKrishna In .NET, use (?s) to make . match any chars. Do not use (s|S) that will slow down performance.

– Wiktor Stribiżew
Sep 14 '18 at 20:35

add a comment | 

0

generally . doesn't match newlines, so try ((.|n)*)<foobar>

share|improve this answer

answered Oct 1 '08 at 18:52

tloachtloach

7,53112943

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  No, don't do that. If you need to match anything including line separators, use the DOTALL (a.k.a. /s or SingleLine) modifier. Not only does the (.|n) hack make the regex less efficient, it's not even correct. At the very least, it should match r (carriage return) as well as n (linefeed). There are other line separator characters, too, albeit rarely used. But if you use the DOTALL flag, you don't have to worry about them.
  
  – Alan Moore
  Apr 26 '09 at 3:17
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  R is the platform-independent match for newlines in Eclipse.
  
  – opyate
  Nov 30 '09 at 11:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @opyate You should post this as an answer as this little gem is incredibly useful.
  
  – jeckhart
  Oct 15 '12 at 21:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  You could try this instead. It won't match the inner brackets and also consider the optionalr.: ((?:.|r?n)*)<foobar>
  
  – ssc-hrep3
  Nov 29 '16 at 9:52
  

  
  
  
  

add a comment | 

0

generally . doesn't match newlines, so try ((.|n)*)<foobar>

share|improve this answer

answered Oct 1 '08 at 18:52

tloachtloach

7,53112943

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  No, don't do that. If you need to match anything including line separators, use the DOTALL (a.k.a. /s or SingleLine) modifier. Not only does the (.|n) hack make the regex less efficient, it's not even correct. At the very least, it should match r (carriage return) as well as n (linefeed). There are other line separator characters, too, albeit rarely used. But if you use the DOTALL flag, you don't have to worry about them.
  
  – Alan Moore
  Apr 26 '09 at 3:17
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  R is the platform-independent match for newlines in Eclipse.
  
  – opyate
  Nov 30 '09 at 11:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @opyate You should post this as an answer as this little gem is incredibly useful.
  
  – jeckhart
  Oct 15 '12 at 21:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  You could try this instead. It won't match the inner brackets and also consider the optionalr.: ((?:.|r?n)*)<foobar>
  
  – ssc-hrep3
  Nov 29 '16 at 9:52
  

  
  
  
  

add a comment | 

0

0

0

generally . doesn't match newlines, so try ((.|n)*)<foobar>

share|improve this answer

answered Oct 1 '08 at 18:52

tloachtloach

7,53112943

generally . doesn't match newlines, so try ((.|n)*)<foobar>

share|improve this answer

answered Oct 1 '08 at 18:52

tloachtloach

7,53112943

share|improve this answer

share|improve this answer

answered Oct 1 '08 at 18:52

tloachtloach

7,53112943

answered Oct 1 '08 at 18:52

tloachtloach

7,53112943

answered Oct 1 '08 at 18:52

tloachtloach

7,53112943

7,53112943

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  No, don't do that. If you need to match anything including line separators, use the DOTALL (a.k.a. /s or SingleLine) modifier. Not only does the (.|n) hack make the regex less efficient, it's not even correct. At the very least, it should match r (carriage return) as well as n (linefeed). There are other line separator characters, too, albeit rarely used. But if you use the DOTALL flag, you don't have to worry about them.
  
  – Alan Moore
  Apr 26 '09 at 3:17
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  R is the platform-independent match for newlines in Eclipse.
  
  – opyate
  Nov 30 '09 at 11:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @opyate You should post this as an answer as this little gem is incredibly useful.
  
  – jeckhart
  Oct 15 '12 at 21:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  You could try this instead. It won't match the inner brackets and also consider the optionalr.: ((?:.|r?n)*)<foobar>
  
  – ssc-hrep3
  Nov 29 '16 at 9:52
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  No, don't do that. If you need to match anything including line separators, use the DOTALL (a.k.a. /s or SingleLine) modifier. Not only does the (.|n) hack make the regex less efficient, it's not even correct. At the very least, it should match r (carriage return) as well as n (linefeed). There are other line separator characters, too, albeit rarely used. But if you use the DOTALL flag, you don't have to worry about them.
  
  – Alan Moore
  Apr 26 '09 at 3:17
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  R is the platform-independent match for newlines in Eclipse.
  
  – opyate
  Nov 30 '09 at 11:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @opyate You should post this as an answer as this little gem is incredibly useful.
  
  – jeckhart
  Oct 15 '12 at 21:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  You could try this instead. It won't match the inner brackets and also consider the optionalr.: ((?:.|r?n)*)<foobar>
  
  – ssc-hrep3
  Nov 29 '16 at 9:52
  

  
  
  
  

1

1

No, don't do that. If you need to match anything including line separators, use the DOTALL (a.k.a. /s or SingleLine) modifier. Not only does the (.|n) hack make the regex less efficient, it's not even correct. At the very least, it should match r (carriage return) as well as n (linefeed). There are other line separator characters, too, albeit rarely used. But if you use the DOTALL flag, you don't have to worry about them.

– Alan Moore
Apr 26 '09 at 3:17

No, don't do that. If you need to match anything including line separators, use the DOTALL (a.k.a. /s or SingleLine) modifier. Not only does the (.|n) hack make the regex less efficient, it's not even correct. At the very least, it should match r (carriage return) as well as n (linefeed). There are other line separator characters, too, albeit rarely used. But if you use the DOTALL flag, you don't have to worry about them.

– Alan Moore
Apr 26 '09 at 3:17

1

1

R is the platform-independent match for newlines in Eclipse.

– opyate
Nov 30 '09 at 11:13

R is the platform-independent match for newlines in Eclipse.

– opyate
Nov 30 '09 at 11:13

@opyate You should post this as an answer as this little gem is incredibly useful.

– jeckhart
Oct 15 '12 at 21:29

@opyate You should post this as an answer as this little gem is incredibly useful.

– jeckhart
Oct 15 '12 at 21:29

You could try this instead. It won't match the inner brackets and also consider the optionalr.: ((?:.|r?n)*)<foobar>

– ssc-hrep3
Nov 29 '16 at 9:52

You could try this instead. It won't match the inner brackets and also consider the optionalr.: ((?:.|r?n)*)<foobar>

– ssc-hrep3
Nov 29 '16 at 9:52

add a comment | 

0

I wanted to match a particular if block in java

   ...

   ...

   if(isTrue){

       doAction();



   }

...

...

}

If I use the regExp

if (isTrue(.|n)*}

it included the closing brace for the method block so I used

if (!isTrue([^}.]|n)*}

to exclude the closing brace from the wildcard match.

share|improve this answer

answered Jan 18 '11 at 9:31

SpangenSpangen

1,89031824

add a comment | 

0

I wanted to match a particular if block in java

   ...

   ...

   if(isTrue){

       doAction();



   }

...

...

}

If I use the regExp

if (isTrue(.|n)*}

it included the closing brace for the method block so I used

if (!isTrue([^}.]|n)*}

to exclude the closing brace from the wildcard match.

share|improve this answer

answered Jan 18 '11 at 9:31

SpangenSpangen

1,89031824

add a comment | 

0

0

0

I wanted to match a particular if block in java

   ...

   ...

   if(isTrue){

       doAction();



   }

...

...

}

If I use the regExp

if (isTrue(.|n)*}

it included the closing brace for the method block so I used

if (!isTrue([^}.]|n)*}

to exclude the closing brace from the wildcard match.

share|improve this answer

answered Jan 18 '11 at 9:31

SpangenSpangen

1,89031824

I wanted to match a particular if block in java

   ...

   ...

   if(isTrue){

       doAction();



   }

...

...

}

If I use the regExp

if (isTrue(.|n)*}

it included the closing brace for the method block so I used

if (!isTrue([^}.]|n)*}

to exclude the closing brace from the wildcard match.

share|improve this answer

answered Jan 18 '11 at 9:31

SpangenSpangen

1,89031824

share|improve this answer

share|improve this answer

answered Jan 18 '11 at 9:31

SpangenSpangen

1,89031824

answered Jan 18 '11 at 9:31

SpangenSpangen

1,89031824

answered Jan 18 '11 at 9:31

SpangenSpangen

1,89031824

1,89031824

add a comment | 

add a comment | 

0

Often we have to modify a substring with a few keywords spread across lines preceding the substring. Consider an xml element:

<TASK>

  <UID>21</UID>

  <Name>Architectural design</Name>

  <PercentComplete>81</PercentComplete>

</TASK>

Suppose we want to modify the 81, to some other value, say 40. First identify .UID.21..UID., then skip all characters including n till .PercentCompleted.. The regular expression pattern and the replace specification are:

String hw = new String("<TASK>n  <UID>21</UID>n  <Name>Architectural design</Name>n  <PercentComplete>81</PercentComplete>n</TASK>");

String pattern = new String ("(<UID>21</UID>)((.|n)*?)(<PercentComplete>)(\d+)(</PercentComplete>)");

String replaceSpec = new String ("$1$2$440$6");

//note that the group (<PercentComplete>) is $4 and the group ((.|n)*?) is $2.



String  iw = hw.replaceFirst(pattern, replaceSpec);

System.out.println(iw);



<TASK>

  <UID>21</UID>

  <Name>Architectural design</Name>

  <PercentComplete>40</PercentComplete>

</TASK>

The subgroup (.|n) is probably the missing group $3. If we make it non-capturing by (?:.|n) then the $3 is (<PercentComplete>). So the pattern and replaceSpec can also be:

pattern = new String("(<UID>21</UID>)((?:.|n)*?)(<PercentComplete>)(\d+)(</PercentComplete>)");

replaceSpec = new String("$1$2$340$5")

and the replacement works correctly as before.

share|improve this answer

edited Oct 26 '12 at 8:49

deadly

1,1431223

answered Apr 21 '12 at 20:05

user1348737user1348737

92

add a comment | 

0

Often we have to modify a substring with a few keywords spread across lines preceding the substring. Consider an xml element:

<TASK>

  <UID>21</UID>

  <Name>Architectural design</Name>

  <PercentComplete>81</PercentComplete>

</TASK>

Suppose we want to modify the 81, to some other value, say 40. First identify .UID.21..UID., then skip all characters including n till .PercentCompleted.. The regular expression pattern and the replace specification are:

String hw = new String("<TASK>n  <UID>21</UID>n  <Name>Architectural design</Name>n  <PercentComplete>81</PercentComplete>n</TASK>");

String pattern = new String ("(<UID>21</UID>)((.|n)*?)(<PercentComplete>)(\d+)(</PercentComplete>)");

String replaceSpec = new String ("$1$2$440$6");

//note that the group (<PercentComplete>) is $4 and the group ((.|n)*?) is $2.



String  iw = hw.replaceFirst(pattern, replaceSpec);

System.out.println(iw);



<TASK>

  <UID>21</UID>

  <Name>Architectural design</Name>

  <PercentComplete>40</PercentComplete>

</TASK>

The subgroup (.|n) is probably the missing group $3. If we make it non-capturing by (?:.|n) then the $3 is (<PercentComplete>). So the pattern and replaceSpec can also be:

pattern = new String("(<UID>21</UID>)((?:.|n)*?)(<PercentComplete>)(\d+)(</PercentComplete>)");

replaceSpec = new String("$1$2$340$5")

and the replacement works correctly as before.

share|improve this answer

edited Oct 26 '12 at 8:49

deadly

1,1431223

answered Apr 21 '12 at 20:05

user1348737user1348737

92

add a comment | 

0

0

0

Often we have to modify a substring with a few keywords spread across lines preceding the substring. Consider an xml element:

<TASK>

  <UID>21</UID>

  <Name>Architectural design</Name>

  <PercentComplete>81</PercentComplete>

</TASK>

Suppose we want to modify the 81, to some other value, say 40. First identify .UID.21..UID., then skip all characters including n till .PercentCompleted.. The regular expression pattern and the replace specification are:

String hw = new String("<TASK>n  <UID>21</UID>n  <Name>Architectural design</Name>n  <PercentComplete>81</PercentComplete>n</TASK>");

String pattern = new String ("(<UID>21</UID>)((.|n)*?)(<PercentComplete>)(\d+)(</PercentComplete>)");

String replaceSpec = new String ("$1$2$440$6");

//note that the group (<PercentComplete>) is $4 and the group ((.|n)*?) is $2.



String  iw = hw.replaceFirst(pattern, replaceSpec);

System.out.println(iw);



<TASK>

  <UID>21</UID>

  <Name>Architectural design</Name>

  <PercentComplete>40</PercentComplete>

</TASK>

The subgroup (.|n) is probably the missing group $3. If we make it non-capturing by (?:.|n) then the $3 is (<PercentComplete>). So the pattern and replaceSpec can also be:

pattern = new String("(<UID>21</UID>)((?:.|n)*?)(<PercentComplete>)(\d+)(</PercentComplete>)");

replaceSpec = new String("$1$2$340$5")

and the replacement works correctly as before.

share|improve this answer

edited Oct 26 '12 at 8:49

deadly

1,1431223

answered Apr 21 '12 at 20:05

user1348737user1348737

92

Often we have to modify a substring with a few keywords spread across lines preceding the substring. Consider an xml element:

<TASK>

  <UID>21</UID>

  <Name>Architectural design</Name>

  <PercentComplete>81</PercentComplete>

</TASK>

Suppose we want to modify the 81, to some other value, say 40. First identify .UID.21..UID., then skip all characters including n till .PercentCompleted.. The regular expression pattern and the replace specification are:

String hw = new String("<TASK>n  <UID>21</UID>n  <Name>Architectural design</Name>n  <PercentComplete>81</PercentComplete>n</TASK>");

String pattern = new String ("(<UID>21</UID>)((.|n)*?)(<PercentComplete>)(\d+)(</PercentComplete>)");

String replaceSpec = new String ("$1$2$440$6");

//note that the group (<PercentComplete>) is $4 and the group ((.|n)*?) is $2.



String  iw = hw.replaceFirst(pattern, replaceSpec);

System.out.println(iw);



<TASK>

  <UID>21</UID>

  <Name>Architectural design</Name>

  <PercentComplete>40</PercentComplete>

</TASK>

The subgroup (.|n) is probably the missing group $3. If we make it non-capturing by (?:.|n) then the $3 is (<PercentComplete>). So the pattern and replaceSpec can also be:

pattern = new String("(<UID>21</UID>)((?:.|n)*?)(<PercentComplete>)(\d+)(</PercentComplete>)");

replaceSpec = new String("$1$2$340$5")

and the replacement works correctly as before.

share|improve this answer

edited Oct 26 '12 at 8:49

deadly

1,1431223

answered Apr 21 '12 at 20:05

user1348737user1348737

92

share|improve this answer

share|improve this answer

edited Oct 26 '12 at 8:49

deadly

1,1431223

edited Oct 26 '12 at 8:49

deadly

1,1431223

edited Oct 26 '12 at 8:49

deadly

1,1431223

1,1431223

answered Apr 21 '12 at 20:05

user1348737user1348737

92

answered Apr 21 '12 at 20:05

user1348737user1348737

92

answered Apr 21 '12 at 20:05

user1348737user1348737

92

92

add a comment | 

add a comment | 

0

In Javascript you can use [^]* to search for zero to infinite characters, including line breaks.

$("#find_and_replace").click(function() {

  var text = $("#textarea").val();

  search_term = new RegExp("[^]*<Foobar>", "gi");;

  replace_term = "Replacement term";

  var new_text = text.replace(search_term, replace_term);

  $("#textarea").val(new_text);

});

<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>

<button id="find_and_replace">Find and replace</button>

<br>

<textarea ID="textarea">abcde

fghij&lt;Foobar&gt;</textarea>

share|improve this answer

answered yesterday

Paul JonesPaul Jones

329311

add a comment | 

0

In Javascript you can use [^]* to search for zero to infinite characters, including line breaks.

$("#find_and_replace").click(function() {

  var text = $("#textarea").val();

  search_term = new RegExp("[^]*<Foobar>", "gi");;

  replace_term = "Replacement term";

  var new_text = text.replace(search_term, replace_term);

  $("#textarea").val(new_text);

});

<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>

<button id="find_and_replace">Find and replace</button>

<br>

<textarea ID="textarea">abcde

fghij&lt;Foobar&gt;</textarea>

share|improve this answer

answered yesterday

Paul JonesPaul Jones

329311

add a comment | 

0

0

0

In Javascript you can use [^]* to search for zero to infinite characters, including line breaks.

$("#find_and_replace").click(function() {

  var text = $("#textarea").val();

  search_term = new RegExp("[^]*<Foobar>", "gi");;

  replace_term = "Replacement term";

  var new_text = text.replace(search_term, replace_term);

  $("#textarea").val(new_text);

});

<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>

<button id="find_and_replace">Find and replace</button>

<br>

<textarea ID="textarea">abcde

fghij&lt;Foobar&gt;</textarea>

share|improve this answer

answered yesterday

Paul JonesPaul Jones

329311

In Javascript you can use [^]* to search for zero to infinite characters, including line breaks.

$("#find_and_replace").click(function() {

  var text = $("#textarea").val();

  search_term = new RegExp("[^]*<Foobar>", "gi");;

  replace_term = "Replacement term";

  var new_text = text.replace(search_term, replace_term);

  $("#textarea").val(new_text);

});

<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>

<button id="find_and_replace">Find and replace</button>

<br>

<textarea ID="textarea">abcde

fghij&lt;Foobar&gt;</textarea>

$("#find_and_replace").click(function() {

  var text = $("#textarea").val();

  search_term = new RegExp("[^]*<Foobar>", "gi");;

  replace_term = "Replacement term";

  var new_text = text.replace(search_term, replace_term);

  $("#textarea").val(new_text);

});

<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>

<button id="find_and_replace">Find and replace</button>

<br>

<textarea ID="textarea">abcde

fghij&lt;Foobar&gt;</textarea>

$("#find_and_replace").click(function() {

  var text = $("#textarea").val();

  search_term = new RegExp("[^]*<Foobar>", "gi");;

  replace_term = "Replacement term";

  var new_text = text.replace(search_term, replace_term);

  $("#textarea").val(new_text);

});

<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>

<button id="find_and_replace">Find and replace</button>

<br>

<textarea ID="textarea">abcde

fghij&lt;Foobar&gt;</textarea>

share|improve this answer

answered yesterday

Paul JonesPaul Jones

329311

share|improve this answer

share|improve this answer

answered yesterday

Paul JonesPaul Jones

329311

answered yesterday

Paul JonesPaul Jones

329311

answered yesterday

Paul JonesPaul Jones

329311

329311

add a comment | 

add a comment | 

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