Mixing
How to match multiple occurrences of a substring
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box;
}
If I have an HTML string such as:
<div><p>£20<span class="abc" /><span class="def">56</span></p></div>
And I want the text:
20<span class="abc" /><span class="def">56
How do I define a regular expression to match the target sections multiple times. So far I have:
str.match(/d*<[^>]*>d*/)
But this will only return the first number section 20<span class="abc" />
I need this to be flexible to match multiple tag / numeric sections while trimming anything leading or trailing the first / last digit in the string.
regex
edited Jan 3 at 9:28
Josh Habdas
3,1492634
asked Aug 12 '11 at 14:09
gb2dgb2d
3,41264183
add a comment |
If I have an HTML string such as:
<div><p>£20<span class="abc" /><span class="def">56</span></p></div>
And I want the text:
20<span class="abc" /><span class="def">56
How do I define a regular expression to match the target sections multiple times. So far I have:
str.match(/d*<[^>]*>d*/)
But this will only return the first number section 20<span class="abc" />
I need this to be flexible to match multiple tag / numeric sections while trimming anything leading or trailing the first / last digit in the string.
regex
edited Jan 3 at 9:28
Josh Habdas
3,1492634
asked Aug 12 '11 at 14:09
gb2dgb2d
3,41264183
add a comment |
If I have an HTML string such as:
<div><p>£20<span class="abc" /><span class="def">56</span></p></div>
And I want the text:
20<span class="abc" /><span class="def">56
How do I define a regular expression to match the target sections multiple times. So far I have:
str.match(/d*<[^>]*>d*/)
But this will only return the first number section 20<span class="abc" />
I need this to be flexible to match multiple tag / numeric sections while trimming anything leading or trailing the first / last digit in the string.
regex
edited Jan 3 at 9:28
Josh Habdas
3,1492634
asked Aug 12 '11 at 14:09
gb2dgb2d
3,41264183
If I have an HTML string such as:
<div><p>£20<span class="abc" /><span class="def">56</span></p></div>
And I want the text:
20<span class="abc" /><span class="def">56
How do I define a regular expression to match the target sections multiple times. So far I have:
str.match(/d*<[^>]*>d*/)
But this will only return the first number section 20<span class="abc" />
I need this to be flexible to match multiple tag / numeric sections while trimming anything leading or trailing the first / last digit in the string.
regex
regex
edited Jan 3 at 9:28
Josh Habdas
3,1492634
asked Aug 12 '11 at 14:09
gb2dgb2d
3,41264183
edited Jan 3 at 9:28
Josh Habdas
3,1492634
asked Aug 12 '11 at 14:09
gb2dgb2d
3,41264183
edited Jan 3 at 9:28
Josh Habdas
3,1492634
edited Jan 3 at 9:28
Josh Habdas
3,1492634
edited Jan 3 at 9:28
Josh Habdas
3,1492634
3,1492634
asked Aug 12 '11 at 14:09
gb2dgb2d
3,41264183
asked Aug 12 '11 at 14:09
gb2dgb2d
3,41264183
asked Aug 12 '11 at 14:09
gb2dgb2d
3,41264183
3,41264183
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
To match multiple times use to need use the global option
str.match(/your_expression_here/g)
^
edited Jun 29 '18 at 13:29
Wiktor Stribiżew
329k16149228
answered Aug 12 '11 at 14:23
James KyburzJames Kyburz
9,26512530
4
This is not answering his question.
– Tim Pietzcker
Aug 12 '11 at 14:56
With 5 upvotes in 7 years Tim is probably right. I added the correct answer.
– Josh Habdas
Jan 3 at 9:23
add a comment |
Simply adding /g and calling it done isn't enough. Matching a substring within a string multiple times is trivial once you're aware of reluctant quantifiers—explained here with the solution to your problem.
Given the string:
<div><p>£20<span class="abc" /><span class="def">56</span></p></div>
You will arrive at the text you wanted using:
d+.*>d+
But given the same string repeated two times:
<div><p>£20<span class="abc" /><span class="def">56</span></p></div><div><p>£20<span class="abc" /><span class="def">56</span></p></div>
You will not find the target selection multiple times. You'll only find it once due to the greedy nature of .*. To make .* non-greedy, or reluctant, simply add a ? after the * and you will arrive at:
d+.*?>d+
Which will find both occurrences of the substring you asked for as proven here.
answered Jan 3 at 9:21
Josh HabdasJosh Habdas
3,1492634
add a comment |
Just allow the group to be repeated: (?:...)+ means "Match ... 1 or more times:
str.match(/d+(?:<[^>]*>)+d+/)
As per Alan Moore's suggestion, I've also changed the d* into d+, making the numbers required instead of optional.
answered Aug 12 '11 at 14:56
Tim PietzckerTim Pietzcker
252k43381462
While you're at it, change eachd*tod+; that's got to be a mistake.
– Alan Moore
Aug 13 '11 at 2:38
@Alan: Sure, you're right. Thanks!
– Tim Pietzcker
Aug 13 '11 at 5:09
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
To match multiple times use to need use the global option
str.match(/your_expression_here/g)
^
edited Jun 29 '18 at 13:29
Wiktor Stribiżew
329k16149228
answered Aug 12 '11 at 14:23
James KyburzJames Kyburz
9,26512530
4
This is not answering his question.
– Tim Pietzcker
Aug 12 '11 at 14:56
With 5 upvotes in 7 years Tim is probably right. I added the correct answer.
– Josh Habdas
Jan 3 at 9:23
add a comment |
To match multiple times use to need use the global option
str.match(/your_expression_here/g)
^
edited Jun 29 '18 at 13:29
Wiktor Stribiżew
329k16149228
answered Aug 12 '11 at 14:23
James KyburzJames Kyburz
9,26512530
4
This is not answering his question.
– Tim Pietzcker
Aug 12 '11 at 14:56
With 5 upvotes in 7 years Tim is probably right. I added the correct answer.
– Josh Habdas
Jan 3 at 9:23
add a comment |
To match multiple times use to need use the global option
str.match(/your_expression_here/g)
^
edited Jun 29 '18 at 13:29
Wiktor Stribiżew
329k16149228
answered Aug 12 '11 at 14:23
James KyburzJames Kyburz
9,26512530
To match multiple times use to need use the global option
str.match(/your_expression_here/g)
^
edited Jun 29 '18 at 13:29
Wiktor Stribiżew
329k16149228
answered Aug 12 '11 at 14:23
James KyburzJames Kyburz
9,26512530
edited Jun 29 '18 at 13:29
Wiktor Stribiżew
329k16149228
edited Jun 29 '18 at 13:29
Wiktor Stribiżew
329k16149228
edited Jun 29 '18 at 13:29
Wiktor Stribiżew
329k16149228
329k16149228
answered Aug 12 '11 at 14:23
James KyburzJames Kyburz
9,26512530
answered Aug 12 '11 at 14:23
James KyburzJames Kyburz
9,26512530
answered Aug 12 '11 at 14:23
James KyburzJames Kyburz
9,26512530
9,26512530
4
This is not answering his question.
– Tim Pietzcker
Aug 12 '11 at 14:56
With 5 upvotes in 7 years Tim is probably right. I added the correct answer.
– Josh Habdas
Jan 3 at 9:23
add a comment |
4
This is not answering his question.
– Tim Pietzcker
Aug 12 '11 at 14:56
With 5 upvotes in 7 years Tim is probably right. I added the correct answer.
– Josh Habdas
Jan 3 at 9:23
4
4
This is not answering his question.
– Tim Pietzcker
Aug 12 '11 at 14:56
This is not answering his question.
– Tim Pietzcker
Aug 12 '11 at 14:56
With 5 upvotes in 7 years Tim is probably right. I added the correct answer.
– Josh Habdas
Jan 3 at 9:23
With 5 upvotes in 7 years Tim is probably right. I added the correct answer.
– Josh Habdas
Jan 3 at 9:23
add a comment |
Simply adding /g and calling it done isn't enough. Matching a substring within a string multiple times is trivial once you're aware of reluctant quantifiers—explained here with the solution to your problem.
Given the string:
<div><p>£20<span class="abc" /><span class="def">56</span></p></div>
You will arrive at the text you wanted using:
d+.*>d+
But given the same string repeated two times:
<div><p>£20<span class="abc" /><span class="def">56</span></p></div><div><p>£20<span class="abc" /><span class="def">56</span></p></div>
You will not find the target selection multiple times. You'll only find it once due to the greedy nature of .*. To make .* non-greedy, or reluctant, simply add a ? after the * and you will arrive at:
d+.*?>d+
Which will find both occurrences of the substring you asked for as proven here.
answered Jan 3 at 9:21
Josh HabdasJosh Habdas
3,1492634
add a comment |
Simply adding /g and calling it done isn't enough. Matching a substring within a string multiple times is trivial once you're aware of reluctant quantifiers—explained here with the solution to your problem.
Given the string:
<div><p>£20<span class="abc" /><span class="def">56</span></p></div>
You will arrive at the text you wanted using:
d+.*>d+
But given the same string repeated two times:
<div><p>£20<span class="abc" /><span class="def">56</span></p></div><div><p>£20<span class="abc" /><span class="def">56</span></p></div>
You will not find the target selection multiple times. You'll only find it once due to the greedy nature of .*. To make .* non-greedy, or reluctant, simply add a ? after the * and you will arrive at:
d+.*?>d+
Which will find both occurrences of the substring you asked for as proven here.
answered Jan 3 at 9:21
Josh HabdasJosh Habdas
3,1492634
add a comment |
Simply adding /g and calling it done isn't enough. Matching a substring within a string multiple times is trivial once you're aware of reluctant quantifiers—explained here with the solution to your problem.
Given the string:
<div><p>£20<span class="abc" /><span class="def">56</span></p></div>
You will arrive at the text you wanted using:
d+.*>d+
But given the same string repeated two times:
<div><p>£20<span class="abc" /><span class="def">56</span></p></div><div><p>£20<span class="abc" /><span class="def">56</span></p></div>
You will not find the target selection multiple times. You'll only find it once due to the greedy nature of .*. To make .* non-greedy, or reluctant, simply add a ? after the * and you will arrive at:
d+.*?>d+
Which will find both occurrences of the substring you asked for as proven here.
answered Jan 3 at 9:21
Josh HabdasJosh Habdas
3,1492634
Simply adding /g and calling it done isn't enough. Matching a substring within a string multiple times is trivial once you're aware of reluctant quantifiers—explained here with the solution to your problem.
Given the string:
<div><p>£20<span class="abc" /><span class="def">56</span></p></div>
You will arrive at the text you wanted using:
d+.*>d+
But given the same string repeated two times:
<div><p>£20<span class="abc" /><span class="def">56</span></p></div><div><p>£20<span class="abc" /><span class="def">56</span></p></div>
You will not find the target selection multiple times. You'll only find it once due to the greedy nature of .*. To make .* non-greedy, or reluctant, simply add a ? after the * and you will arrive at:
d+.*?>d+
Which will find both occurrences of the substring you asked for as proven here.
answered Jan 3 at 9:21
Josh HabdasJosh Habdas
3,1492634
answered Jan 3 at 9:21
Josh HabdasJosh Habdas
3,1492634
answered Jan 3 at 9:21
Josh HabdasJosh Habdas
3,1492634
answered Jan 3 at 9:21
Josh HabdasJosh Habdas
3,1492634
3,1492634
add a comment |
add a comment |
Just allow the group to be repeated: (?:...)+ means "Match ... 1 or more times:
str.match(/d+(?:<[^>]*>)+d+/)
As per Alan Moore's suggestion, I've also changed the d* into d+, making the numbers required instead of optional.
answered Aug 12 '11 at 14:56
Tim PietzckerTim Pietzcker
252k43381462
While you're at it, change eachd*tod+; that's got to be a mistake.
– Alan Moore
Aug 13 '11 at 2:38
@Alan: Sure, you're right. Thanks!
– Tim Pietzcker
Aug 13 '11 at 5:09
add a comment |
Just allow the group to be repeated: (?:...)+ means "Match ... 1 or more times:
str.match(/d+(?:<[^>]*>)+d+/)
As per Alan Moore's suggestion, I've also changed the d* into d+, making the numbers required instead of optional.
answered Aug 12 '11 at 14:56
Tim PietzckerTim Pietzcker
252k43381462
While you're at it, change eachd*tod+; that's got to be a mistake.
– Alan Moore
Aug 13 '11 at 2:38
@Alan: Sure, you're right. Thanks!
– Tim Pietzcker
Aug 13 '11 at 5:09
add a comment |
Just allow the group to be repeated: (?:...)+ means "Match ... 1 or more times:
str.match(/d+(?:<[^>]*>)+d+/)
As per Alan Moore's suggestion, I've also changed the d* into d+, making the numbers required instead of optional.
answered Aug 12 '11 at 14:56
Tim PietzckerTim Pietzcker
252k43381462
Just allow the group to be repeated: (?:...)+ means "Match ... 1 or more times:
str.match(/d+(?:<[^>]*>)+d+/)
As per Alan Moore's suggestion, I've also changed the d* into d+, making the numbers required instead of optional.
answered Aug 12 '11 at 14:56
Tim PietzckerTim Pietzcker
252k43381462
edited Aug 13 '11 at 5:09
answered Aug 12 '11 at 14:56
Tim PietzckerTim Pietzcker
252k43381462
answered Aug 12 '11 at 14:56
Tim PietzckerTim Pietzcker
252k43381462
answered Aug 12 '11 at 14:56
Tim PietzckerTim Pietzcker
252k43381462
252k43381462
While you're at it, change eachd*tod+; that's got to be a mistake.
– Alan Moore
Aug 13 '11 at 2:38
@Alan: Sure, you're right. Thanks!
– Tim Pietzcker
Aug 13 '11 at 5:09
add a comment |
While you're at it, change eachd*tod+; that's got to be a mistake.
– Alan Moore
Aug 13 '11 at 2:38
@Alan: Sure, you're right. Thanks!
– Tim Pietzcker
Aug 13 '11 at 5:09
While you're at it, change each
d* to d+; that's got to be a mistake.– Alan Moore
Aug 13 '11 at 2:38
While you're at it, change each
d* to d+; that's got to be a mistake.– Alan Moore
Aug 13 '11 at 2:38
@Alan: Sure, you're right. Thanks!
– Tim Pietzcker
Aug 13 '11 at 5:09
@Alan: Sure, you're right. Thanks!
– Tim Pietzcker
Aug 13 '11 at 5:09
add a comment |
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