Mixing
Regex: don't match string ending with newline (n) with end-of-line anchor ($)
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box;
}
I can't figure out how to match a string but not if it has a trailing newline character (n), which seems automatically stripped:
import re
print(re.match(r'^foobar$', 'foobar'))
# <_sre.SRE_Match object; span=(0, 6), match='foobar'>
print(re.match(r'^foobar$', 'foobarn'))
# <_sre.SRE_Match object; span=(0, 6), match='foobar'>
print(re.match(r'^foobar$', 'foobarnn'))
# None
For me, the second case should also return None.
When we set the end of a pattern with $, like ^foobar$, it should only match a string like foobar, not foobarn.
What am I missing?
python regex python-3.x match newline
edited Feb 11 '18 at 10:15
DeepSpace
40.6k44779
asked Feb 11 '18 at 10:02
Arthur WhiteArthur White
2817
add a comment |
I can't figure out how to match a string but not if it has a trailing newline character (n), which seems automatically stripped:
import re
print(re.match(r'^foobar$', 'foobar'))
# <_sre.SRE_Match object; span=(0, 6), match='foobar'>
print(re.match(r'^foobar$', 'foobarn'))
# <_sre.SRE_Match object; span=(0, 6), match='foobar'>
print(re.match(r'^foobar$', 'foobarnn'))
# None
For me, the second case should also return None.
When we set the end of a pattern with $, like ^foobar$, it should only match a string like foobar, not foobarn.
What am I missing?
python regex python-3.x match newline
edited Feb 11 '18 at 10:15
DeepSpace
40.6k44779
asked Feb 11 '18 at 10:02
Arthur WhiteArthur White
2817
1
What exactly is a single line string and how are you reading this? Are you reading a binary file and parsing this yourself?
– Roger Credlin
Feb 11 '18 at 10:06
I removed the "single line" term. It was misleading and adds no value to the question. Otherwise, no, I'm not parsing a file. I just need to check the pattern of a simple string, like in my example.
– Arthur White
Feb 11 '18 at 10:11
add a comment |
I can't figure out how to match a string but not if it has a trailing newline character (n), which seems automatically stripped:
import re
print(re.match(r'^foobar$', 'foobar'))
# <_sre.SRE_Match object; span=(0, 6), match='foobar'>
print(re.match(r'^foobar$', 'foobarn'))
# <_sre.SRE_Match object; span=(0, 6), match='foobar'>
print(re.match(r'^foobar$', 'foobarnn'))
# None
For me, the second case should also return None.
When we set the end of a pattern with $, like ^foobar$, it should only match a string like foobar, not foobarn.
What am I missing?
python regex python-3.x match newline
edited Feb 11 '18 at 10:15
DeepSpace
40.6k44779
asked Feb 11 '18 at 10:02
Arthur WhiteArthur White
2817
I can't figure out how to match a string but not if it has a trailing newline character (n), which seems automatically stripped:
import re
print(re.match(r'^foobar$', 'foobar'))
# <_sre.SRE_Match object; span=(0, 6), match='foobar'>
print(re.match(r'^foobar$', 'foobarn'))
# <_sre.SRE_Match object; span=(0, 6), match='foobar'>
print(re.match(r'^foobar$', 'foobarnn'))
# None
For me, the second case should also return None.
When we set the end of a pattern with $, like ^foobar$, it should only match a string like foobar, not foobarn.
What am I missing?
python regex python-3.x match newline
python regex python-3.x match newline
edited Feb 11 '18 at 10:15
DeepSpace
40.6k44779
asked Feb 11 '18 at 10:02
Arthur WhiteArthur White
2817
edited Feb 11 '18 at 10:15
DeepSpace
40.6k44779
asked Feb 11 '18 at 10:02
Arthur WhiteArthur White
2817
edited Feb 11 '18 at 10:15
DeepSpace
40.6k44779
edited Feb 11 '18 at 10:15
DeepSpace
40.6k44779
edited Feb 11 '18 at 10:15
DeepSpace
40.6k44779
40.6k44779
asked Feb 11 '18 at 10:02
Arthur WhiteArthur White
2817
asked Feb 11 '18 at 10:02
Arthur WhiteArthur White
2817
asked Feb 11 '18 at 10:02
Arthur WhiteArthur White
2817
2817
1
What exactly is a single line string and how are you reading this? Are you reading a binary file and parsing this yourself?
– Roger Credlin
Feb 11 '18 at 10:06
I removed the "single line" term. It was misleading and adds no value to the question. Otherwise, no, I'm not parsing a file. I just need to check the pattern of a simple string, like in my example.
– Arthur White
Feb 11 '18 at 10:11
add a comment |
1
What exactly is a single line string and how are you reading this? Are you reading a binary file and parsing this yourself?
– Roger Credlin
Feb 11 '18 at 10:06
I removed the "single line" term. It was misleading and adds no value to the question. Otherwise, no, I'm not parsing a file. I just need to check the pattern of a simple string, like in my example.
– Arthur White
Feb 11 '18 at 10:11
1
1
What exactly is a single line string and how are you reading this? Are you reading a binary file and parsing this yourself?
– Roger Credlin
Feb 11 '18 at 10:06
What exactly is a single line string and how are you reading this? Are you reading a binary file and parsing this yourself?
– Roger Credlin
Feb 11 '18 at 10:06
I removed the "single line" term. It was misleading and adds no value to the question. Otherwise, no, I'm not parsing a file. I just need to check the pattern of a simple string, like in my example.
– Arthur White
Feb 11 '18 at 10:11
I removed the "single line" term. It was misleading and adds no value to the question. Otherwise, no, I'm not parsing a file. I just need to check the pattern of a simple string, like in my example.
– Arthur White
Feb 11 '18 at 10:11
add a comment |
3 Answers
3
active
oldest
votes
You more likely don't need $ but rather Z:
>>> print(re.match(r'^foobarZ', 'foobarn'))
None
Zmatches only at the end of the string.
answered Feb 11 '18 at 10:43
revorevo
34.3k135188
1
That's the way to go. I suspected the negative lookahead to be a dirty hack for this. Thank you!
– Arthur White
Feb 11 '18 at 11:39
add a comment |
This is the defined behavior of $, as can be read in the docs that @zvone linked to or even on https://regex101.com:
$ asserts position at the end of the string, or before the line terminator right at the end of the string (if any)
You can use an explicit negative lookahead to counter this behavior:
import re
print(re.match(r'^foobar(?!n)$', 'foobar'))
# <_sre.SRE_Match object; span=(0, 6), match='foobar'>
print(re.match(r'^foobar(?!n)$', 'foobarn'))
# None
print(re.match(r'^foobar(?!n)$', 'foobarnn'))
# None
answered Feb 11 '18 at 10:09
DeepSpaceDeepSpace
40.6k44779
add a comment |
The documentation says this about the $ character:
Matches the end of the string or just before the newline at the end of
the string, and in MULTILINE mode also matches before a newline.
So, without the MULTILINE option, it matches exactly the first two strings you tried: 'foobar' and 'foobarn', but not 'foobarnn', because that is not a newline at the end of the string.
On the other hand, if you choose MULTILINE option, it will match the end of any line:
>>> re.match(r'^foobar$', 'foobarnn', re.MULTILINE)
<_sre.SRE_Match object; span=(0, 6), match='foobar'>
Of course, this will also match in the following case, which may or may not be what you want:
>>> re.match(r'^foobar$', 'foobarnanother linen', re.MULTILINE)
<_sre.SRE_Match object; span=(0, 6), match='foobar'>
In order to NOT match the ending newline, use the negative lookahead as DeepSpace wrote.
answered Feb 11 '18 at 10:10
zvonezvone
10.2k12448
@DeepSpace You are right, I did not see that point...
– zvone
Feb 11 '18 at 10:15
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
You more likely don't need $ but rather Z:
>>> print(re.match(r'^foobarZ', 'foobarn'))
None
Zmatches only at the end of the string.
answered Feb 11 '18 at 10:43
revorevo
34.3k135188
1
That's the way to go. I suspected the negative lookahead to be a dirty hack for this. Thank you!
– Arthur White
Feb 11 '18 at 11:39
add a comment |
You more likely don't need $ but rather Z:
>>> print(re.match(r'^foobarZ', 'foobarn'))
None
Zmatches only at the end of the string.
answered Feb 11 '18 at 10:43
revorevo
34.3k135188
1
That's the way to go. I suspected the negative lookahead to be a dirty hack for this. Thank you!
– Arthur White
Feb 11 '18 at 11:39
add a comment |
You more likely don't need $ but rather Z:
>>> print(re.match(r'^foobarZ', 'foobarn'))
None
Zmatches only at the end of the string.
answered Feb 11 '18 at 10:43
revorevo
34.3k135188
You more likely don't need $ but rather Z:
>>> print(re.match(r'^foobarZ', 'foobarn'))
None
Zmatches only at the end of the string.
answered Feb 11 '18 at 10:43
revorevo
34.3k135188
answered Feb 11 '18 at 10:43
revorevo
34.3k135188
answered Feb 11 '18 at 10:43
revorevo
34.3k135188
answered Feb 11 '18 at 10:43
revorevo
34.3k135188
34.3k135188
1
That's the way to go. I suspected the negative lookahead to be a dirty hack for this. Thank you!
– Arthur White
Feb 11 '18 at 11:39
add a comment |
1
That's the way to go. I suspected the negative lookahead to be a dirty hack for this. Thank you!
– Arthur White
Feb 11 '18 at 11:39
1
1
That's the way to go. I suspected the negative lookahead to be a dirty hack for this. Thank you!
– Arthur White
Feb 11 '18 at 11:39
That's the way to go. I suspected the negative lookahead to be a dirty hack for this. Thank you!
– Arthur White
Feb 11 '18 at 11:39
add a comment |
This is the defined behavior of $, as can be read in the docs that @zvone linked to or even on https://regex101.com:
$ asserts position at the end of the string, or before the line terminator right at the end of the string (if any)
You can use an explicit negative lookahead to counter this behavior:
import re
print(re.match(r'^foobar(?!n)$', 'foobar'))
# <_sre.SRE_Match object; span=(0, 6), match='foobar'>
print(re.match(r'^foobar(?!n)$', 'foobarn'))
# None
print(re.match(r'^foobar(?!n)$', 'foobarnn'))
# None
answered Feb 11 '18 at 10:09
DeepSpaceDeepSpace
40.6k44779
add a comment |
This is the defined behavior of $, as can be read in the docs that @zvone linked to or even on https://regex101.com:
$ asserts position at the end of the string, or before the line terminator right at the end of the string (if any)
You can use an explicit negative lookahead to counter this behavior:
import re
print(re.match(r'^foobar(?!n)$', 'foobar'))
# <_sre.SRE_Match object; span=(0, 6), match='foobar'>
print(re.match(r'^foobar(?!n)$', 'foobarn'))
# None
print(re.match(r'^foobar(?!n)$', 'foobarnn'))
# None
answered Feb 11 '18 at 10:09
DeepSpaceDeepSpace
40.6k44779
add a comment |
This is the defined behavior of $, as can be read in the docs that @zvone linked to or even on https://regex101.com:
$ asserts position at the end of the string, or before the line terminator right at the end of the string (if any)
You can use an explicit negative lookahead to counter this behavior:
import re
print(re.match(r'^foobar(?!n)$', 'foobar'))
# <_sre.SRE_Match object; span=(0, 6), match='foobar'>
print(re.match(r'^foobar(?!n)$', 'foobarn'))
# None
print(re.match(r'^foobar(?!n)$', 'foobarnn'))
# None
answered Feb 11 '18 at 10:09
DeepSpaceDeepSpace
40.6k44779
This is the defined behavior of $, as can be read in the docs that @zvone linked to or even on https://regex101.com:
$ asserts position at the end of the string, or before the line terminator right at the end of the string (if any)
You can use an explicit negative lookahead to counter this behavior:
import re
print(re.match(r'^foobar(?!n)$', 'foobar'))
# <_sre.SRE_Match object; span=(0, 6), match='foobar'>
print(re.match(r'^foobar(?!n)$', 'foobarn'))
# None
print(re.match(r'^foobar(?!n)$', 'foobarnn'))
# None
answered Feb 11 '18 at 10:09
DeepSpaceDeepSpace
40.6k44779
edited Feb 11 '18 at 10:19
answered Feb 11 '18 at 10:09
DeepSpaceDeepSpace
40.6k44779
answered Feb 11 '18 at 10:09
DeepSpaceDeepSpace
40.6k44779
answered Feb 11 '18 at 10:09
DeepSpaceDeepSpace
40.6k44779
40.6k44779
add a comment |
add a comment |
The documentation says this about the $ character:
Matches the end of the string or just before the newline at the end of
the string, and in MULTILINE mode also matches before a newline.
So, without the MULTILINE option, it matches exactly the first two strings you tried: 'foobar' and 'foobarn', but not 'foobarnn', because that is not a newline at the end of the string.
On the other hand, if you choose MULTILINE option, it will match the end of any line:
>>> re.match(r'^foobar$', 'foobarnn', re.MULTILINE)
<_sre.SRE_Match object; span=(0, 6), match='foobar'>
Of course, this will also match in the following case, which may or may not be what you want:
>>> re.match(r'^foobar$', 'foobarnanother linen', re.MULTILINE)
<_sre.SRE_Match object; span=(0, 6), match='foobar'>
In order to NOT match the ending newline, use the negative lookahead as DeepSpace wrote.
answered Feb 11 '18 at 10:10
zvonezvone
10.2k12448
@DeepSpace You are right, I did not see that point...
– zvone
Feb 11 '18 at 10:15
add a comment |
The documentation says this about the $ character:
Matches the end of the string or just before the newline at the end of
the string, and in MULTILINE mode also matches before a newline.
So, without the MULTILINE option, it matches exactly the first two strings you tried: 'foobar' and 'foobarn', but not 'foobarnn', because that is not a newline at the end of the string.
On the other hand, if you choose MULTILINE option, it will match the end of any line:
>>> re.match(r'^foobar$', 'foobarnn', re.MULTILINE)
<_sre.SRE_Match object; span=(0, 6), match='foobar'>
Of course, this will also match in the following case, which may or may not be what you want:
>>> re.match(r'^foobar$', 'foobarnanother linen', re.MULTILINE)
<_sre.SRE_Match object; span=(0, 6), match='foobar'>
In order to NOT match the ending newline, use the negative lookahead as DeepSpace wrote.
answered Feb 11 '18 at 10:10
zvonezvone
10.2k12448
@DeepSpace You are right, I did not see that point...
– zvone
Feb 11 '18 at 10:15
add a comment |
The documentation says this about the $ character:
Matches the end of the string or just before the newline at the end of
the string, and in MULTILINE mode also matches before a newline.
So, without the MULTILINE option, it matches exactly the first two strings you tried: 'foobar' and 'foobarn', but not 'foobarnn', because that is not a newline at the end of the string.
On the other hand, if you choose MULTILINE option, it will match the end of any line:
>>> re.match(r'^foobar$', 'foobarnn', re.MULTILINE)
<_sre.SRE_Match object; span=(0, 6), match='foobar'>
Of course, this will also match in the following case, which may or may not be what you want:
>>> re.match(r'^foobar$', 'foobarnanother linen', re.MULTILINE)
<_sre.SRE_Match object; span=(0, 6), match='foobar'>
In order to NOT match the ending newline, use the negative lookahead as DeepSpace wrote.
answered Feb 11 '18 at 10:10
zvonezvone
10.2k12448
The documentation says this about the $ character:
Matches the end of the string or just before the newline at the end of
the string, and in MULTILINE mode also matches before a newline.
So, without the MULTILINE option, it matches exactly the first two strings you tried: 'foobar' and 'foobarn', but not 'foobarnn', because that is not a newline at the end of the string.
On the other hand, if you choose MULTILINE option, it will match the end of any line:
>>> re.match(r'^foobar$', 'foobarnn', re.MULTILINE)
<_sre.SRE_Match object; span=(0, 6), match='foobar'>
Of course, this will also match in the following case, which may or may not be what you want:
>>> re.match(r'^foobar$', 'foobarnanother linen', re.MULTILINE)
<_sre.SRE_Match object; span=(0, 6), match='foobar'>
In order to NOT match the ending newline, use the negative lookahead as DeepSpace wrote.
answered Feb 11 '18 at 10:10
zvonezvone
10.2k12448
edited Feb 11 '18 at 10:20
answered Feb 11 '18 at 10:10
zvonezvone
10.2k12448
answered Feb 11 '18 at 10:10
zvonezvone
10.2k12448
answered Feb 11 '18 at 10:10
zvonezvone
10.2k12448
10.2k12448
@DeepSpace You are right, I did not see that point...
– zvone
Feb 11 '18 at 10:15
add a comment |
@DeepSpace You are right, I did not see that point...
– zvone
Feb 11 '18 at 10:15
@DeepSpace You are right, I did not see that point...
– zvone
Feb 11 '18 at 10:15
@DeepSpace You are right, I did not see that point...
– zvone
Feb 11 '18 at 10:15
add a comment |
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1
What exactly is a single line string and how are you reading this? Are you reading a binary file and parsing this yourself?
– Roger Credlin
Feb 11 '18 at 10:06
I removed the "single line" term. It was misleading and adds no value to the question. Otherwise, no, I'm not parsing a file. I just need to check the pattern of a simple string, like in my example.
– Arthur White
Feb 11 '18 at 10:11