How to handle " in Regex Python
Question:
I am trying to grab fary_trigger_post in the code below using Regex. However, I don’t understand why it always includes " in the end of the matched pattern, which I don’t expect.
Any idea or suggestion?
re.match(
r'-instance[ "']*(.+)[ "']*$',
'-instance "fary_trigger_post" '.strip(),
flags=re.S).group(1)
'fary_trigger_post"'
Thank you.
Answers:
Here’s what I’d use in your matching string, but it’s hard to provide a better answer without knowing all your cases:
r'-instances+"(.+)"s*$'
The (.+) is greedy and grabs ANY character until the end of the input. If you modified your input to include characters after the final double quote (e.g. '-instance "fary_trigger_post" asdf') you would find the double quote and the remaining characters in the output (e.g. fary_trigger_post" asdf). Instead of .+ you should try [^"']+ to capture all characters except the quotes. This should return what you expect.
re.match(r'-instance[ "']*([^"']+)[ "'].*$', '-instance "fary_trigger_post" '.strip(), flags=re.S).group(1)
Also, note that I modified the end of the expression to use .* which will match any characters following the last quote.
When you try to get group 1 (i.e. (.+)) regex will follow this match to the end of string, as it can match . (any character) 1 or more times (but it will take maximum amount of times). I would suggest use the following pattern:
'-instance[ "']*(.+)["']+ *$'
This will require regex to match all spaces in the end and all quoutes separatelly, so that it won’t be included into group 1
I am trying to grab fary_trigger_post in the code below using Regex. However, I don’t understand why it always includes " in the end of the matched pattern, which I don’t expect.
Any idea or suggestion?
re.match(
r'-instance[ "']*(.+)[ "']*$',
'-instance "fary_trigger_post" '.strip(),
flags=re.S).group(1)
'fary_trigger_post"'
Thank you.
Here’s what I’d use in your matching string, but it’s hard to provide a better answer without knowing all your cases:
r'-instances+"(.+)"s*$'
The (.+) is greedy and grabs ANY character until the end of the input. If you modified your input to include characters after the final double quote (e.g. '-instance "fary_trigger_post" asdf') you would find the double quote and the remaining characters in the output (e.g. fary_trigger_post" asdf). Instead of .+ you should try [^"']+ to capture all characters except the quotes. This should return what you expect.
re.match(r'-instance[ "']*([^"']+)[ "'].*$', '-instance "fary_trigger_post" '.strip(), flags=re.S).group(1)
Also, note that I modified the end of the expression to use .* which will match any characters following the last quote.
When you try to get group 1 (i.e. (.+)) regex will follow this match to the end of string, as it can match . (any character) 1 or more times (but it will take maximum amount of times). I would suggest use the following pattern:
'-instance[ "']*(.+)["']+ *$'
This will require regex to match all spaces in the end and all quoutes separatelly, so that it won’t be included into group 1