Regular expression to return all characters between two special characters

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If you’re new to REG(gular) EX(pressions) you learn about them at Python Docs. Or, if you want a gentler introduction, you can check out the HOWTO. They use Perl-style syntax.

Regex

The expression that you need is .*?\[(.*)\].*. The group that you want will be \1.
– .*?: . matches any character but a newline. * is a meta-character and means Repeat this 0 or more times. ? makes the * non-greedy, i.e., . will match up as few chars as possible before hitting a ‘[‘.
– \[: \ escapes special meta-characters, which in this case, is [. If we didn’t do that, [ would do something very weird instead.
– (.*): Parenthesis ‘groups’ whatever is inside it and you can later retrieve the groups by their numeric IDs or names (if they’re given one).
– \].*: You should know enough by now to know what this means.

Implementation

First, import the re module — it’s not a built-in — to where-ever you want to use the expression.

Then, use re.search(regex_pattern, string_to_be_tested) to search for the pattern in the string to be tested. This will return a MatchObject which you can store to a temporary variable. You should then call it’s group() method and pass 1 as an argument (to see the ‘Group 1’ we captured using parenthesis earlier). I should now look like:

>>> import re
>>> pat = r'.*?\[(.*)].*'             #See Note at the bottom of the answer
>>> s = "foobar['infoNeededHere']ddd"
>>> match = re.search(pat, s)
>>> match.group(1)
"'infoNeededHere'"

An Alternative

You can also use findall() to find all the non-overlapping matches by modifying the regex to (?>=\[).+?(?=\]).
– (?<=\[): (?<=) is called a look-behind assertion and checks for an expression preceding the actual match.
– .+?: + is just like * except that it matches one or more repititions. It is made non-greedy by ?.
– (?=\]): (?=) is a look-ahead assertion and checks for an expression following the match w/o capturing it.
Your code should now look like:

>>> import re
>>> pat = r'(?<=\[).+?(?=\])'  #See Note at the bottom of the answer
>>> s = "foobar['infoNeededHere']ddd[andHere] [andOverHereToo[]"
>>> re.findall(pat, s)
["'infoNeededHere'", 'andHere', 'andOverHereToo[']

Note: Always use raw Python strings by adding an ‘r’ before the string (E.g.: r'blah blah blah').

10x for reading! I wrote this answer when there were no accepted ones yet, but by the time I finished it, 2 ore came up and one got accepted. 🙁 x<

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