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<
Answer from Yatharth Agarwal on Stack OverflowIf 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<
^.*\['(.*)'\].*$ will match a line and capture what you want in a group.
You have to escape the [ and ] with \
The documentation at the rubular.com proof link will explain how the expression is formed.
Python/Regex: Get all strings between any two characters - Stack Overflow
python - Match text between two strings with regular expression - Stack Overflow
Python Regex Get String Between Two Substrings - Stack Overflow
getting string between 2 characters in python - Stack Overflow
Use re.search
>>> import re
>>> s = 'Part 1. Part 2. Part 3 then more text'
>>> re.search(r'Part 1\.(.*?)Part 3', s).group(1)
' Part 2. '
>>> re.search(r'Part 1(.*?)Part 3', s).group(1)
'. Part 2. '
Or use re.findall, if there are more than one occurances.
With regular expression:
>>> import re
>>> s = 'Part 1. Part 2. Part 3 then more text'
>>> re.search(r'Part 1(.*?)Part 3', s).group(1)
'. Part 2. '
Without regular expression, this one works for your example:
>>> s = 'Part 1. Part 2. Part 3 then more text'
>>> a, b = s.find('Part 1'), s.find('Part 3')
>>> s[a+6:b]
'. Part 2. '
If the string contains only one instance, use re.search() instead:
>>> import re
>>> s = "api('randomkey123xyz987', 'key', 'text')"
>>> match = re.search(r"api\('([^']*)'", s).group(1)
>>> print match
randomkey123xyz987
You want the string between the ( and ,, you are catching everything between the parens:
match = re.findall("api\((.*?),", string)
print match
["'randomkey123xyz987'"]
Or match between the '':
match = re.findall("api\('(.*?)'", string)
print match
['randomkey123xyz987']
If that is how your strings actually look you can split:
string = "api('randomkey123xyz987', 'key', 'text')"
print(string.split(",",1)[0][4:])
Use this regex:
period_1_(.*)\.ssa
For example, in Perl you would extract it like this:
my ($substr) = ($string =~ /period_1_(.*)\.ssa/);
For Python, use this code:
m = re.match(r"period_1_(.*)\.ssa", my_long_string)
print m.group(1)
Last print will print string you are looking for (if there is a match).
(?<=period_1_)(.*)(?=.ssa)
This extracts the part between "period_1_" and ".ssa".
https://regexr.com/3jtbm
Tweeky way!
>>> char1 = '('
>>> char2 = ')'
>>> mystr = "mystring(123234sample)"
>>> print mystr[mystr.find(char1)+1 : mystr.find(char2)]
123234sample
$ is a special character in regex (it denotes the end of the string). You need to escape it:
>>> re.findall(r'\$(.*?)\$', '$sin (x)$ is an function of x')
['sin (x)']