python capture from regex

Question

I am writing a regex to find a ? and capture everything before that. Here's what I have so far.

f = 'file.doc?gs_f'
f1 = re.search(r'(\?)',f)
print(f[:f1.start()]) #prints "file.doc"

How do I use the regex to capture everything before the ?. The is something called a look behind but I am unsure how to use it.

Answer 1

This would work:

^[^\?]+

And use findall:

f = 'file.doc?gs_f'
f1 = re.findall("^[^\?]+",f)

You can see a regex working below for any kind of string: Regex_Working

Answer 2

Multiple possibilities:

1. Either everything not a question mark: [^?]+
2. Or a lazy quantifier with capturing groups: (.+?)\?
3. Or a positive lookahead: .+?(?=\?)

See the corresponding demos on regex101.com (1, 2 and 3).

Answer 3

You do not need look behind.

f = 'file.doc?gs_f'
re.search(r'([\w\.]*(?=\?))',f).group(0)

A good here Regex lookahead, lookbehind and atomic groups

Answer 4

Yes, it is called Positive Look Ahead. You have to look ahead till ? comes. Here is the Regex which is performing your reqt. let me know if you have any question.

Regex: '.*(?=\?)'

Here,
.* will pull everything.
(?=) is a positive look ahead
\ escape character for question mark symbol. As question mark symbol is Quantifier in Regex.