regex to match any character or none?

Question

I have the two following peices of strings;

line1 = [16/Aug/2016:06:13:25 -0400] "GET /file/ HTTP/1.1" 302 random stuff ignore

line2 = [16/Aug/2016:06:13:25 -0400] "" 400 random stuff ignore

I'm trying to grab these two parts;

"GET /file/ HTTP/1.1" 302
"" 400

Basically any character in between the two "" or nothing in between "". So far I've tried this;

regex_example = re.search("\".+?\" [0-9]{3}", line1)
print regex_example.group()

This will work with line1, but give an error for line2. This is due to the '.' matching any character, but giving an error if no character exists.

Is there any way for it to match any character or nothing in between the two ""?

Answer 1

Use .*? instead of .+?.

+ means "1 or more"

* means "0 or more"

Regex101 Demo

If you want a more efficient regex, use a negated character class [^"] instead of a lazy quantifier ?. You should also use the raw string flag r and \d for digits.

r'"[^"]*" \d{3}'

Answer 2

You can use:

import re

lines = ['[16/Aug/2016:06:13:25 -0400] "GET /file/ HTTP/1.1" 302 random stuff ignore', '[16/Aug/2016:06:13:25 -0400] "" 400 random stuff ignore']

rx = re.compile(r'''
        "[^"]*" # ", followed by anything not a " and a "
        \       # a space
        \d+     # at least one digit
        ''', re.VERBOSE)

matches = [m.group(0) \
            for line in lines \
            for m in rx.finditer(line)]

print(matches)
# ['"GET /file/ HTTP/1.1" 302', '"" 400']

---

See a demo on ideone.com.