regex for finding file paths

Question

I used this regex(\\/.*\\.[\\w:]+) to find all file paths and directories. But in a line like this \"file path /log/file.txt some lines /log/var/file2.txt

Answer 1

You can use python re

something like this:

import re
msg="file path /log/file.txt some lines /log/var/file2.txt"
matches = re.findall("(/[a-zA-Z\./]*[\s]?)", msg)
print(matches)

Ref: https://docs.python.org/2/library/re.html#finding-all-adverbs

Answer 2

Use regex(\/.*?\.[\w:]+) to make regex non-greedy. If you want to find multiple matches in the same line, you can use re.findall().

Update: Using this code and the example provided, I get:

import re
re.findall(r'(\/.*?\.[\w:]+)', "file path /log/file.txt some lines /log/var/file2.txt")
['/log/file.txt', '/log/var/file2.txt']

Answer 3

Your regex (\/.*\.[\w:]+) uses .* which is greedy and would match [\w:]+ after the last dot in file2.txt. You could use .*? instead.

But it would also match /log////var////.txt

As an alternative you might use a repeating non greedy pattern that would match the directory structure (?:/[^/]+)+? followed by a part that matches the filename /\w+\.\w+

(?:/[^/]+)+?/\w+\.\w+

import re
s = "file path /log/file.txt some lines /log/var/file2.txt or /log////var////.txt"
print(re.findall(r'(?:/[^/]+)+?/\w+\.\w+', s))

That would result in:

['/log/file.txt', '/log/var/file2.txt']

Demo