Regex Issue with Python re

Question

Trying to use the python re package to look for filenames with a certain pattern. Got a wee test script which just has all values hardcoded in but this isn't the normal use:

#!/usr/bin/env python

try:
    import re2 as re
except ImportError:
    import re

filepath1 = "C:\Users\Administrator\AppData\Local\Temp\77ce4ba2a605e22b8699eef874d075fb585d259ed6cade2e503e6dbf58020aa0.exe:Zone.Identifier"
filepath2 = "C:\Users\Administrator\AppData\Local\Temp\svchost.exe:Zone.Identifier"
re_pattern = re.compile("C\:\\\\Users\\\\[^\\\\]*\\\\AppData\\\\Local\\\\Temp\\\\[^.]*\.exe\:Zone\.Identifier")

print "1: " + str(re_pattern.search(filepath1))
print "2: " + str(re_pattern.search(filepath2))

For some reason this returns None for 1 and a match for 2, but as far as I can work out they should both be matching. Probably just a stupid mistake but if someone can spot it that would be awesome.

Basically the pattern should match any .exe with a Zone ID in the %TEMP% directory, regardless of username

ever heard of raw prefix? that would avoid doubling the backslashes. — Jean-François Fabre
– Jean-François Fabre ♦, Commented May 28, 2018 at 16:00

Jean-François Fabre · Accepted Answer · 2018-05-28 16:06:12Z

The issue is that one of the paths starts with 7, and if you try \7 in the console, you'll see that it's interpreted as a code, because you're not using raw prefix for literals.

>>> print("\7")
<some garbage char, bell?>
>>> print(r"\7")
\7

That explains that your regex doesn't work for that particular path (for the other path you were "lucky" because you're using Python 2 and \+any upper char isn't a particular escape sequence so it's not changed (In python 3, \U is interpreted!)

Now, for paths, in this simple case, you could use fnmatch instead to match wildcards not regexes:

import fnmatch

filepath1 = r"C:\Users\Administrator\AppData\Local\Temp\77ce4ba2a605e22b8699eef874d075fb585d259ed6cade2e503e6dbf58020aa0.exe:Zone.Identifier"
filepath2 = r"C:\Users\Administrator\AppData\Local\Temp\svchost.exe:Zone.Identifier"
filepath3 = r"C:\Urs\Administrator\AppData\Local\Temp\svchost.exe:Zone.Identifier"

for f in (filepath1,filepath2,filepath3):
    print(f,fnmatch.fnmatch(f,r"C:\Users\*\AppData\*\Temp\*.exe:Zone.Identifier"))

prints:

C:\Users\Administrator\AppData\Local\Temp\77ce4ba2a605e22b8699eef874d075fb585d259ed6cade2e503e6dbf58020aa0.exe:Zone.Identifier True
C:\Users\Administrator\AppData\Local\Temp\svchost.exe:Zone.Identifier True
C:\Urs\Administrator\AppData\Local\Temp\svchost.exe:Zone.Identifier False

Rakesh · Accepted Answer · 2018-05-28 16:05:56Z

1

Sorry i misunderstood your question.

import re
filepath1 = r"C:\Users\Administrator\AppData\Local\Temp\77ce4ba2a605e22b8699eef874d075fb585d259ed6cade2e503e6dbf58020aa0.exe:Zone.Identifier"
filepath2 = r"C:\Users\Administrator\AppData\Local\Temp\svchost.exe:Zone.Identifier"

print(re.search(r"C\:\\Users\\(.*)\\AppData\\Local\\Temp\\[a-zA-Z0-9]+\.exe\:Zone\.Identifier$", filepath1))
print(re.search(r"C\:\\Users\\(.*)\\AppData\\Local\\Temp\\[a-zA-Z0-9]+\.exe\:Zone\.Identifier$", filepath2))

Output:

<_sre.SRE_Match object at 0x03176AE0>
<_sre.SRE_Match object at 0x03176AE0>

Please note the raw string(r) at the beginning of your filepath

answered May 28, 2018 at 16:05

Rakesh

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1 Comment

Brae Over a year ago

Thanks! This is what I was looking for. Marked the other as the answer though since the detail might help others :)

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