Python: If statement "If not none" handling

Question

I am using Python's regex with an if-statement: if the match is None, then it should go to the else clause. But it shows this error:

AttributeError: 'NoneType' object has no attribute 'group'

The script is:

import string
chars = re.escape(string.punctuation)
sub='FW: Re: 29699' 
if re.search("^FW: (\w{10})",sub).group(1) is not None :
    d=re.search("^FW: (\w{10})",sub).group(1)
else:
    a=re.sub(r'['+chars+']', ' ',sub)
    d='_'.join(a.split())

Every help is great help!

You've written is not None rather than is None, which seems to be what you need. — spies006
– spies006, Commented May 22, 2017 at 5:00

Arya McCarthy · Accepted Answer · 2017-05-22 05:04:49Z

6

Your problem is this: if your search doesn't find anything, it will return None. You can't do None.group(1), which is what your code amounts to. Instead, check whether the search result is None—not the search result's first group.

import re
import string

chars = re.escape(string.punctuation)
sub='FW: Re: 29699' 
search_result = re.search(r"^FW: (\w{10})", sub)

if search_result is not None:
    d = search_result.group(1)
else:
    a = re.sub(r'['+chars+']', ' ', sub)
    d = '_'.join(a.split())

print(d)
# FW_RE_29699

answered May 22, 2017 at 5:04

Arya McCarthy

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1 Comment

Manu Sharma Over a year ago

that's so logical, Thanks!

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