CODEWITHSUNDEEP
pythonregexpython-3.x
Dani
Dani

Reputation: 544

How can I match a regular expression in Python up to a certain character, and return None if not match?

I'd want to know if it's possible to match a regular expression until a particular character (':') but avoiding negative logical expressions like [^:]* cause I'd like it to return None if not matches encountered. See an example of what it's expected:

import re
string='AbcS:sdaf'
pattern='whatever needed' # match all up to : 
re.search(pattern, string).group()

'AbcS'

string2='AbcSsdaf'
pattern2='whatever needed' # match all up to : 
re.search(pattern2, string2).group()

None

In other post i've seen some answers mencioning pattern='[^:]*' but this is not what I want because returns all the string if there is not a ':' in the string.

Thank you all

Upvotes: 2

Views: 2726

Answers (4)

Rahul
Rahul

Reputation: 402

Try this: re.search(pattern, string.split(":")[0]).group()

Upvotes: 0

Wiktor Stribiżew
Wiktor Stribiżew

Reputation: 626835

A non-regex way is to split with : and count the number of splits. If it is more than 1, get the first item, else, return None:

splits = s.split(':')
if len(splits) > 1:
    print(splits[0])
else:
    print(None)

See this Python demo.

You may also use a regex approach like

^([^:]*):

and grab Group 1 if there is a match. See the regex demo.

Details

  • ^ - start of string
  • ([^:]*) - Capturing group 1: any 0+ chars other than :
  • : - a : char.

Python demo:

import re
strs= ['AbcS:sdaf', 'AbcSsdaf']
pattern= re.compile(r'([^:]*):') # re.match requires a match at  string start, ^ is redundant
for s in strs:
    m = re.match(pattern, s)
    if m:
        print(m.group(1))
    else:
        print(m)

Upvotes: 0

Borisu
Borisu

Reputation: 848

You need a capturing group:

pattern = '(.*):'  # this will match any and all characters up untill ":"
pattern = '([a-zA-Z])*:'  # this will match only letters
pattern = '([a-zA-Z0-9])*:'  # matches alphanumericals

If you speak the search criteria, you'll easily remember how to recreate it on your own: capture "()" anything "." as many times as possible "*" until you see a semicolon ":".

Most importantly re.search will return None if there is no ":" in the search string.

Upvotes: 0

Gurmanjot Singh
Gurmanjot Singh

Reputation: 10360

Try this regex:

^(?=.*:)[^:]*

Click for Demo

Explanation:

  • ^ - asserts the start of the string
  • (?=.*:) - positive lookahead to make sure that the line contains a :
  • [^:]* - matches 0+ occurrences of any character that is not a :

Upvotes: 3

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