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I'm trying to extract a pattern from string using python regex. But it is not working with the below pattern

headerRegex = re.compile(r'^[^ ]*\s+\d*') 
        mo = headerRegex.search(string) 
        return mo.group()

My requirment is regular expression that should start with anything except white space and followed by one or more whitespace then digits occurence one or more

Example
i/p: test  7895  => olp:7895(correct)
i/p:  8545 ==> Not matching
i/p: @#@#  3453 ==>3453

May I know what is missing in my regex to implement this requirement?

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  • Change your *s to +s. Commented May 25, 2020 at 17:18
  • 1
    Or use ^\S+[^\S\r\n]+\d+$ regex101.com/r/IEHGSe/1 Commented May 25, 2020 at 17:27
  • Do you want to extract a pattern or do you want to match a pattern? If you want to search a pattern inside a string what do you mean by regex shouldn't start with white space. Please explain. (or) If you want to find if the input matches with the pattern then try pattern=re.compile('\S+\s+\d+') re.match(pattern,string_to_be_matched) Commented May 25, 2020 at 17:41
  • I'm trying to extract the number from the string Commented May 26, 2020 at 2:02

1 Answer 1

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In the pattern that you tried, only matching whitespace chars is mandatory, and you might possibly also match only newlines.

Change the quantifiers to + to match 1+ times, and if you don't want to match newlines as well use [^\S\r\n]+ instead.

If that exact match is only allowed, add an anchor $ to assert the end of the string, or add \Z if there is no newline following allowed.

^\S+[^\S\r\n]+\d+$
  • ^ Start of string
  • \S+ Match 1+ times a non whitespace char
  • [^\S\r\n]+ Match 1+ times a whitespace char except newlines
  • \d+ Match 1+ digits
  • $ End of string

Regex demo

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