Grab numbers that are 6 long except those after a specific phrase in Python

Question

I am using Python in case it matters.

I am trying to grab all 6-long numbers from a file: ([0-9]{6}) which works fine.

But I want to ignore any such numbers if they are immediately preceded by "Obsolete #:".

So for example Obsolete #:748275 would get ignored, but not something else 957252.

I am trying stuff like [^Obsolete #:]([0-9]{6}) but it simply doesn't work / grabs the number anyway.

Does this answer your question? [Python Regex Negative Lookbehind](https://stackoverflow.com/questions/13947933/python-regex-negative-lookbehind). More info here: https://www.regular-expressions.info/lookaround.html — Pranav Hosangadi, Oct 06 '20 at 16:29

Pranav Hosangadi · Accepted Answer · 2020-10-06T16:35:27.140

1

(?<!Obsolete #:)\b(\d{6})\b
Try it: https://regex101.com/r/RsMoi9/1

Explanation:

(?<!Obsolete #:): Negative lookbehind. Do not match anything that comes after this.
\b: Word boundary
(\d{6}): Capture six digits
\b: Another word boundary

edited Oct 06 '20 at 16:35

answered Oct 06 '20 at 16:34

Pranav Hosangadi

I had been trying `(?!Obsolete #:)(\d{6})` - why does this fail but yours works? – user525966 Oct 06 '20 at 16:34
1

@user525966 `(?!Obsolete #:)` is a negative look _ahead_, useful if you want to match something that is _not followed_ by `Obsolete #:` – Pranav Hosangadi Oct 06 '20 at 16:36

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