I've got a regex issue, I'm trying to ignore just the number '41', I want 4, 1, 14 etc to all match.
I've got this [^\b41\b] which is effectively what I want but this also ignores all single iterations of the values 1 and 4.
As an example, this matches "41", but I want it to NOT match: \b41\b
Try something like:
\b(?!41\b)(\d+)
The (?!...) construct is a negative lookahead so this means: find a word boundary that is not followed by "41" and capture a sequence of digits after it.
You could use a negative look-ahead assertion to exclude 41:
/\b(?!41\b)\d+\b/
This regular expression is to be interpreted as: At any word boundary \b, if it is not followed by 41\b ((?!41\b)), match one or more digits that are followed by a word boundary.
Or the same with a negative look-behind assertion:
/\b\d+\b(?<!\b41)/
This regular expression is to be interpreted as: Match one or more digits that are surrounded by word boundaries, but only if the substring at the end of the match is not preceded by \b41 ((?<!\b41)).
Or can even use just basic syntax:
/\b(\d|[0-35-9]\d|\d[02-9]|\d{3,})\b/
This matches only sequences of digits surrounded by word boundaries of either:
4 at the first position or not a 1 at the second position
This is similar to the question "Regular expression that doesn’t contain certain string", so I'll repeat my answer from there:
^((?!41).)*$
This will work for an arbitrary string, not just 41. See my response there for an explanation.
