Regular expression to match last number in a string

Your regex \d+(?!\d+) says

match any number if it is not immediately followed by a number.

which is incorrect. A number is last if it is not followed (following it anywhere, not just immediately) by any other number.

When translated to regex we have:

(\d+)(?!.*\d)

Rubular Link

I took it this way: you need to make sure the match is close enough to the end of the string; close enough in the sense that only non-digits may intervene. What I suggest is the following:

/(\d+)\D*\z/

1. \z at the end means that that is the end of the string.
2. \D* before that means that an arbitrary number of non-digits can intervene between the match and the end of the string.
3. (\d+) is the matching part. It is in parenthesis so that you can pick it up, as was pointed out by Cameron.

You can use

.*(?:\D|^)(\d+)

to get the last number; this is because the matcher will gobble up all the characters with .*, then backtrack to the first non-digit character or the start of the string, then match the final group of digits.

Your negative lookahead isn't working because on the string "1 3", for example, the 1 is matched by the \d+, then the space matches the negative lookahead (since it's not a sequence of one or more digits). The 3 is never even looked at.

Note that your example regex doesn't have any groups in it, so I'm not sure how you were extracting the number.