Python regex negative lookahead - only include result text, not whole text

Question

r = r'(?!.*66)get.*on'
txt = "get your kicks on Route 66"
res = re.search(r, txt)

I only want res=None if the exclusion text "66" would be included in the result text i.e. between "get" and "on". In this case it's outside the resulting text so I want to see a result but instead it's checking if "66" is anywhere in the whole text and returns res=None. Is there a way to do what I'm trying to do?

You call that "lookahead"? I think what's in your code should be called "lookbehind" — iBug, Jan 29 '18 at 12:23

score 1 · Accepted Answer · answered Jan 29 '18 at 12:23

To match a string between two strings excluding a third string, you can use a tempered greedy token (see more details about this construct at rexegg.com):

r = r'get(?:(?!66).)*on'
txt = "get your kicks on Route 66"
res = re.search(r, txt)

Here, (?:(?!66).)* matches any char (other than a line break char if re.DOTALL is not specified), 0 or more repetitions/consecutive occurrences, that does not start a 66 char sequence.

See the regex demo.

Note that you might want to tune it to get the shortest substring using r'get(?:(?!get|66).)*?on' expression.

See an updated regex demo.

If anyone knows a better approach, please post as an answer, and please let me know why this answer is not helpful. — Wiktor Stribiżew, Feb 03 '18 at 18:48

Python regex negative lookahead - only include result text, not whole text

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