find all occurrences between 2 values in non default pattern-CodePudding

I am stumbling into an issue with a regex search in python

So I have:

testVariable = re.findall(r'functest(.*?)1', 'functest exampleOne [2] functest exampleTwo [1] functest exampleOne throw [2] functest exampleThree [1]')

Current Output is:

[' exampleOne [2] functest exampleTwo [', ' exampleOne throw [2] functest exampleThree [']

But what I want is to find all occurences between ‘functest’ & 1' <or 2, or 3 based on need> so output should be like:

['exampleTwo [, exampleThree [']

this because both above are between functest & 1 as I need. Anyone have any idea?

CodePudding user response：

If there can not be any digits in between matching the first occurrence of 1 or 3:

\bfunctest\b\s*(\D*)[13]\b

The pattern matches:

\bfunctest\b\s* Match the word functest followed by optional whitespace chars
(\D*) Capture Optional non digits in group 1
[13] Match either 1 or 3
\b A word boundary

See a regex demo.

Or you can exclude matching the square brackets before matching a digit using a negated character class:

\bfunctest\b\s*([^][]*\[)[13]]

See another regex demo.

Example

import re

pattern = r"\bfunctest\b\s*([^][]*\[)239]"

s = "functest exampleOne [2] functest exampleTwo [239] functest exampleOne throw [2] functest exampleThree [1] functest exampleFour [2] functest exampleFive [239]"

print(re.findall(pattern, s))

Output

['exampleTwo [', 'exampleFive [']

CodePudding user response：

Found a way by using the following. It still includes functest, but at least does the job

testVariable = re.findall(r'functest(?:(?!functest).)*?239', 'functest exampleOne [2] functest exampleTwo [239] functest exampleOne throw [2] functest exampleThree [1] functest exampleFour [2] functest exampleFive [239]')

Output: ['functest exampleTwo [239', 'functest exampleFive [239']