there is one \s and it is outside of the lookaround ... so its checking every other number
BINGO! Thank you.
If I put the first space into a lookbehind: s[(?<=\s)(\d+)\d\K\s(?=(\d+)\d\s)]{
Or substitute the zero length \b--which will serve the same purpose: s[\b(\d+)\d\K\s(?=(\d+)\d\s)]{
The substitution works as I wanted it to.
But damn I could not see that for looking.
In reply to Re^2: Why doesn't this regex work? ( lookaround backtracking pos)
by BrowserUk
in thread Why doesn't this regex work? (Solved!)
by BrowserUk
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