That is actually the usual behavior. The regex is usually implemented as a DFA (Finite State Machine), and occasionally they have to backtrack to complete the match. Your code is executed at the time the DFA moves through your (?{}) block even if it doesn't make it to the end which is when the DFA "matches."

In this case, it is your lookbehind that's causing the DFA to remember something, wander forward to see if it's relevant, and wander back when it isn't... If you put your (?{}) block at the very end it would only print when it actually finds a match.

-Paul

In reply to Re: Regex code block executes twice per match using look-arounds by jettero
in thread Regex code block executes twice per match using look-arounds by johngg

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