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equivalent of the (\k<a>) regex
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​​​​​​​_s.replacerx("(?s)str (.)=\s*(.+)\s*str \1v=\s*(.+)\s*\1v\.findreplace\(''\[\]'' ''┋''\)\s*","$2-[]$3---[]")
_s.replacerx("(?s)\s*---\s*str kv=.+|(?m)^ +","")
out _s


Messages In This Thread
equivalent of the (\k<a>) regex - by Davider - 11-21-2022, 11:18 PM
RE: equivalent of the (\k<a>) regex - by Davider - 11-22-2022, 11:25 AM

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