postgresql - Seemingly incorrect regex evaluation in regexp_replace -


i stumbled upon curious behavior of regexp_replace postgresql function. looks bug doubt myself first. when run

select regexp_replace(e'1%2_3', '([_%])', e'\\ \\1', 'g')

it correctly prefixes either underscore or percent backslash+space , produces "1\ %2\ _3". when remove space (it doesn't have space, can character)

select regexp_replace(e'1%2_3', '([_%])', e'\\\\1', 'g')

it stops using captured parenthesized expression in substitution , produces "1\12\13" instead of "1\%2\_3". appreciate if tell me doing wrong. need add backslash before characters in string.

update: able achieve desired behavior running

select regexp_replace(e'1%2_3', '([_%])', e'\\\\\\1', 'g')

my original example still seems bit illogical , inconsistent. inconsistency using same e'...' syntax 4 backslashes may produce different result.

in second query, after backslash escapes processed @ string level, have replacement string \\1.

what's happening escaped backslash prevents \1 being recognized back-reference. need set of backslashes, replacement string \\\1 literal backslash , back-reference. since every literal backslash needs escaped, need double of them.

select regexp_replace(e'1%2_3', '([_%])', e'\\\\\\1', 'g')

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