Postgres Regular Expression Positive Lookbehind with Repetition

Question

Working for the first time with Postgres flavor regular expressions and stumped on it's behavior here.

Input: 'xxxaaabc'

Expected: 'xxxaaazzzbc'

Logic: Match M a's, capture group, and add 'zzz' directly after the sequence

Attempts:

select regexp_replace('aaabc','(?!\Sa+)','zzz','i')

Result: zzzxxxaaabc

select regexp_replace('aaabc','(?!\Sa+)','zzz','i')

Result: xxxazzzaabc

select select regexp_replace('xxxaaabc','(?!=a+)','zzz','i')

Result: zzzxxxaaabc

This one seems to get pretty close regexp_replace('aaabc','(?!\Sa+)','zzz','i'), but the repetition of a's isn't working.

I think you want select regexp_replace('xxxaaabc','(a+)','\1zzz','i') AS Result; (demo) — Wiktor Stribiżew
– Wiktor Stribiżew, Commented Oct 18, 2022 at 16:14

Wiktor Stribiżew · Accepted Answer · 2022-10-18 16:19:03Z

2

Note the (?!\Sa+) pattern matches an empty location that is not immediately followed with a single non-whitespace char and then one or more a chars.

The (?!=a+) pattern matches a location in a string, that is not immediately followed with = and then one or more as.

You can capture one or more as and then use a backreference to the match and then just append the zzz:

select regexp_replace('xxxaaabc','(a+)','\1zzz','i') AS Result;

See the DB fiddle.

answered Oct 18, 2022 at 16:19

Wiktor Stribiżew

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