regexp_replace replaces wrong part of the string

Question

When running:

select regexp_replace('( (test :Name (x) :Table (y) )','\s+\:Name \(.*?\)',' avner ');

I get:

"( (test avner "

But if I run:

select regexp_replace('( (test :Name (x) :Table (y) )','\:Name \(.*?\)',' avner ');

I get:

"( (test  avner  :Table (y) )"

Why is the \s+ at the start cause matching till the end of the string?

I tested this in Regex101. The first regex with \s+ doesn't match to the end of the string. Have you set any flags? — Farhad Alizadeh Noori
– Farhad Alizadeh Noori, Commented Sep 9, 2014 at 16:59
@Wolph the "x" is enough for that and then we have the ")". Why should it match all the way? it's lazy... — Farhad Alizadeh Noori
– Farhad Alizadeh Noori, Commented Sep 9, 2014 at 17:02
@Erwin Brandstetter The desired result is replacing the spaces before name as well. Something like: "( (testavner :Table (y) )" — Avner Levy
– Avner Levy, Commented Sep 10, 2014 at 7:43

Erwin Brandstetter · Accepted Answer · 2014-09-10 04:23:14Z

1

The reason is (per documentation):

A branch — that is, an RE that has no top-level | operator — has the same greediness as the first quantified atom in it that has a greediness attribute.

Bold emphasis mine. Simplifying your question to:

SELECT substring('( (test :Name (x) :Table (y) )', '\s+\:Name \(.*?\)')
      ,substring('( (test :Name (x) :Table (y) )',    '\:Name \(.*?\)')

If you want the second quantifier to be non-greedy, change the first to non-greedy as well. Especially, since this doesn't change anything:

SELECT substring('( (test :Name (x) :Table (y) )', '\s+?:Name \(.*?\)')

And there is no need to escape the colon (:).

SQL Fiddle.

edited Sep 10, 2014 at 4:23

answered Sep 10, 2014 at 2:00

Erwin Brandstetter

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