On Jan 28, 2010, at 21:59 , Alvaro Herrera wrote:
Hi Michael,
Michael Glaesemann wrote:
We came across a regexp that takes very much longer than expected.
PostgreSQL 8.4.1 on x86_64-unknown-linux-gnu, compiled by GCC gcc
(GCC) 4.1.2 20080704 (Red Hat 4.1.2-44), 64-bit
SELECT 'ooo...' ~ $r$Z(Q)[^Q]*A.*?(\1)$r$; -- omitted for email
brevity
The ? after .* is pointless.
Interesting. I would expect that *? would be the non-greedy version of
*, meaning match up to the first \1 (in this case the first Q
following A), rather than as much as possible.
For example, in Perl:
$ perl -e " if ('oooZQoooAoooQooQooQooo' =~ /Z(Q)[^Q]*A.*(\1)/)
{ print \$&; } else { print 'NO'; }" && echo
ZQoooAoooQooQooQ
$ perl -e " if ('oooZQoooAoooQooQooQooo' =~ /Z(Q)[^Q]*A.*?(\1)/)
{ print \$&; } else { print 'NO'; }" && echo
ZQoooAoooQ
If I'm reading the docs right, Postgres does support non-greedy * as *?:
<http://www.postgresql.org/docs/8.4/interactive/functions-matching.html#POSIX-QUANTIFIERS-TABLE
>
However, as you point out, Postgres doesn't appear to take this into
account:
postgres=# select regexp_replace('oooZQoooAoooQooQooQooo', $r$(Z(Q)
[^Q]*A.*(\2))$r$, $s$X$s$);
regexp_replace
----------------
oooXooo
(1 row)
postgres=# select regexp_replace('oooZQoooAoooQooQooQooo', $r$(Z(Q)
[^Q]*A.*?(\2))$r$, $s$X$s$);
regexp_replace
----------------
oooXooo
(1 row)
Michael Glaesemann
michael.glaesem...@myyearbook.com
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