On Sat, 2017-07-22 at 10:49 -0400, Tom Lane wrote: > Igor Korot <ikorot01@xxxxxxxxx> writes: > > But it works incorrectly - it should return: > > 9.5.7 on x86_64-redhat-linux-gnu, compiled by gcc (GCC) 6.3.1 > > 20161221 > > (Red Hat 6.3.1-1), 64-bit > > i.e. without the word "PosgreSQL", since '\s' should match the > > (first) > > space in the version(). > > position() is not a regex operation, it's just a plain substring > match. > > regression=# SELECT position( '\s' in version() ) ; > position > ---------- > 0 > (1 row) > > You hardly need any flexibility for this anyway, so I'd just do > > regression=# SELECT position( ' ' in version() ) ; > position > ---------- > 11 > (1 row) > > Although possibly what you really want is split_part(). > > regression=# select split_part(version(), ' ', 2); > split_part > ------------ > 9.5.7 > (1 row) > > regards, tom lane > > An alternative select:- SELECT version(), (regexp_split_to_array( version(), E'\\s+'))[2] Cheers, Rob -- Sent via pgsql-general mailing list (pgsql-general@xxxxxxxxxxxxxx) To make changes to your subscription: http://www.postgresql.org/mailpref/pgsql-general
