Обсуждение: BUG #4574: LIKE fails on non-varying character with no wildcards

The following bug has been logged online:

Bug reference:      4574
Logged by:          Nat!
Email address:      nat@mulle-kybernetik.com
PostgreSQL version: 8.3.5
Operating system:   Mac OS X 10.4
Description:        LIKE fails on non-varying character with no wildcards
Details:

The documentation claims:

http://www.postgresql.org/docs/8.3/interactive/functions-matching.html#FUNCT
IONS-LIKE

"If pattern does not contain percent signs or underscore, then the pattern
only represents the string itself; in that case LIKE acts like the equals
operator."


But:

create temporary table foo ( nummer character(12) );

insert into foo ( nummer) values( '1848' ) ;

select count(*) from foo where nummer = '1848' ;
-- returns 1

select count(*) from foo where nummer like '1848' ;
-- returns 0

drop table foo;


Whereas:


create temporary table bar ( nummer character varying );

insert into bar ( nummer) values( '1848' ) ;

select count(*) from bar where nummer = '1848' ;
-- returns 1

select count(*) from bar where nummer like '1848' ;
-- returns 1

drop table bar;


OTOH. Oracle has the same behaviour.

"Nat!" <nat@mulle-kybernetik.com> writes:
> create temporary table foo ( nummer character(12) );

> insert into foo ( nummer) values( '1848' ) ;

> select count(*) from foo where nummer = '1848' ;
> -- returns 1

> select count(*) from foo where nummer like '1848' ;
> -- returns 0

The trailing spaces are significant when doing LIKE on a char(n) column.

regression=# select '1848'::character(12) like '1848';
 ?column?
----------
 f
(1 row)

regression=# select '1848'::character(12) like '1848        ';
 ?column?
----------
 t
(1 row)

(By and large, my advice for all such cases is "don't use char(n)".
It has no redeeming social value whatever.)

            regards, tom lane