I am just getting to this message. I have overhauled the LIKE handling
for the upcoming 6.5 beta(soon to start).
I am attaching a patch that represents my changes.
FYI, the SQL standard say LIKE '%%' is the same as LIKE '%'. To match a
literal '%', you have to do '\%'.
Let me know if you see any LIKE problems. Thanks.
> CC'd to the PostgreSQL Hackers list.
>
> I performed the same test as Brian on the cvs version of 6.4 and it
> exhibits the same behavior. could we get a fix in for the release.
> -DEJ
>
> > Hehehe...
> >
> > Try using the "*", as I posted to your early send... see if it makes
> > any
> > difference.
> Doing a like '*' will search for the character '*';
>
> > > er sorry about that last "early send" message...
> > >
> > > well, at first I though you were correct, but it turns out that
> > > postgresql
> > > is also inconsistent. Consider a table with a field username. 3
> > records:
> > >
> > > "Brian % Schaffner"
> > > "Brian T Schaffner"
> > > "%"
> > >
> > > select * from table where username='%' gets all rows (expected)
> > > select * from table where username='%%' gets the row with "%"
> > (expected)
> > > select * from table where username='%%%' gets the row with "%"
> > > (expected)
> > > select * from table where username='%%%%' gets no rows (expected)
> > > select * from table where username='% %' gets the 2 name rows
> > (expected)
> > > select * from table where username='% %%' gets the 2 name rows (NOT
> > > expected)
> > > select * from table where username='%% %' gets no rows (NOT
> > expected)
> > > select * from table where username='% % %' gets the 2 name rows
> > > (expected)
> > > select * from table where username='% %% %' gets the 2 name rows
> > (NOT
> > > expected)
> > > select * from table where username='% % % %' gets no rows (expected)
> > >
> > > so, if %% is the LIKE representation for a literal %, then why does
> > '%
> > > %%' return
> > > the 2 name rows, and '%% %' return no rows, and '% %% %' not return
> > the
> > > row with the
> > > embedded literal %?
> > >
> > > I could not get postgres to single out the row with the embedded %
> > using
> > > LIKE.
> > >
> > > why is this not getting any easier to define?
> > >
> > > -----Original Message-----
> > > From: Daniel J. Lashua [mailto:djl@stftx9.irngtx.tel.gte.com]
> > > Sent: Friday, September 25, 1998 10:58 AM
> > > To: Brian Schaffner
> > > Cc: 'Rasmus Lerdorf'; 'bourbon@bourbon.netvision.net.il';
> > > php3@lists.php.net
> > > Subject: RE: [PHP3] ARGH!! strstr() changed?
> > >
> > >
> > > On Fri, 25 Sep 1998, Brian Schaffner wrote:
> > >
> > > > in PostgreSQL (6.3.2 on FreeBSD 2.2.6):
> > > >
> > > > select * from table where field like '%%';
> > > >
> > > > returns NO rows;
> > > >
> > > > select * from table where field like '%';
> > > >
> > > > returns ALL rows;
> > > >
> > > > -brian-
> > > >
> > >
> > >
> > > I am not in any way doubting what you say... but that doesn't seem
> > > right.
> > > They should both reuturn all rows. Maybe in Postgres %% is the way
> > to
> > > state you actually want to search for ONE "%"?
> > >
> > > Out of curiosity, if you have time, could you test that. Make a
> > table
> > > with
> > > a field and in one row of the table in the field insert "%". Then do
> > > your
> > > select * from table where somefield LIKE '%%' and see if it returns
> > the
> > > one row?
> > >
> > > Am I offbase, or does that sound like incorrect behavior to you too?
> > >
> > > Daniel
> > >
> > >
> >
> >
> > --
> > PHP 3 Mailing List http://www.php.net/
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> > For help: php3-help@lists.php.net Archive:
> > http://www.php.net/mailsearch.php3
> >
>
>
--
Bruce Momjian | http://www.op.net/~candle
maillist@candle.pha.pa.us | (610) 853-3000
+ If your life is a hard drive, | 830 Blythe Avenue
+ Christ can be your backup. | Drexel Hill, Pennsylvania 19026
Index: like.c
===================================================================
RCS file: /usr/local/cvsroot/pgsql/src/backend/utils/adt/like.c,v
retrieving revision 1.22
retrieving revision 1.23
diff -c -r1.22 -r1.23
*** like.c 1999/03/15 02:18:36 1.22
--- like.c 1999/03/15 13:45:09 1.23
***************
*** 150,159 ****
{
int matched;
! for (; *p; text ++, p++)
{
- if (*text == '\0' && *p != '%')
- return LIKE_ABORT;
switch (*p)
{
case '\\':
--- 150,157 ----
{
int matched;
! for (; *p && *text; text++, p++)
{
switch (*p)
{
case '\\':
***************
*** 161,187 ****
p++;
/* FALLTHROUGH */
default:
! if (*text !=*p)
return LIKE_FALSE;
! continue;
case '_':
/* Match anything. */
! continue;
case '%':
! while (*++p == '%')
! /* Consecutive percents act just like one. */
! continue;
if (*p == '\0')
/* Trailing percent matches everything. */
return LIKE_TRUE;
while (*text)
! if ((matched = DoMatch(text ++, p)) != LIKE_FALSE)
return matched;
return LIKE_ABORT;
}
}
! return *text == '\0';
}
--- 159,201 ----
p++;
/* FALLTHROUGH */
default:
! if (*text != *p)
return LIKE_FALSE;
! break;
case '_':
/* Match anything. */
! break;
case '%':
! /* %% is the same as % according to the SQL standard */
! /* Advance past all %'s */
! while (*p == '%')
! p++;
if (*p == '\0')
/* Trailing percent matches everything. */
return LIKE_TRUE;
while (*text)
! {
! /* Optimization to prevent most recursion */
! if ((*text == *p ||
! *p == '\\' || *p == '%' || *p == '_') &&
! (matched = DoMatch(text, p)) != LIKE_FALSE)
return matched;
+ text++;
+ }
return LIKE_ABORT;
}
}
! if (*text != '\0')
! return LIKE_ABORT;
! else
! {
! /* End of input string. Do we have matching string remaining? */
! if (p[0] == '\0' || (p[0] == '%' && p[1] == '\0'))
! return LIKE_TRUE;
! else
! return LIKE_ABORT;
! }
}
