RE: [PHP3] [OFF-TOPIC] POSTGRES

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

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