HEllo,
I think this will better work. What Jeremy wants to do is
to OUTER joing the tables:
SELECT distinct users.uid , users.username
FROM users LEFT OUTER JOIN picks using(uid)
may do the trick.
If you say letf join and the use ids not equal in the
where, then I think it's same as saying:
where uid = pid
and uid != pid
and will always be false.
REF:
http://www.mysql.com/doc/J/O/JOIN.html
--- Andrey Hristov <[EMAIL PROTECTED]> wrote:
> I don't know if this will help but why not to try
> $sql = "SELECT distinct users.uid , users.username
> FROM users LEFT JOIN picks USING(uid)
> WHERE picks.users_uid <> users.uid
>
> Andrey Hristov
> IcyGEN Corporation
> http://www.icygen.com
> 99%
>
> ----- Original Message -----
> From: "Jeremy Morano" <[EMAIL PROTECTED]>
> To: <[EMAIL PROTECTED]>
> Sent: Monday, August 27, 2001 9:52 PM
> Subject: [PHP] clause
>
>
> > Hi,
> >
> > Can somebody help me out?
> > My where clause is completely being ignored.
> > More specifically the <>. I tried to use != and that
> didn't work either.
> > However, when I substitute it with an = , It fuctions
> correctly.
> >
> > Right now, the output is all the users.uid and all the
> users.username from
> > the table users.
> > H E L P !
> >
> >
> >
> > $connection = @mysql_connect("l", "c", "c") or
> die("Couldn't connect.");
> >
> > $db = @mysql_select_db($db_name, $connection) or
> die("Couldn't select
> > database.");
> >
> > $sql = "SELECT distinct users.uid , users.username
> > FROM users, picks
> > WHERE picks.users_uid <> users.uid
> > ";
> >
> > $result = @mysql_query($sql,$connection) or
> die("Couldn't execute query.");
> >
> >
> > while ($row = mysql_fetch_array($result)) {
> > $uid = $row['uid'];
> > $username = $row['username'];
> >
> > $option_block .= "<option
> value=\"$uid\">$username</option>";
> > }
> >
> >
> > --
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> >
> >
>
>
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=====
Mehmet Erisen
http://www.erisen.com
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