[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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Re: [PHP] clause
Hello Jeremy,
you are using two tables, but you are not joining them.
What's you goal?
--- Jeremy Morano [EMAIL PROTECTED] wrote:
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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[EMAIL PROTECTED]
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[EMAIL PROTECTED]
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=
Mehmet Erisen
http://www.erisen.com
__
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Re: [PHP] clause
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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To unsubscribe, e-mail: [EMAIL PROTECTED]
For additional commands, e-mail: [EMAIL PROTECTED]
To contact the list administrators, e-mail: [EMAIL PROTECTED]
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Re: [PHP] clause
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;
}
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, e-mail:
[EMAIL PROTECTED]
For additional commands, e-mail:
[EMAIL PROTECTED]
To contact the list administrators, e-mail:
[EMAIL PROTECTED]
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, e-mail:
[EMAIL PROTECTED]
For additional commands, e-mail:
[EMAIL PROTECTED]
To contact the list administrators, e-mail:
[EMAIL PROTECTED]
=
Mehmet Erisen
http://www.erisen.com
__
Do You Yahoo!?
Make international calls for as low as $.04/minute with Yahoo! Messenger
http://phonecard.yahoo.com/
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