Re: Arraying JOINED tables
| From: | Denis Arh | Date: | Sun, 16 Jun 2002 07:28:17 +0000 |
| Subject: | Re: Arraying JOINED tables | ||
| References: | 1 | Groups: | php.db |
| Request: | Send a blank email to php-db+get-19829@lists.php.net to get a copy of this message | ||
Try find out if you have an error in your query, like this:
$query = "SELECT * FROM logins,auth WHERE logins.authname = $variable1
AND auth.authid = $variable2";
if ( $result = @mysql_query($query) ) {
$row = mysql_fetch_array($result);
}
else {
echo mysql_error();
}
and the other thing... maybe you have fields in both table logins and auth
with the same name
Denis Arh
----- Original Message -----
From: "César Aracena" <caracena@infovia.com.ar>
To: "PHP DB List" <php-db@lists.php.net>
Sent: Sunday, June 16, 2002 8:59 AM
Subject: [PHP-DB] Arraying JOINED tables
Hi all. Hope you're all alright since I don't see any of you writing for
some time now ;-)
This should be an easy one for all of you. I want to make a basic SELECT
query from two tables and fetch all the results into one array. I'm
doing it like this:
$query = "SELECT * FROM logins,auth WHERE logins.authname = $variable1
AND auth.authid = $variable2";
$result = mysql_query($query);
$row = mysql_fetch_array($result);
and I get: Warning: Supplied argument is not a valid MySQL result
resource in blah blah.
What am I doing wrong? Thanks in advance,
Cesar Aracena <mailto:webmaster@icaam.com.ar>
CE / MCSE+I
Neuquen, Argentina
+54.299.6356688
+54.299.4466621