Re: [PHP-DB] Arraying JOINED tables From: Denis Arh (denis <email protected>)
Date: 06/16/02

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 ( $result =  <email protected>($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 <email protected>>
To: "PHP DB List" <php-db <email protected>>
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 <email protected>>
CE / MCSE+I
Neuquen, Argentina
+54.299.6356688
+54.299.4466621 

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