It looks like mysql_query is failing, add these two lines to your code, and
see what message pops up:
...
$dbdo = mysql_query($query,$dbconnect);
if(false === $dbdo) echo mysql_errno(),': ',mysql_error();
else echo 'The query worked, $dbdo value is:',$dbdo;
while($row= mysql_fetch_array($dbdo)){
...
Chris
-----Original Message-----
From: tony [mailto:[EMAIL PROTECTED]
Sent: Saturday, December 27, 2003 4:18 PM
To: [EMAIL PROTECTED]
Subject: [PHP] newbie mysql and php
hello
I'm new with php just learning and i have just a problem with the following
code
$dbconnect = mysql_connect("localhost", "prog_tony","PASSWORD");
mysql_select_db("prog_dealer", $dbconnect);
$stop = 0;
$counter = 1;
while(!$stop){
$query = "SELECT name FROM category WHERE id = $counter";
$dbdo = mysql_query($query,$dbconnect);
while($row= mysql_fetch_array($dbdo)){
if($row[0]){
$counter++;
}else{
$stop=1;
}
$row[0] = "";
}
}
I'm getting an error with mysql_fetch_array() which is line 14 because I
didn't show the other lines since it is not relevant.
here is the error
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result
resource in /home/prog/public_html/php/cateadded.php on line 14
Thank you
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