RE: [PHP] newbie mysql and php
| From: | Chris | Date: | Sun, 28 Dec 2003 08:18:11 +0000 |
| Subject: | RE: [PHP] newbie mysql and php | ||
| References: | 1 | Groups: | php.general |
| Request: | Send a blank email to php-general+get-173394@lists.php.net to get a copy of this message | ||
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:awardd@essex.ac.uk]
Sent: Saturday, December 27, 2003 4:18 PM
To: php-general@lists.php.net
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
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php