Tony wrote:
> 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
Try checking if the db returned anything before you fetch it.


$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,$dbconnect);
############ Added this line ###########
if (mysql_num_rows($dbdo) > 0) {
########################################
while($row= mysql_fetch_array($dbdo)){
if($row[0]){
$counter++;
}else{
$stop=1;
}
$row[0] = "";
}
}
}