Re: newbie mysql and php
| From: | Brad Pauly | Date: | Sun, 28 Dec 2003 04:20:44 +0000 |
| Subject: | Re: newbie mysql and php | ||
| References: | 1 | Groups: | php.general |
| Request: | Send a blank email to php-general+get-173390@lists.php.net to get a copy of this message | ||
On Sat, 2003-12-27 at 17:17, 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
Try checking your query results. mysql_query might return false which is
not a valid result.
$dbdo = mysql_query($query, $dbconnect)
or die(mysql_error());
That should let you know what is going on.
- Brad