Posted by Andy Hassall on 12/19/05 21:54

On 19 Dec 2005 11:50:43 -0800, briansmccabe@gmail.com wrote:

>my code:
>
>$query6 = "SELECT COUNT (movie_id) as quant FROM movies WHERE divx = 1
>AND format = 'reg'";
>$result6 = mysql_fetch_array($query6);
>
>later on the page:
>
>echo $result6[quant]
>
>result:
>
>Warning: mysql_fetch_array(): supplied argument is not a valid MySQL
>result resource in /home/tgupc/public_html/admin/divxreport.php on line
>19

Kimmo posted code with basic error handling, which you've removed. Put it back
in again and it'll tell you why it failed.

--
Andy Hassall :: andy@andyh.co.uk :: http://www.andyh.co.uk
http://www.andyhsoftware.co.uk/space :: disk and FTP usage analysis tool

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