Posted by briansmccabe on 12/19/05 22:24

$query6 = "SELECT COUNT (movie_id) as quant FROM movies WHERE divx = 1
AND format = 'reg'" or die(mysql_error());
$result6 = mysql_fetch_array($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


see for yourself: http://www.tgupc.com/admin/divxreport.php





Andy Hassall wrote:
> On 19 Dec 2005 12:01:34 -0800, briansmccabe@gmail.com wrote:
> It can't have got here and produced the same error, if you put in the error
> handling Kimmo posted.
>
> Post your revised code for the lines between mysql_query and
> mysql_fetch_array.

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