|
Posted by Richard Conway on 07/12/05 15:36
Luigi Donatello Asero wrote:
> "Richard Conway" <ric@shoe.com> skrev i meddelandet
> news:FaOAe.2$HQ1.0@fe08.ams...
>
>>Try using mysql_fetch_array instead
>
>
> I get:
>
> Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result
> resource in /home/s/scaiecat/www/sv/test20.php on line 7
>
> on
> https://www.scaiecat-spa-gigi.com/sv/test20.php
>
>
In that case, there is something wrong with your query. Try changing
the following line from:
$result = mysql_query("SELECT * FROM name of the table", $db);
to:
$result = mysql_query("SELECT * FROM name of the table", $db) or
die(mysql_error());
This will trap the error and output it for you so you can get some idea
of where you've gone wrong.
Navigation:
[Reply to this message]
|