my guess would be that sgid is a character field. if this isn't the case
then try hardcoding a known id into the query and seeing what happens. It
could also be that the connection has failed, try something like:
if ($connection = mysql_connect())
{
...
$id = rand(1,2);
$query = "SELECT songname FROM mp3 WHERE sgid = " .$id;
if ($result = mysql_query($query,$connection))
{ $mp3d = mysql_fetch_array($result);
...
} else echo("query failed");
...
} else echo("connection failed");
this'll give you an idea of where it's going wrong
Tim Ward
Senior Systems Engineer
Please refer to the following disclaimer in respect of this message:
http://www.stivesdirect.com/e-mail-disclaimer.html
> -----Original Message-----
> From: Chris Cocuzzo [mailto:[EMAIL PROTECTED]]
> Sent: 23 July 2001 23:57
> To: [EMAIL PROTECTED]
> Subject: [PHP]MySQL error, what's wrong here..
>
>
> <?php
> $id = rand(1,2);
> $query = "SELECT songname FROM mp3 WHERE sgid = " .$id;
> $result = mysql_query($query,$connection);
> $mp3d = mysql_fetch_array($result);
> ?>
>
> the server is telling me line 43(which starts with $mp3d) is
> not a valid
> mysql result resource. what am i doing wrong??
>
> chris
>
>
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, e-mail: [EMAIL PROTECTED]
For additional commands, e-mail: [EMAIL PROTECTED]
To contact the list administrators, e-mail: [EMAIL PROTECTED]
