One problem:  you are using a database wrapper for which you have provided
no information.  $db->query is a function of a class $db, but we don't know
what that function does exactly.

Anyway, the $queryid doesn't contain a valid MySQL result resource.  So go
back a few steps.  Is $various_query a valid MySQL result?  Print it.  Keep
adding print statements all over the place until you find out when
$various_query is no longer (or if ever) a valid MySQL result resource.

My bet is that $db->query isn't returning a valid MySQL result.  Do a print
after your query on $various_query and see what it shows.  It should show:

Resource id #2

or something similar when you print it.


On Mon, 28 Oct 2002 [EMAIL PROTECTED] wrote:

> Hi,
> I got a query and I want to jump through the results.Here's my query:
>         $various_query = $db->query( "SELECT
>                                     int_param5,
>                                     char_param1,
>                                     is_admin_pass
>                                     FROM various" );
> And here I seek the data and fetch it into an array with mysql_fetch_row:
> $various = $db->seek( $various_query, 0 );
>             function seek( $result, $row )
>             {
>                 $result = mysql_data_seek( $result, $row );
>                 return $this->row( $result );
>             }
>             function row( $queryid )
>             {
>                 $this->record = mysql_fetch_row( $queryid );
>                 return $this->record;
>             }
> But if I load the page,I get this error:
> Warning: mysql_fetch_row(): supplied argument is not a valid MySQL result
> resource in
> What is wrong???

Peter Beckman            Systems Engineer, Fairfax Cable Access Corporation
[EMAIL PROTECTED]                   

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