second query inside the loop of the first query's results

[eponymous]

Firstly, hello! I'm very glad to have come across this forum, and have discovered several answers to my issues already. I'm hoping somebody can help me with this one.

Relevent bit of code:

$query = "SELECT * FROM SS_League WHERE ID='$session_uid' ORDER BY Name";
$results = mysql_query( $query, $link );
while( $myrow = mysql_fetch_array( $results ) ) {
$lid = $myrow[ "ID" ];
$lname = $myrow[ "Name" ];

$query_league = "SELECT * FROM SS_League" . $lid . " WHERE AgentID = '$session_uid'";
$results_league = mysql_query( $query_league, $link );
while( $myrow_league = mysql_fetch_array( $results_league ) ) { // This is line 33, see error below
$lworth = $myrow_league[ "Worth" ];
}
// Lines which output the variables in some html.
}

This is the error when I load the page:
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in [file name] on line 33.

I've checked, and the problem is not the use of $lid in the second query. The error occurs even when I hard-code a table name into the query.

I feel as though I must be missing something obvious, but I'm totally stumped.

Thanks in advance. :)

[dreamcatcher]

Hello eponymous, welcome to Webmaster World. :)

To receive that error the query must be failing. So we need to look at this line:

$results_league = mysql_query( $query_league, $link );

Change this to the following:

$results_league = mysql_query( $query_league, $link ) or die(mysql_error());

This should output the problem to the browser.

dc

[eponymous]

Ahh, that did indeed show the problem. I made a mistake in my logic.

I've not used die before, but I'll definitely be doing so in future. Very handy.

Thanks for such a swift and helpful response. :)

[dreamcatcher]

Anytime my friend. :)