Member Avatar for Nirmeen Ased 
$sql="select *, (select Faculty from userinfo where Faculty=$college) as category from award where HonoringYear=$year";
               $rs=mysql_query($sql);
               return $rs;
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  • Latest Post Latest Post by Webville312

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Member Avatar for pritaeas

Replace line 2 with:

$rs = mysql_query($sql) or die(mysql_error());

It could be a problem with the sub-select, but hard to diagnose without any sample data.

Member Avatar for TonyG_cyprus

Also single quotes around $year

 AS category FROM award WHERE HonoringYear='$year'";

(also, makes easier reading if you capitalise )

Member Avatar for Webville312

... and also add single quotes to this line;

..select Faculty from userinfo where Faculty='$college'
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