Mysql Query Help needed

I have a query "select cust_fnn, cust_name, agroup_access.group_access_cust from cust, agroup_access where agroup_access.group_access_group='$id' && cust.cust_fnn!=agroup_access.group_access_cust order by cust.cust_name" The 2 tables are as follows agroup_access agroup_access_id agroup_access_group agroup_access_cust 1 9 1610005847 2 9 0265478986 3 9 1610010254 Cust cst_id cust_name cust_fnn (ect....) 1 xxx 1610005847 2 xxxxx 0265478986 And so on.... 200+ records What I am trying to do is pull all records from cust where cust_fnn != agroup_access_cust && agroup_access_group=9.. What I am getting is all cust_fnn records but 3 of each and only 2 in the case of it matching the agroup_access_cust.. Can anyone see what I may have dome wrong.. I am trying to display them in a drop down box, below is the function for the drop down box and the line to call the function.. combo5("fnn","select cust_fnn, cust_name, agroup_access.group_access_cust from cust, agroup_access where agroup_access.group_access_group='$id' && cust.cust_fnn!=agroup_access.group_access_cust order by cust.cust_name",""); function combo5($name, $query, $sel) { $dbq=mysql_query($query); $html="<select name=\"$name\" class=\"longdd\">\n"; while( $data = mysql_fetch_array($dbq) ) { if($data[cust_fnn]==$sel) $html.="<option value=$data[cust_fnn] selected>$data[cust_name] : $data[cust_fnn]</option>\n"; else $html.="<option value=$data[cust_fnn]>$data[cust_name]: $data[cust_fnn]</option>\n"; } $html.="</select>\n"; return $html; } Hope someone can help... ------- Chris Kay, Eleet Internet Services chris@eleet.com.au -------