RE: [PHP] bug in code - can't find it!

> Ok, here is my query and the bit of code I can't figure out: > > $query = "SELECT * from apt_user_t a, apt_company_t b "; > $query .= "WHERE a.user_cd = b.user_cd "; > $query .= "ORDER BY a.username"; > > $search_results = mysql_query($query) or die("Select > Failed!"); > while ($search_result2 = mysql_fetch_array($search_results)) > { > <INPUT > TYPE="checkbox" > NAME="<? echo $search_result2['user_cd'];?>" > SIZE="20" > MAXLENGTH="50" > VALUE="<?if ($search_result2[a.retired_flag] == > 1){?>CHECKED<?}?>"> <?echo $search_result2[a.retired_flag]?> > } > Nothing shows up with the echo or the value. I only included this > checkbox, but all of the other values show up fine. Can someone give me a > hint? Mysql does not prefix returned columns with table_name., so there's probably no "a.retired_flag" index in your array. A simple way to check this would be to print_r($search_result2). If you have a column named retired_flag in both table a and table b and you specifically want the one from table a in your result set you are going to have to alias it to a different name in your query, i.e. "SELECT *, a.retired_flag as r_flag"