Re: create dropdown menu from returned $result

Currently you create 1 selectbox for each loop through the while loop (leaving aside that the <select> is in and the </select> outside the very same). By taking a quick look at it I recommend that you put the echo "<SELECT NAME='$tablename'>"; line outside & above the while loop. ...Hendrik PAUL FERRIE wrote:

What am i overlooking? <? include('common.php'); $result = mysql_list_tables("***_vinrev"); $i = 0; while ($i < mysql_num_rows ($result)) { $tb_names[$i]= mysql_tablename ($result, $i); echo "<SELECT NAME='$tablename'>"; echo "<OPTION VALUE=\"".$tb_names[$i]."\">". $tb_names[$i]." </OPTION> "; echo $tb_names[$i] . "<BR>"; $i++;

     }

echo "</SELECT>"; ?> When tested heres what i get http://thor.ancilenetworks.co.uk/~pferrie/vinrev/adm/table_list.php There are 2 tables within the DB 'reviews', 'albums' Cheers Paul