Re: create dropdown menu from returned $result

PAUL FERRIE wrote:

What am i overlooking?

The part where you tell us what you're actually trying to achieve...

<? 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>";

Why are you echoing data and displaying <br> between <select> and </select>... That's bad HTML.

$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'

So I see a select box with two options, 'reviews' and 'albums' that's made from badly formed HTML. What are you trying to accomplish?? -- ---John Holmes... Amazon Wishlist: www.amazon.com/o/registry/3BEXC84AB3A5E/ php|architect: The Magazine for PHP Professionals – www.phparch.com