Re: create dropdown menu from returned $result

I am trying to create a drop down menu to set varaible $tablrname Cheers paul "John W. Holmes" <holmes072000@charter.net> wrote in message news:3FB7A7C2.8040900@charter.net... > 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