Re: [PHP] create dropdown menu from returned $result

[Duncan]

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
 

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