RE: [PHP] Dynamic Form

echo "<SELECT NAME=\"whatever\">\n"; while($row = mysql_fetch_assoc($yourquery) echo "<OPTION ". ($whatever == $row['value']? "SELECTED ":"") ."VALUE=\"{$row['value']}\">{$row['name']}</OPTION>\n"; echo "</SELECT>\n"; Just insert a ternary operator in there, and check if the submitted value is equal to the database row's value... If so, add "SELECTED" to the OPTION tag. --Matt -----Original Message----- From: Jared Mashburn [mailto:jmashburn@ubtanet.com] Sent: Wednesday, September 12, 2001 6:45 PM To: php-general@lists.php.net Subject: [PHP] Dynamic Form Hell0, I have a MySql database with 3 columns. The first column is "id" second is "name" and third is "value", I have two drop-down lists, with the first filled with an array from the column "name". I would like for the second drop-down list be changed according to the "value" of what has been selected in the first drop-down list. I have fill that I'm going in the right direction, but have run into a wall. Can anyone give me some advice in doing this miraculous feat? Thanks, Jared -- PHP General Mailing List (http://www.php.net/) To unsubscribe, e-mail: php-general-unsubscribe@lists.php.net For additional commands, e-mail: php-general-help@lists.php.net To contact the list administrators, e-mail: php-list-admin@lists.php.net