Dynamic menus

[Jim F]

I did not include my question in the previous message

I have taken this from another example I saw of where one drop down
list is dependent on the selection of the previous. I can't get the
data to show up in the second menu after I select the first. Anyone
see what I am doing wrong?

Thanks, Jim

<?php
include('./common_db.inc') ;
$link_id = db_connect();
mysql_select_db ("test_aces_dat a");
?>

<form name=carform action=
<?php echo $_SERVER['PHP_SELF'];
?> method=post>
<?php


$query =mysql_query("s elect DISTINCT CQ_PROD_PLATFOR M AS make_id from
CQ_data ORDER BY make_id");
echo '<select name=make
onchange="docum ent.carform.sub mit()"'.">"."\n ";
if (!isset($make))
{
echo '<option value=null selected>Choose a Make</option>'."\n";
}
while(list($mak e_id) = mysql_fetch_row ($query))
{
echo '<option value='.$make_i d;
if (isset($make)&& $make_id == $make)
{
echo " selected";
}
echo '>'."$make_id". '</option>'."\n";
}
echo '</select>'."\n";
echo '<select name=model
onchange="docum ent.carform.sub mit()"'.">"."\n ";
if (!isset($make))
{
echo '<option value=null selected>Choose a Make First</option>'."\n";
}
else
{
$result = mysql_query("se lect DISTINCT CQ_SUBSYSTEM AS name from
CQ_data where make_id='$make' ORDER BY NAME");
echo '<option value=null selected>Choose a Model</option>'."\n";
while(list($mod el_id) = mysql_fetch_row ($result))
{
echo '<option value=' . $model_id;
if ($model_id == $model)
{
echo " selected";
}
echo '>$model_id</option>'."\n";
}
} echo '</select>'."\n";
?>

[steef]

Re: Dynamic menus

Jim~

I'm working on a similar problem with dependent dropdowns in php, and
I'm curious if you ever figured out why your second menu wasn't
populating? If you'd share your solution I'd be grateful. Thanks!