By using this site, you agree to our updated Privacy Policy and our Terms of Use. Manage your Cookies Settings.
449,278 Members | 1,136 Online
Bytes IT Community
+ Ask a Question
Need help? Post your question and get tips & solutions from a community of 449,278 IT Pros & Developers. It's quick & easy.

Multiple Dynamic Drop Down Population

P: n/a
Greetings:

I have a table with 3 pieces of data that I would like to use to
dynamically populate 3 drop downs using javascript. The fields are
state, orgname, office. If it's not already obvious, I'd like orgname
drop down to change when a state is selected and I would like office
drop down to change when an orgname is selected. I can do this with
multiple tables but am having difficulty getting it to work when the
data is in the same table. Below is the code to get state and orgname
from separate tables(the code reflects one table and is broken in the
below state). It's the best I can come up with and I can see why it
doesn't work but I know there must be a way to pull all the pieces from
a single table. Advice is much appreciated.

<code>
$list=$_SESSION['list'];
if(isset($list) and strlen($list) 0){
$quer=mysql_query("SELECT DISTINCT orgname,org_id FROM organization
WHERE state=$list ORDER BY orgname");
}else{$quer=mysql_query("SELECT DISTINCT orgname FROM organization ORDER
BY orgname"); }
$quer2=mysql_query("SELECT DISTINCT state FROM organization ORDER BY
state");

//first drop down
echo "<select name='state' onchange=\"reload(this.form)\"><option
value='0'>Select one</option>";
while($state = mysql_fetch_array($quer2)) {
if($state['org_id']==@$list){echo "<option selected
value='$state[state]'>$state[state]</option>"."<BR>";}
else{echo "<option value='$state[state]'>$state[state]</option>";}
}
echo "</select>";

//next drop down
echo "<select name='org'><option value=''>Select one</option>";
while($org = mysql_fetch_array($quer)) {
echo "<option value='$org[org_id]'>$org[orgname]</option>";
}
echo "</select>";
</code>

--

Regards,

Jeff Gardner
___________________________

"Contrary to popular belief, Unix is user friendly. It just happens
to be very selective about who its friends are." --Kyle Hearn
Sep 10 '06 #1
Share this Question
Share on Google+
1 Reply


P: n/a
Personally i think the best way would be ajax, and there is a million
ways to accomplish it in ajax. Here is a really simple bare bones
example. Like i said this is not the only way to do it but it is pretty
straight forward.

Example:
http://developer.apple.com/internet/...e/example.html
Tutorial:
http://developer.apple.com/internet/...mlhttpreq.html
Jeff Gardner wrote:
Greetings:

I have a table with 3 pieces of data that I would like to use to
dynamically populate 3 drop downs using javascript. The fields are
state, orgname, office. If it's not already obvious, I'd like orgname
drop down to change when a state is selected and I would like office
drop down to change when an orgname is selected. I can do this with
multiple tables but am having difficulty getting it to work when the
data is in the same table. Below is the code to get state and orgname
from separate tables(the code reflects one table and is broken in the
below state). It's the best I can come up with and I can see why it
doesn't work but I know there must be a way to pull all the pieces from
a single table. Advice is much appreciated.

<code>
$list=$_SESSION['list'];
if(isset($list) and strlen($list) 0){
$quer=mysql_query("SELECT DISTINCT orgname,org_id FROM organization
WHERE state=$list ORDER BY orgname");
}else{$quer=mysql_query("SELECT DISTINCT orgname FROM organization ORDER
BY orgname"); }
$quer2=mysql_query("SELECT DISTINCT state FROM organization ORDER BY
state");

//first drop down
echo "<select name='state' onchange=\"reload(this.form)\"><option
value='0'>Select one</option>";
while($state = mysql_fetch_array($quer2)) {
if($state['org_id']==@$list){echo "<option selected
value='$state[state]'>$state[state]</option>"."<BR>";}
else{echo "<option value='$state[state]'>$state[state]</option>";}
}
echo "</select>";

//next drop down
echo "<select name='org'><option value=''>Select one</option>";
while($org = mysql_fetch_array($quer)) {
echo "<option value='$org[org_id]'>$org[orgname]</option>";
}
echo "</select>";
</code>

--

Regards,

Jeff Gardner
___________________________

"Contrary to popular belief, Unix is user friendly. It just happens
to be very selective about who its friends are." --Kyle Hearn
Sep 10 '06 #2

This discussion thread is closed

Replies have been disabled for this discussion.