Jquery post method - Auto populate two dropdown fields onchange - php

Question

I have a dropdown select called 'Select Company' on change of which i would auto-populate the 'Locations' dropdown by fetching the company id from the table. Below is my PHP code that returns the results after querying for both the fields that need to be auto-populated :
$output = array();
if(!empty($data))
{
$out = "<select name='location' id='location'>";
$out .="<option value=''>Select a location</option>";
for($i=0;$i<count($data);$i++)
{
$out .="<option value='".$data[$i]['id']."'>".$data[$i]['location']."</option>";
}
$out .= "</select>";
}
else{
$out = "<select name='location' id='location'>";
$out .="<option value=''>No locations added</option>";
$out .="</select>";
}
$sql2 = "select id,username from client24_bcpfm.users where cid = '$cid'";
$data2 = return_results($sql2);
if(!empty($data2))
{
$out2 = "<select name='username' id='username'>";
$out2 .="<option value=''>Select a User</option>";
for($i=0;$i<count($data2);$i++)
{
$out2 .="<option value='".$data2[$i]['id']."'>".$data2[$i]['username']."</option>";
}
$out2 .= "</select>";
}
else{
$out2 = "<select name='username' id='username'>";
$out2 .="<option value=''>No Users</option>";
$out2 .="</select>";
}
$output['0'] = $out;
$output['1'] = $out2;
print_r(json_encode($output));
Part where i am stuck , using the jquery to iterate the two different arrays for two different fields :
$.post('locations2.php',{cid:cid},
function(data)
{
data = $.parseJSON(data);
$("#location").html(data);
return;
});
Please let me know how to return the php values for 2 dropdowns and also iterate through them in the front end jquery .

Answer 1

If you want to return an array, you can also do it like this:
echo json_encode(array(
"out1" => $out,
"out2" => $out2));
and get the values in jquery like this:
$.post('locations2.php',{cid:cid},
function(data)
{
data = $.parseJSON(data);
//first array
$("#location").html(data.out1);
//second array
$("#location").html(data.out2);
return;
});