Update a Select Box on Parent Page from FancyBox2 - php

Question

I'm trying to submit a form in a fancybox where users can add a company to a select box that exists on the modals parent page. Im doing this by submitting the modal information to a script that adds the company to my database. Then I run a query to to get all the updated companies as a group of tags. Then I am trying to pass that group of tags to the parent page as a jquery update. Im not sure if this is the best approach or where I'm going wrong.
I am attempting to use this post as a guide:
Find element on site from a fancybox iframe
But I have two problems with my code.
One: The fancybox is not closing
Two: The select box on the parent page is not updating
I am not sure where I am going wrong with my success call. The code from the Modal page is:
$("#send-message").click(function(){
$(this).closest('form').submit(function(){
return false;
});
var frm = $(this).closest('form');
if($(frm).valid()){
$("#ajax-loading").show();
var data = $(frm).serialize();
$(frm).find('textarea,select,input').attr('disabled', 'disabled');
$.post(
"../forms/company_add.php",
data,
function(data) {
if (data.success) {
// data.redirect contains the string URL to redirect to
$('#companyselect', $(parent.document)).html(data.success);
parent.$.fancybox.close();
}
else {
$("#ajax-loading").hide();
$(frm).find('textarea,select,input').removeAttr('disabled');
$("#send_message_frm").append(data.error);
}
},
"json"
);
}
});
The Code from company_add.php returns all the options tags like such:
if ($_POST) {
// Collect POST data from form
$name = filter($_POST['name']);
$conmail = filter($_POST['conmail']);
$addy = filter($_POST['addy']);
$confax = filter($_POST['confax']);
$city = filter($_POST['city']);
$state = filter($_POST['state']);
$con = filter($_POST['con']);
$conphone = filter($_POST['phone']);
$zip = filter($_POST['zip']);
}
$search1 = mysql_query("SELECT man_name FROM manufacturers WHERE man_name = '$name'");
$outcome1 = mysql_fetch_row($search1);
$num_rows1 = mysql_num_rows($search1);
$imageid1 = $outcome1[0];
$imageid1 = filter($imageid1);
if ($num_rows1 > 0) {
echo json_encode(array(
"error" => '<div class="msg-error">A company by that name already exists.</div>'
));
} else {
$stmnt = mysql_query("INSERT INTO manufacturers (manufacturer_id, man_name, man_address, man_city, man_state,man_zip, man_contact, man_phone, man_fax, man_mail) VALUES ('NULL', '" . $name . "', '" . $addy . "' ,'" . $city . "', '" . $state . "' , '" . $zip . "' , '" . $con . "' , '" . $conphone . "' , '" . $confax . "', '" . $conmail . "' )");
//echo "Duplicate WAS found:" . $answer1;
mysql_query($answer1);
//}
$resp['status'] = 'success';
if (empty($error)) {
$nada = "SELECT man_name FROM manufacturers ORDER BY man_name ASC";
$resulter = mysql_query($nada);
$comp1 = '0';
//Spit out array of companys as select boxes
$select = '<option value="">--Select one--</option>';
while ($result59 = mysql_fetch_array($resulter))
$select .= '<option value="' . $result59['man_name'] . '">' . $result59['man_name'] . '</option>';
echo json_encode(array(
"success" =>$select
));
} else {
echo json_encode(array(
"error" => '<div class="msg-error">Error: Unable to add your company at this time</div>'
));
}
}
I am new to programming and very new to Jquery so I'm hoping someone can see where I'm going wrong. I am using fancybox 2 and php.

Answer 1

Have you checked error console for any JavaScript errors?
You can try this anyway:
parent.$('#companyselect').html(data.success);