Javascript / PHP-将数据传递给模式（弹出窗口）

Question

I currently have a javascript popup div that contains a select object that is populated from a POST variable that was generated from the main page. Is it efficient to pass data with html via post to pass data to a popup? Or is there a better way to do this?

main page

< ?php 
require('testclass.php);
$obj = new testclass();
foreach ($obj->getlist as $listobj)
{
$output .= "<option>" . $listobj['name'] . "</option>";
?>

main page javascript

<script type=text/javascript">
$(document).ready(function () {
$("#a_popup").click(function () {
$("#div_popup").load("popup.php", {"list" : "<?php echo $output ?>"});
});
});

popup.php

<select> <?php echo $_POST['list'] ?> </select>

Answer 1

You can send data back to page and do something like this

   <select name="what ever you want">
                            <option value=""></option>
                            <?php
                            foreach ($values_from_page_you_sent as $key => $value)
                            {

                                    echo '<option value="' . $key . '">' . $value . '</option>';
                            }
                            ?>
                        </select>

Something like this you can modify it according to your use values_from_page_you_sent are values that you send from some X page to this page.

Edit: if you are looking Jquery/JS based solution then that is another scenario which you can do some thing like this

jQuery.each(data, function(key, value) {
            jQuery('select[name="' + populatedElement + '"]')
                    .append(jQuery("<option></option>")
                    .attr("value", key)
                    .text(value));
        });