How to create a Multi-Select Dropdown in HTML using jQuery and PHP?

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PHP

jQuery

<script>
jQuery(document).ready(function()
{
    jQuery.ajax(
    {
        type: 'GET',
        url: 'specificationTypes.php',
        dataType: 'JSON',
        success:function(data)
        {
                    for(i=0;i<jQuery(data).length;i++)
            {
                $('#specificationType').append('<option value = '+ data[i].specification_type_id+'>' + data[i].specification_type_name + '</option>');
            }
        }
    });
});
</script>

<script>
jQuery(document).ready(function()
{
    jQuery('#specificationType').change(function()
        {
            var specificationType = jQuery("#specificationType").val();

            var jsonString = JSON.stringify(specificationType);

            jQuery.ajax(
            {
                type:"POST",
                url:"specifications.php",
                data: {data : jsonString},
                cache: false,
                success:function(data)
                {
                    for(i=0;i<jQuery(data).length;i++)
                    {
                        jQuery("#specifications").append('<option value = '+ data[i].project_specification_details+'>' + data[i].project_specification_details + '</option>');

                        console.log('success');
                    }
                }
            });
    });
});
</script>

my .php file

<?php
    mysql_connect("localhost","root","myfourwalls");

    mysql_select_db("new_project");

    $specificationType = json_decode(stripslashes($_POST['data']));
    $result = mysql_query("SELECT DISTINCT project_specification_details FROM project_specifications WHERE project_specification_type IN(".implode(",",$specificationType).")");

    $specifications = array();

    while ($data = mysql_fetch_array($result,MYSQL_ASSOC)) {

    $data_array['project_specification_details'] = $row['$project_specification_details'];

    array_push($specifications,$data);
    }
    $data = json_encode($specifications);

    echo $data;
?>

I got the JSON data to the variable called data but cant append it to the UI.Some error is coming in the console.