没有Ajax响应后期值

Question

这里我根据selectbox值从数据库返回ajax值。我的脚本工作但是当我试图寻找阵列解决方案时，在ajax中发布值没有响应。为什么我没有获得价值？任何帮助将不胜感激。谢谢高级。

自动完成-ajax.php

<?php
    if(isset($_POST['pr_code'])) {
        $pr_code= $_POST['pr_code'];

        $sql = $mysqli->query("SELECT * FROM purchase_request WHERE pr='$pr_code'");
        while($row = $sql->fetch_assoc())
        {
            $approved= $row['approved'];
            $requested = $row['requested'];
        }
        echo json_encode(array("approved" => $approved, "requested" => $requested));
    }
?>

的index.php

<?php
    $combo = $mysqli->query("SELECT * FROM purchase_request GROUP BY pr ORDER BY id");
    $option = '';
    while($row = $combo->fetch_assoc())
    {
        $option .= '<option value = "'.$row['pr'].'">'.$row['pr'].'</option>';
    }
?>
<select id="tag"><option value="">wala</option><?php echo $option; ?></select>
<input name="requested" id="requested" type="text">
<input name="approved" id="approved" type="text">

<script type="text/javascript">
$(document).ready(function()
{
     $('select[id="tag"]').change(function() 
     {
         var pr_code = $("#tag").val();

         $.ajax({
             type: "POST",
             url: "autocomplete-ajax.php",
             data :"pr_code="+pr_code,
             dataType:'text',
             type:'POST',
             success:function(data){
                 var approved=data.approved;
                 var requested=data.requested;
                 alert(approved);
                 $('#approved').val()=approved;
                 $('#requested').val()=requested;
            }
       });
       return false;
    });
});
</script>