错误的响应 ajax

我尝试从我的数据库中填充文本框。我正在使用 onChange 事件。我可以看到代码正在运行，但我得到错误的响应。

我正在使用以下代码：

索引.php

<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "database";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
     die("Connection failed: " . $conn->connect_error);
} 
$sql = "SELECT * FROM cus";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
     echo "<select class='form-control select2' name='cus_id' onChange='getCus(this.value)' style='width: 100%;'>";
              echo "<option selected disabled hidden value=''></option>";
     // output data of each row
     while($row = $result->fetch_assoc()) {
                      echo "<option value='" . $row["cus_id"]. "'>" . $row["cus_id"].  " | " . $row["cus_name"]. "</option>";
     }                   
echo "</select>";
} else {
     echo "0 results";
}
$conn->close();
?>  
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
<script>
function getCus() {
    // getting the selected id in combo
    var selectedItem = jQuery('#cus_id option:selected').val();
    // Do an Ajax request to retrieve the product price
jQuery.ajax({
    url: 'get5.php',
    method: 'POST',
    data: {'cus_id': selectedItem}, 
    success: function(response){
        // and put the price in text field
        jQuery('#cus_id').val(response);
    },
    error: function (request, status, error) {
        alert(request.responseText);
    },
}); 
}
</script>

获取5.php

<?php
// Turn off all error reporting
error_reporting(0);
?>
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "database";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname) ;
// Check connection
if ($conn->connect_error) 
    {
    die('Connection failed: ' . $conn->connect_error) ;
    } 
else 
    {
    $cus_name = isset($_POST['cus_name'])?$_POST['cus_name']:'';
    $cus_ad = isset($_POST['cus_ad'])?$_POST['cus_ad']:'';
    $cus_pos = isset($_POST['cus_pos'])?$_POST['cus_pos']:'';
    $cus_pla = isset($_POST['cus_pla'])?$_POST['cus_pla']:'';
    $cus_cou = isset($_POST['cus_cou'])?$_POST['cus_cou']:'';
    $query = 'SELECT * FROM cus WHERE cus_id=' . $cus_id . ' ';
    $res = mysqli_query($conn, $query) ;
if (mysqli_num_rows($res) > 0) 
{
    $result = mysqli_fetch_assoc($res) ;
    echo $result['cus_name'];
    echo $result['cus_ad'];
    echo $result['cus_pos'];
    echo $result['cus_pla'];
    echo $result['cus_cou'];
}else{
    $result = mysqli_fetch_assoc($res) ;
    echo "0 res";
}
    }
?>

我看不出我的代码有什么问题。当我运行这个时，我得到响应"0 res"。谁能看出出了什么问题？