提交无需页面刷新php的ajax表单

我在index.php-中有此表单和ajax调用

 <form method='post' id='saleForm' >
    <input type='text' name='addCustomer' class='addInput' id='adlastname' placeholder='customer last name'>
    <input type='text' name='addYear' class='addInput'  id='adYear' placeholder='year'>
    <input type='text' name='addMake' class='addInput'  id='adMake' placeholder='make'>
    <input type='text' name='addModel' class='addInput'  id='adModel' placeholder='model'>
    <input type='text' name='addGross' class='addInput'  id='adGross' placeholder='front gross'>
    <input type='hidden' name='Id' value='<?php echo $id;?>'>
    <input type='submit' name='subSale'  id='subSale' value='Save' >
</form>

<script>
    $("form").submit(function(e) {
    var url = "add.php"; // the script where you handle the form input.

$.ajax({
       type: "POST",
       url: url,
       data: $("form").serialize(), // serializes the form's elements.
       success: function(data)
       {
           alert(data); // show response from the php script.
       }
     });

e.preventDefault(); // avoid to execute the actual submit of the form.
});
</script>

我正在尝试使用add.php将其提交给数据库-

$cust = $_POST['addCustomer'];
$addMake = $_POST['addMake'];
$addModel = $_POST['addModel'];
$addYear = $_POST['addYear'];
$addGross = $_POST['addGross'];
$subSale = $_POST['subSale'];
$soldDate = date('m/d/yy');

if(isset($subSale)){
    $sql = "INSERT INTO `$sold_table` ( `Id`,`Customer`, `Year`, `Make`, `Model`, `FrontGross`, `SoldDate` ) VALUES (NULL,'$cust', '$addYear', '$addMake', '$addModel', '$addGross', '$soldDate')";
    if (mysqli_query($conn, $sql)) {
        echo 'success!';
    } else {
       echo "Error: " . $sql . "<br>" . mysqli_error($conn);
    }
}

当我单击“保存”按钮时，它显示为空警报，并且由于某种原因数据没有被推送。我知道插入是正确的，因为当我不使用ajax调用而只使用表单操作时，它是正确的。