如何将AJAX响应值传递给PHP变量

我想使用结果值并将其传递给php变量，这是我的代码…

billingCoffee.php

$("#linkAddSize").click(function(e){
            e.preventDefault();
            var txtCoffeeName = document.getElementById("txtCoffeeName").value;
            var cmbSizes = document.getElementById("cmbSizes").value;
            var txtPrice = document.getElementById("txtPrice").value;
            $.ajax({
                url: "addSizeandPrice.php",
                type: "POST",
                data: {coffeename: txtCoffeeName, sizes: cmbSizes, price: txtPrice},
                datatype: "json",
                success: function (result){
                    //set it php variable
                }
            });
        });

addSizeandPrice.php

    if($tableresult){
        $query = "INSERT INTO tbl$CoffeeName (CoffeeSize, Price) VALUES ('$Size', '$Price');";
        $insertresult = mysqli_query($con, $query);
        if($insertresult){
            SESSION_START();
            $_SESSION['nameCoffee'] = $CoffeeName;
            echo $_SESSION['nameCoffee'];
        }
        else{
            echo "Something went wrong!";
        }
    }

我想在不刷新页面的情况下使用变量…我有这个想法使用AJAX，但不知道如何设置它在php变量。