js 跳转页面并且POST提交表单，C#后台获取参数值

要求：需要有 action，name，method

document.write("<form action='../../payment.aspx' method='post' name='formx1'>");
                            document.write("<input type='text' name='SCENERY_NAME' value='" + $(".itemName").text() + "' />");
                            document.write("<input type='text' name='orderNo' value='" + data + "' />");
                            document.write("<input type='text' name='price' value='" + price + "' />");
                            document.write("</form>");
                            document.formx1.submit();