传递值jquery.ajax [英] passing values jquery.ajax

问题描述

亲爱的所有人，

我是MVC2和jquery的新手，尝试使用jquery.ajax在Model中传递值.
但是我收到一个错误"[object xmlHttpRequest] 403"

这是我的jQuery代码

Dear All,

I''m novice to MVC2 and jquery, trying to use jquery.ajax to pass values in Model.
But I''m getting an error " [object xmlHttpRequest] 403 "

Here is my jquery code

var nameVal = $("#txtName").val();
        var ageVal = $("#txtAge").val();
        var phoneVal = $("#txtPhone").val();
        alert(nameVal + " " + ageVal + " " + phoneVal + " ");
        $.ajax({
            type: "POST",
            url: "/Models/OrderClass.cs/Insert",
            data: "{name:''" + nameVal + "'',age:''" + ageVal + "'',phone:''" + phoneVal + "''}",
            contentType: "application/json",
            dataType: "json",
            success: function(data) {
                alert(data.d);
            },
            error: function(xmlHttpRequest, textStatus, errorThrown) {
                if (xmlHttpRequest.readyState == 0 || xmlHttpRequest.status == 0) {
                    alert(xmlHttpRequest + " " + xmlHttpRequest.status);
                } else {
                alert("ELSE :-) "+xmlHttpRequest + " " + xmlHttpRequest.status);
                }
            }
        });

这是模型代码

here is model code

[System.Web.Services.WebMethod]
        [HttpPost]
        public static int Insert(string name,string age, string phone)
        {
            int retVal = 0;
            if (!string.IsNullOrEmpty(name) && !string.IsNullOrEmpty(age) && !string.IsNullOrEmpty(phone))
            {
                try
                {
                    using(MySqlConnection con = new MySqlConnection(ConfigurationManager.AppSettings["mvcIceCream"]))
                    {
                        MySqlCommand cmd = new MySqlCommand();
                        cmd.Connection = con;
                        cmd.CommandText = "INSERT INTO student(Name,Age,Phone) VALUES(''" + name + "'',''" + age + "'',''" + phone + "'')";
                        cmd.CommandType = System.Data.CommandType.Text;
                        con.Open();
                        cmd.ExecuteNonQuery();
                        con.Close();
                        retVal = 1;
                    }
                }
                catch (Exception ex) { }
            }
            return retVal;
        }

请指导我.
谢谢

Please guide me.
Thanks

推荐答案

(#txtName").val(); var ageVal =

("#txtName").val(); var ageVal =

(#txtAge").val(); var phoneVal =

("#txtAge").val(); var phoneVal =

(#txtPhone").val(); alert(nameVal +" + ageVal +" + phoneVal +");

("#txtPhone").val(); alert(nameVal + " " + ageVal + " " + phoneVal + " ");

这篇关于传递值jquery.ajax的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！