对 Json 响应感到沮丧

Question

所以这是我的问题。我正在使用 Jquery 的 $.ajax 将一系列值传递回 Web 方法。 Web 方法获取值，创建一个对象，然后将其作为 json 发送回调用页面。收到回复后，我将无法访问回复并显示其值。

谁能解释我需要做什么才能完成这项工作？

jquery 脚本：

$(document).ready(function() {
    $("#create").click(function() {
        var name = $('#name').val();
        var company = $('#company').val();
        var location = $('#location').val();
        var phonenumber = $('#phonenumber').val();
        var country = $('#country').val();

        $.ajax({
            type: "POST",
            url: "WebService.asmx/MakeEmployee",
            data: "{name:'" + name +
                          "',company:'" + company +
                          "',location:'" + location +
                          "',phonenumber:'" + phonenumber +
                          "',country:'" + country +
                          "'}",
            contentType: "application/json; charset=utf-8",
            dataType: "json",
            success: function(msg) {
                AjaxSucceeded(msg.d);
            }
        });
    });

    function AjaxSucceeded(data) {
        //var item = jQuery.parseJSON(data) // this doesn't work for me.
        $("#response").html(
            "<ul><li> " + data.Name +
            "</li><li> " + data.Company +
            "</li><li> " + data.Address +
            "</li><li> " + data.Phone +
            "</li><li> " + data.Country +
            "</ul> "
            );
        };
  });

网络方法：

[WebMethod]
[ScriptMethod(ResponseFormat = ResponseFormat.Json)]
public string MakeEmployee(string name, string company, 
              string location, string phoneNumber, string country)
{
    Employee e = new Employee(name, company, location, phoneNumber, country);
    return new JavaScriptSerializer().Serialize(e);
}

我得到的回应是：

{"d":"\"Name\":\"bob\",
          \"Company\":\"google\",
          \"Address\":\"home\",
          \"Phone\":\"123\",
          \"Country\":\"usa\"}"}

这是我认为我应该回来的：

{"Name":"bob",
     "Company":"google",
      "Address":"home",
      "Phone":"123",
      "Country":"usa"}

页面再次呈现后我得到的错误是：

•undefined
•undefined
•undefined
•undefined
•undefined

Answer 1

你的响应已经被解析为 JSON，所以它已经是一个对象了……不需要再次解析，直接使用它，像这样：

function AjaxSucceeded(data) {
    $("#response").html(
        "<ul><li> " + data.Name +
        "</li><li> " + data.Company +
        "</li><li> " + data.Address +
        "</li><li> " + data.Phone +
        "</li><li> " + data.Country +
        "</ul> "
    );
}

{ d: ... } 包装器由 ASP.Net 添加，这是正常行为。之后您的问题是元素未正确返回，您需要从 ASP.Net 返回一个 object 而不是 string，最好是这样：

[WebMethod, ScriptMethod(ResponseFormat = ResponseFormat.Json)]
public Employee MakeEmployee(string name, string company, 
  string location, string phoneNumber, string country) {
    return new Employee(name, company, location, phoneNumber, country);
}

...Employee 在 JavaScript 端具有您想要的属性。让 ASP.Net 在这里处理序列化而不是直接进行，整体上您会得到更清晰的响应。

Answer 2

尝试将此 ajax initaliazer 函数用于 asp.net ajax。它设置了大多数默认值，因此您只需提供 url/params 只需先调用您的 document.ready() 函数，然后再调用。

function jqueryInit() {
$.ajaxSetup({
    type: "POST",
    contentType: "application/json; charset=utf-8",
    data: "{}",
    dataFilter: function (data) {
        var msg;
        if (typeof (JSON) !== 'undefined' &&
                typeof (JSON.parse) === 'function')
            msg = JSON.parse(data);
        else
            msg = eval('(' + data + ')');

        if (msg.hasOwnProperty('d'))
            return msg.d;
        else
            return msg;
    }
});

}

Answer 3

从清理您的服务方法开始。你真的不需要这个构造函数和所有这些属性。你已经有一个Employee 类型，所以使用它：

[WebMethod]
[ScriptMethod(ResponseFormat = ResponseFormat.Json)]
public Employee MakeEmployee(Employee e)
{
    // Maybe do something more useful here with this employee
    // like raise his salary
    return e;
}

然后清理你的javascript：

$.ajax({
    type: 'POST',
    url: 'WebService.asmx/MakeEmployee',
    data: JSON.stringify({
        // All those correspond to Employee properties you would like to pass
        Name: $('#name').val(),
        Company: $('#company').val(),
        Location: $('#location').val(),
        PhoneNumber: $('#phonenumber').val(),
        Country: $('#country').val()
    }),
    contentType: 'application/json; charset=utf-8',
    dataType: 'json',
    success: function(msg) {
        // msg.d is gonna be the returned employee
        AjaxSucceeded(msg.d);
    }
});