使用jQuery ajax中的参数调用WebMethod失败

时间:2014-04-15 07:18:46

标签: jquery ajax webforms webmethod

我对以下问题感到困惑。

我有'WebForm1.aspx'和'WebService1.asmx'。 当我在WebService WITHOUT参数('HelloWorld')中调用WebMethod时,它工作正常。当我调用Method WITH参数('SayHello')时,它失败了。

它甚至没有命中方法(我没有达到方法中设置的断点)。 xmlHttpRequest错误是“内部服务器错误”

[WebService(Namespace = "http://tempuri.org/")]
[WebServiceBinding(ConformsTo = WsiProfiles.BasicProfile1_1)]
[System.ComponentModel.ToolboxItem(false)]
// To allow this Web Service to be called from script, using ASP.NET AJAX, uncomment the following line. 
[System.Web.Script.Services.ScriptService]
public class WebService1 : System.Web.Services.WebService
{
    [WebMethod]
    public string HelloWorld()
    {
        return "Hello World";
    }

    [WebMethod]
    public string SayHello(string firstName, string lastName)
    {
        return "Hello " + firstName + " " + lastName;
    }
}



我的WebForms1.aspx:

    <div><br />No Parameters </div>
<div id="NoParameters"></div>
<div><br />With Parameters</div>
<div id="WithParameters"></div>


<script type="text/javascript">
    $(document).ready(function () {
        // SayHello returns a string we want to display.  Examples A, B and C show how you get the data in native
        // format (xml wrapped) as well as in JSON format.  Also how to send the parameters in form-encoded format,
        // JSON format and also JSON objects.  To get JSON back you need to send the params in JSON format.

        // Test - call a function that returns a string.
        // No Parameters
        $.ajax({
            type: "POST",
            url: "WebService1.asmx/HelloWorld",
            data: "{}",
            dataType: "text",
            success: function (data) {
                $("#NoParameters").html(data); // show the string that was returned, this will be the data inside the xml wrapper
            }
        });

        // Example A - call a function that returns a string.
        // Params are sent as form-encoded, data that comes back is text
        $.ajax({
            type: "POST",
            url: "WebService1.asmx/SayHello",

            //data: "firstName=Aidy&lastName=F", // the data in form-encoded format, ie as it would appear on a querystring
            //contentType: "application/x-www-form-urlencoded; charset=UTF-8", // if you are using form encoding, this is default so you don't need to supply it

            data: "{firstName:'Aidy', lastName:'F'}", // the data in JSON format.  Note it is *not* a JSON object, is is a literal string in JSON format
            contentType: "application/json; charset=utf-8", // we are sending in JSON format so we need to specify this

            dataType: "text", // the data type we want back, so text.  The data will come wrapped in xml
            success: function (data) {
                $("#WithParameters").html(data); // show the string that was returned, this will be the data inside the xml wrapper
            }
            , error: function(xmlHttpRequest, status, err) {alert(err);}
        });
    });

我正在使用的代码来自互联网上的示例。我尝试将参数传递为表单编码(conmmented out)和JSON。似乎没什么用。

怎么办?

  • 它可能是jQuery 1.8.2的版本吗?
  • 我应该使用'Fiddler'吗? (尚未使用它,学习曲线)
  • IE11中的F12开发人员工具会有帮助吗?
  • IIS Express是否在我的开发PC上本地运行?

非常感谢任何帮助。

我的浏览器有什么问题吗?在我的开发者机器上,有很多'上面有一个带有波形符号的A出现在googled页面上,我在IE中提交这个问题时遇到了问题。在我的普通PC上,提交很好。有些编码问题吗?

1 个答案:

答案 0 :(得分:9)

在ajax调用中更改&#34; data&#34; 参数 数据:JSON.stringify({Msg:&#39; Hello Client&#39;}), 并定义您的网络方法,如: - 的 [的WebMethod] public static string GetArray(string Msg) {

返回消息;

<强>}

在ajax调用中传入数据参数时,必须相同方法参数名称。

您可以在下面看到以下示例: -

        $(document).ready(function() {
    debugger;
    $.ajax({
        type: "POST",
        url: "flight-result-online.aspx/GetArray",
        data: JSON.stringify({ title: 'MP3', songname: 'Gulabi Ankhe' }),
        contentType: "application/json; charset=utf-8",
        dataType: "json",
        success: function(msg) {
            alert(msg.d);
        },
        error: function(msg) {
            alert(msg.d);
        }
    });

});

并且您的Web方法定义如下: -

[WebMethod]
public static string GetArray(string title, string songname)
{

return title+" "+songname;

}