将JSON对象从Android客户端传递到c#中的WCF Restful服务

本文关键字:中的 WCF 服务 Restful 对象 JSON Android 客户端 | 更新日期: 2023-09-27 18:15:44

我有一个android客户端以JSON对象的形式发送数据到我的服务器。我的服务器是由WCF实现的,作为一个用c#编写的RESTful服务。我有一个类叫"用户"在我的WCF,我想在安卓客户端执行登录操作。但是当我以JSON格式发布我的对象到WCF服务时,我得到一个空对象(在Wrapped配置中),或者我得到一个字段为空的对象(在Bare配置中)。有人能解决这个问题吗?

下面是我的客户端生成的JSON示例:

{"用户":{"用户名":"123","通过":"123","设备":" 123 "}}

这是我的WCF接口代码:

    [OperationContract]
    [WebInvoke(Method = "POST",
        UriTemplate = "Login",
        BodyStyle = WebMessageBodyStyle.Wrapped,
        ResponseFormat = WebMessageFormat.Json,
        RequestFormat = WebMessageFormat.Json
      )]
    string Login(User user);
这是我的App.Config:
<system.serviceModel>
<services>
  <service behaviorConfiguration="CityManService.TrackingBehavior"
    name="CityManService.Tracking">
    <endpoint address="" behaviorConfiguration="json" binding="webHttpBinding"
      contract="CityManService.ITrackingService">
      <identity>
        <dns value="localhost" />
      </identity>
    </endpoint>
    <host>
      <baseAddresses>
        <add baseAddress="http://localhost:8731/CityManService/Tracking/" />
      </baseAddresses>
    </host>
  </service>
</services>
<behaviors>
  <endpointBehaviors>
    <behavior name="json">
      <webHttp />
    </behavior>
  </endpointBehaviors>
  <serviceBehaviors>
    <behavior name="CityManService.TrackingBehavior">
      <serviceMetadata httpGetEnabled="true" />
      <serviceDebug includeExceptionDetailInFaults="false" />
    </behavior>
  </serviceBehaviors>
</behaviors>

最后这是我的客户端(android)代码:
HttpPost request = new HttpPost("http://localhost:8731/CityManService/Tracking/Login");
request.setHeader("Accept", "application/json");
request.setHeader("Content-type", "application/json");
// Build JSON string
JSONStringer userJson = new JSONStringer()
    .object()
       .key("User")
          .object()
         .key("UserName").value(username.getText().toString())                                                                                                                                                        .key("Pass").value(password.getText().toString())                                                     .key("Device").value(password.getText().toString())
          .endObject()
       .endObject();
StringEntity entity = new StringEntity(userJson.toString(),"UTF-8");                                                
entity.setContentEncoding(new BasicHeader(HTTP.CONTENT_TYPE, "application/json"));
entity.setContentType("application/json");
request.setEntity(entity);
// Send request to WCF service
DefaultHttpClient httpClient = new DefaultHttpClient();             
HttpResponse response = httpClient.execute(request);

将JSON对象从Android客户端传递到c#中的WCF Restful服务

包装器的名称应该是参数名称,而不是类型名称。在您的服务中,操作定义为

[WebInvoke(...)] 
string Login(User user);

所以输入应该作为

传递
{"user":{"UserName":"123","Pass":"123","Device":"123"}}

(注意小写的"user"对象名)