android - 从 Android 使用 Wcf，错误请求错误

Question

我正在尝试从 Android 中使用 Wcf，我的代码

private final static String SERVICE_URI = "http://188.59.2.211:8081/Survey.svc";

String email = "xx@gmail.com";
String password = "123";

public void onResume() {
        super.onResume();

        Login(email,password);
    }

private void Login(String email, String password)
    {
        try {

        URL url = new URL(SERVICE_URI + "/Login/email="+email+"&password="+password);
            HttpGet request = new HttpGet(url.toURI());
            request.setHeader("Accept", "application/json");
            request.setHeader("Content-type", "application/json");

            HttpClient httpClient = new DefaultHttpClient();
            HttpResponse response = (HttpResponse) httpClient.execute(request);
            if ( response != null )
            {
              Log.i( "login", "received " + response.toString());
            }
            else
            {
              Log.i( "login", "got a null response" );
            }

            HttpEntity responseEntity = response.getEntity();

            // Read response data into buffer
            char[] buffer = new char[(int)responseEntity.getContentLength()];
            InputStream stream = responseEntity.getContent();
            InputStreamReader reader = new InputStreamReader(stream);
            reader.read(buffer);
            stream.close();

        } catch (Exception e) {
            e.printStackTrace();
        }
    }

和 wcf 服务端，

[ServiceContract(Namespace = "http://saas.com")]
public interface ISurvey
{
    [OperationContract]
    [WebGet(
       UriTemplate = "/Login/email={email}&password={password}",
       BodyStyle = WebMessageBodyStyle.WrappedRequest,
       ResponseFormat = WebMessageFormat.Json,
       RequestFormat = WebMessageFormat.Json)]
    LoginResult Login(string email, string password);
}

和我的配置，

<system.serviceModel>
<behaviors>
  <endpointBehaviors>
    <behavior name="httpBehavior">
      <webHttp />
    </behavior>
  </endpointBehaviors>
  <serviceBehaviors>
    <behavior>
      <serviceMetadata httpGetEnabled="true"/>
      <serviceDebug includeExceptionDetailInFaults="false"/>
    </behavior>
  </serviceBehaviors>
</behaviors>
<serviceHostingEnvironment multipleSiteBindingsEnabled="true" />
<services>
  <service name="SAASService.Survey">
    <endpoint address=""
              behaviorConfiguration="httpBehavior"
              binding="webHttpBinding"
              contract="SAASService.ISurvey" />
    <endpoint address="mex"
              binding="mexHttpBinding"
              contract="IMetadataExchange" />
  </service>
</services>

问题是，我使用下面的 url，

“http://188.59.2.211:8081/Survey.svc/Login/email=xx@gmail.com&password=123”

它总是给出“Bad Request”错误，statusCode = 400

怎么了？

Answer 1

从 Android 调用这样的网络服务是行不通的。另外，SOAP 是 xml，不是 json，所以你的内容类型是错误的。

要在 Android 中使用 Web 服务，最好使用 ksoap-android图书馆。你可以查看this tutorial示例代码。

这是我在最近的一个项目中使用的示例代码。

private static final String NAMESPACE = "http://webService.protocols.users.mydomain.com/";
private static final String URL = "http://1.1.1.1/domain/ws/userServices?wsdl";
private static final String SOAP_ACTION = "";

public boolean authenticate(String username, String password) {
    String METHOD_NAME = "authenticate";

    SoapObject request = new SoapObject(NAMESPACE, METHOD_NAME);
    request.addProperty("arg0", username);
    request.addProperty("arg1", password);

    SoapSerializationEnvelope envelope = new SoapSerializationEnvelope(SoapEnvelope.VER11);
    envelope.setOutputSoapObject(request);

    HttpTransportSE httpTransport = new HttpTransportSE(URL);
    httpTransport.debug = true;
    try {
        httpTransport.call(SOAP_ACTION, envelope);
    } catch (Exception e) {
        return false;
    }

    SoapObject result = null;
    try {
        result = (SoapObject) envelope.getResponse();
    } catch (SoapFault soapFault) {
        return false;
    }

    // anything below this line relates to the service return type, may not apply to you
    boolean success = Boolean.valueOf(result.getProperty("success").toString());
    if (success)
        sessionId.set(result.getProperty("sessionId").toString());
    else
        error_message.set(result.getProperty("errorMessage").toString());

    return success;
}