4

我已经创建了这个服务（使用 WCF、Azure）：

[ServiceBehavior(AddressFilterMode = AddressFilterMode.Any)]
public class Service1 : IPMPService
{
    public int Dummy()
    {
        return 0;
    }
}

IPMPService 在哪里：

[ServiceContract]
public interface IPMPService
{
    [WebGet()]
    [OperationContract]
    int Dummy();
}

并试图在我的 android 应用程序中使用它：
第一次尝试：

String METHOD_NAME = "DummyRequest";
String NAMESPACE = "http://tempuri.org/";
String URL = "http://tyty.cloudapp.net/Service1.svc";
String SOAP_ACTION = "http://tyty.cloudapp.net/Service1.svc/Dummy";

String res = "";
try {
    SoapObject request = new SoapObject(NAMESPACE, METHOD_NAME);
    SoapSerializationEnvelope envelope = new SoapSerializationEnvelope(SoapEnvelope.VER11);
    envelope.dotNet = true;
    envelope.setOutputSoapObject(request);
    HttpTransportSE androidHttpTransport = new HttpTransportSE(URL);
    androidHttpTransport.call(SOAP_ACTION, envelope);
    SoapPrimitive result = (SoapPrimitive) envelope.getResponse();
    // to get the data
    String resultData = result.toString();
    res = resultData;
    // 0 is the first object of data
    } catch (Exception e) {
        res = e.getMessage();
    }

结果：SoapFault - faultcode: 'a:ActionNotSupported' faultstring: 由于 EndpointDispatcher 的 ContractFilter 不匹配，接收方无法处理带有 Action 'http://tyty.cloudapp.net/Service1.svc/Dummy' 的消息. 这可能是因为合约不匹配（发送方和接收方之间的操作不匹配）或发送方和接收方之间的绑定/安全不匹配。检查发送方和接收方是否具有相同的合同和相同的绑定（包括安全要求，例如消息、传输、无）。faultactor：'null' 详细信息：null

第二次尝试（取自此处）：

String SERVER_HOST = "http://tyty.cloudapp.net";
int SERVER_PORT = 8080;
String URL1 = "/Service1.svc/Dummy";
String keywords = null;
HttpEntity entity = null;
HttpClient client = new DefaultHttpClient();
HttpGet get = new HttpGet(URL1);
try {
HttpResponse response = client.execute(target, get);
entity = response.getEntity();
keywords = EntityUtils.toString(entity);
} catch (Exception e) {
e.printStackTrace();
   } finally {
if (entity != null)
        try {
        entity.consumeContent();
    } catch (IOException e) {
  }
}

结果：java.net.UnknownHostException：无法解析主机“http://tyty.cloudapp.net”：没有与主机名关联的地址

我还在 c# 中创建了一个（工作）客户端，这是它的 App.config：

<configuration>
  <system.serviceModel>
    <bindings>
      <basicHttpBinding>
        <binding name="BasicHttpBinding_IPMPService" closeTimeout="00:01:00"
            openTimeout="00:01:00" receiveTimeout="00:10:00" sendTimeout="00:01:00"
            allowCookies="false" bypassProxyOnLocal="false" hostNameComparisonMode="StrongWildcard"
            maxBufferSize="65536" maxBufferPoolSize="524288" maxReceivedMessageSize="65536"
            messageEncoding="Text" textEncoding="utf-8" transferMode="Buffered"
            useDefaultWebProxy="true">
          <readerQuotas maxDepth="32" maxStringContentLength="8192" maxArrayLength="16384"
              maxBytesPerRead="4096" maxNameTableCharCount="16384" />
          <security mode="None">
            <transport clientCredentialType="None" proxyCredentialType="None"
                realm="" />
            <message clientCredentialType="UserName" algorithmSuite="Default" />
          </security>
        </binding>
      </basicHttpBinding>
    </bindings>
    <client>
      <endpoint address="http://tyty.cloudapp.net/Service1.svc"
          binding="basicHttpBinding" bindingConfiguration="BasicHttpBinding_IPMPService"
          contract="IPMPService" name="BasicHttpBinding_IPMPService" />
    </client>
  </system.serviceModel>
</configuration>

如前所述，Android 中的两次尝试都失败了。谁能告诉我我做错了什么，以及如何做对？
谢谢。

4

2 回答 2

8

您的第一次尝试非常接近。只需要两个修复：

String METHOD_NAME = "Dummy";
String NAMESPACE = "http://tempuri.org/";
String URL = "http://tyty.cloudapp.net/Service1.svc";
String SOAP_ACTION = "http://tempuri.org/IPMPService/Dummy";

String res = "";
try {
    SoapObject request = new SoapObject(NAMESPACE, METHOD_NAME);
    SoapSerializationEnvelope envelope = new SoapSerializationEnvelope(SoapEnvelope.VER11);
    envelope.dotNet = true;
    envelope.setOutputSoapObject(request);
    HttpTransportSE androidHttpTransport = new HttpTransportSE(URL);
    androidHttpTransport.call(SOAP_ACTION, envelope);
    SoapPrimitive result = (SoapPrimitive) envelope.getResponse();
    // to get the data
    String resultData = result.toString();
    res = resultData;
    // 0 is the first object of data
} catch (Exception e) {
    res = e.getMessage();
}

第一个是方法名称：它只是Dummy（而不是DummyRequest）。第二个是 SOAP 动作：它是http://tempuri.org/IPMPService/Dummy.

两者都可以从 WSDL 派生。您的操作定义为：

<wsdl:operation name="Dummy">
    <wsdl:input wsaw:Action="http://tempuri.org/IPMPService/Dummy" message="tns:IPMPService_Dummy_InputMessage"/>
    <wsdl:output wsaw:Action="http://tempuri.org/IPMPService/DummyResponse" message="tns:IPMPService_Dummy_OutputMessage"/>
</wsdl:operation>

如果然后查看IPMPService_Dummy_InputMessage消息类型，则该元素称为Dummy：

<wsdl:message name="IPMPService_Dummy_InputMessage">
    <wsdl:part name="parameters" element="tns:Dummy"/>
</wsdl:message>

在这里可以找到 SOAP 操作：

<wsdl:operation name="Dummy">
    <soap:operation soapAction="http://tempuri.org/IPMPService/Dummy" style="document"/>
    ...
于 2012-05-27T09:03:45.650 回答
0 
public class RESTConnection extends Global 
{


    Global objglobal=Global.getInstance();

    private static String SERVICE_URI = "your service path";

    public JSONObject GetCashierLogin(String UserName, String Password)
    {
        new ArrayList<String>();  

        JSONObject jObj = null;  
        try 
        {

            SERVICE_URI=objglobal.getConnectionString();
            DefaultHttpClient httpClient = new DefaultHttpClient(); 
            HttpGet request = new HttpGet(SERVICE_URI+ "/LoginByUserIdAndPassword?Username=" + UserName + "&password=" + Password + "");
            request.setHeader("Accept", "application/json");
            request.setHeader("Content-type", "application/json");
            HttpResponse response = httpClient.execute(request);
            HttpEntity responseEntity = response.getEntity();
            String bufferLogin = EntityUtils.toString(responseEntity, HTTP.UTF_8);
            JSONArray jsonArray = new JSONArray(bufferLogin);

            if (jsonArray != null)
            {
                for (int i = 0; i < jsonArray.length(); i++)
                {
                    jObj=(JSONObject) jsonArray.get(i);

                }
            }

        } 
        catch (Exception e) 
        {
            e.printStackTrace();
        }
        return jObj;
    }
于 2013-06-14T13:44:38.413 回答