我做了一个简单的项目，使用ksoap2调用wcf web服务。但是当它调用envelope.getResponse（）;它给错误说
错误： SoapFault - faultcode：'a：InternalServiceFault'faultstring：'由于内部错误，服务器无法处理请求。有关错误的更多信息，请在服务器上启用IncludeExceptionDetailInFaults（来自ServiceBehaviorAttribute或配置行为）以将异常信息发送回客户端，或者根据Microsoft .NET Framework 3.0 SDK文档启用跟踪并检查服务器跟踪日志。 faultactor：'null'detail：null
package testing.wcf;
import java.io.IOException;
import org.ksoap2.SoapEnvelope;
import org.ksoap2.serialization.SoapObject;
import org.ksoap2.serialization.SoapPrimitive;
import org.ksoap2.serialization.SoapSerializationEnvelope;
import org.ksoap2.transport.HttpTransportSE;
import org.xmlpull.v1.XmlPullParserException;
import org.xmlpull.v1.XmlSerializer;
import android.os.Bundle;
import android.os.StrictMode;
import android.widget.TextView;
import android.annotation.SuppressLint;
import android.app.Activity;
public class MainActivity extends Activity
{
private static final String strNAMESPACE = "http://www.testing.co.in/TestingService/";
private static final String strURL = "http://www.testing.co.in/TestingService/UserDetails.svc";
private static final String strSOAP_ACTION = "http://testing.co.in/TestingService/UserDetails/LoginDetails";
private static final String strMETHOD_NAME = "LoginDetails";
TextView tv;
StringBuilder sb;
String strInputXML;
@Override
public void onCreate(Bundle savedInstanceState)
{
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
tv = (TextView)findViewById(R.id.testing);
sb = new StringBuilder();
Call();
tv.setText(sb.toString());
//setContentView(tv);
}
public void Call()
{
try
{
SoapObject request = new SoapObject(strNAMESPACE, strMETHOD_NAME);
String inputxml = "<?xml version="+"\""+"1.0"+"\""+" encoding="+"\""+"utf-8"+"\""+" ?>" +"<MOB> \n <PIN>0000</PIN> \n <LOGINID>TEST</LOGINID> \n <PNUMBER>112233</pNUMBER> \n <REQUESTID>LoginVal</REQUESTID> \n </MOB>";
request.addAttribute("strInputXML", inputxml);
request.addAttribute("strOutputXML","");
SoapSerializationEnvelope envelope = new SoapSerializationEnvelope(SoapEnvelope.VER11);
envelope.dotNet = true;
envelope.setOutputSoapObject(request);
HttpTransportSE androidHttpTransport = new HttpTransportSE(strURL);
androidHttpTransport.call(strSOAP_ACTION, envelope);
SoapPrimitive result = (SoapPrimitive)envelope.getResponse();
String resultData = result.toString();
sb.append(resultData + "\n");
}
catch(Exception e)
{
sb.append("Error:\n" + e.getMessage() + "\n");
}
}
}
这里我想发送这样的请求
<?xml version="1.0" encoding="utf-8" ?>
<PhoneData>
<PINNO>0000</PINNO>
<LOGINID>HELLO</LOGINID>
<PASSWORD>1234</PASSWORD>
<REQID>0</REQID>
</PhoneData>
我的回复XML应该是
<?xml version="1.0" encoding="utf-8" ?>
<PhoneData>
<OTA>1234</OTA>
</PhoneData>
答案 0 :(得分：4)
我发布了使用WCF的工作代码（WCF的绑定必须是basicHttpBinding！）：
private static final String NAMESPACE = "http://tempuri.org/";
private static String URL="your url";
private static final String SOAP_ACTION_VALIDATION = "IValidateUser_wcf/ValidateUser";
private static final String VALIDATION_METHOD = "ValidateUser";
public boolean validateUser_WCF(String username, String password){
SoapSerializationEnvelope envelope = null;
SoapObject request = null;
HttpTransportSE httpTransportSE = null;
try {
request = new SoapObject(NAMESPACE, VALIDATION_METHOD);
request.addProperty("username", username);
request.addProperty("password", password);
envelope = new SoapSerializationEnvelope(SoapEnvelope.VER11);
envelope.dotNet = true;
envelope.setOutputSoapObject(request);
//////////////////////////////
// here you can add a HEADER element if you want
Element[] header = new Element[1];
header[0] = new Element().createElement(NAMESPACE_INFOCAD, "a1");
header[0].addChild(Node.TEXT, "HeaderTextContent");
envelope.headerOut = header;
//////////////////////////////
httpTransportSE = new HttpTransportSE(URL+VALIDATION_URI, 10*10000); // second parameter is timeout
httpTransportSE.debug = true;
httpTransportSE.setXmlVersionTag("<?xml version=\"1.0\" encoding=\"utf-8\"?>");
httpTransportSE.call(NAMESPACE+SOAP_ACTION_VALIDATION, envelope);
// if response is a simple text result, you can call SoapPrimitive, if not, you have to call SoapObject result and navigate in response's tree like an xml file
SoapPrimitive result = (SoapPrimitive)envelope.getResponse();
//To get the data.
String textResult = result.toString();
Log.i("textResult", textResult);
return true;
} catch (Exception e) {
// TODO: handle exception
e.printStackTrace();
}finally{
// here you can see in LOG what is request you send and what is response received
Log.i(getClass().getSimpleName(),"requestDump : "+httpTransportSE.requestDump);
Log.i(getClass().getSimpleName(),"responseDump : "+httpTransportSE.responseDump);
}
return false;
}
希望我的代码可以帮助你:)。它工作在100％
答案 1 :(得分：1)
我认为你应该研究一下名为FaultContract的东西。如果从服务中抛出异常，则错误消息将永远不会到达客户端。所以这里是FaultContract的用武之地。比如说，对服务的调用会导致服务异常，客户端应该如何知道捕获的错误？
以下是有关如何实施FaulContract：
//An interface for your service
[ServiceContract()]
public interface IMultiply
{
[OperationContract()]
int Multiply(int number1, int number2);
}
“服务类”中的方法应implement interface
//Service implementation
public class DoSomeMath: IMultiply
{
public int Multiply(int number1, int number2)
{
try
{
int k = number1 * number2;
}
catch (Exception e)
{
MyFaultException theFault = new MyFaultException();
theFault.Reason = "Some Error " + e.Message.ToString();
throw new FaultException<MyFaultException>(theFault);
}
return k;
}
}
在客户端，您现在可以捕获异常。如需进一步阅读，我建议您阅读：
Exception Handling in WCF using Fault Contract
我希望这会指出你正确的方向。
答案 2 :(得分：0)
对于Android模拟器使用此ip：10.0.2.2否则它将无法正常工作!! private final String SOAP_URL =“http://10.0.2.2:15398/Service1.svc”;