我在调用webservice时遇到问题。我在服务器上有一个.NET Web服务,我在Android中使用KSOAP2(ksoap2-j2se-full-2.1.2)。在运行程序时,我得到了一个运行时异常,如“org.ksoap2.serialization.SoapPrimitive”。我该怎么办?
这是我的代码。
package projects.ksoap2sample;
import org.ksoap2.SoapEnvelope;
import org.ksoap2.serialization.SoapObject;
import org.ksoap2.serialization.SoapSerializationEnvelope;
import org.ksoap2.transport.HttpTransportSE;
import android.app.*;
import android.os.*;
import android.widget.TextView;
public class ksoap2sample extends Activity {
/** Called when the activity is first created. */
private static final String SOAP_ACTION = "http://tempuri.org/HelloWorld";
private static final String METHOD_NAME = "HelloWorld";
private static final String NAMESPACE = "http://tempuri.org/";
private static final String URL = "http://192.168.1.19/TestWeb/WebService.asmx";
TextView tv;
@Override
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.main);
tv=(TextView)findViewById(R.id.text1);
try {
SoapObject request = new SoapObject(NAMESPACE, METHOD_NAME);
//request.addProperty("prop1", "myprop");
SoapSerializationEnvelope envelope = new SoapSerializationEnvelope(SoapEnvelope.VER11);
envelope.dotNet=true;
envelope.setOutputSoapObject(request);
HttpTransportSE androidHttpTransport = new HttpTransportSE(URL);
androidHttpTransport.call(SOAP_ACTION, envelope);
Object result = (Object)envelope.getResponse();
String[] results = (String[]) result;
tv.setText( ""+results[0]);
}
catch (Exception e) {
tv.setText(e.getMessage());
}
}
}
答案 0 :(得分:24)
这很简单。您将结果导入Object
,这是一个原始的。
您的代码:
Object result = (Object)envelope.getResponse();
正确的代码:
SoapObject result=(SoapObject)envelope.getResponse();
//To get the data.
String resultData=result.getProperty(0).toString();
// 0 is the first object of data.
我认为这绝对有用。
答案 1 :(得分:14)
您的.NET Webservice如何?
我使用ksoap 2.3 from code.google.com获得了相同的效果。我按照The Code Project上的教程(这是很棒的BTW。)
每次我使用
Integer result = (Integer)envelope.getResponse();
获取我的webservice的结果(无论类型如何,我都尝试了Object,String,int)我遇到了org.ksoap2.serialization.SoapPrimitive
异常。
我找到了解决方案(解决方法)。我要做的第一件事就是从我的webservice方法中删除“SoapRpcMethod()属性。
[SoapRpcMethod(), WebMethod]
public Object GetInteger1(int i)
{
// android device will throw exception
return 0;
}
[WebMethod]
public Object GetInteger2(int i)
{
// android device will get the value
return 0;
}
然后我将我的Android代码更改为:
SoapPrimitive result = (SoapPrimitive)envelope.getResponse();
但是,我得到一个SoapPrimitive对象,它有一个私有的“值”字段。幸运的是,该值通过toString()
方法传递,因此我使用Integer.parseInt(result.toString())
来获取我的值,这对我来说已经足够了,因为我没有任何需要从Web获取的复杂类型服务。
以下是完整资料来源:
private static final String SOAP_ACTION = "http://tempuri.org/GetInteger2";
private static final String METHOD_NAME = "GetInteger2";
private static final String NAMESPACE = "http://tempuri.org/";
private static final String URL = "http://10.0.2.2:4711/Service1.asmx";
public int GetInteger2() throws IOException, XmlPullParserException {
SoapObject request = new SoapObject(NAMESPACE, METHOD_NAME);
PropertyInfo pi = new PropertyInfo();
pi.setName("i");
pi.setValue(123);
request.addProperty(pi);
SoapSerializationEnvelope envelope =
new SoapSerializationEnvelope(SoapEnvelope.VER11);
envelope.dotNet = true;
envelope.setOutputSoapObject(request);
AndroidHttpTransport androidHttpTransport = new AndroidHttpTransport(URL);
androidHttpTransport.call(SOAP_ACTION, envelope);
SoapPrimitive result = (SoapPrimitive)envelope.getResponse();
return Integer.parseInt(result.toString());
}
答案 2 :(得分:7)
向SoapPrimitive输入信封:
SoapPrimitive result = (SoapPrimitive)envelope.getResponse();
String strRes = result.toString();
它会起作用。
答案 3 :(得分:4)
您可以使用以下代码调用Web服务并获得响应。请确保您的Web服务以数据表格式返回响应..如果您使用 SQL Server 数据库。如果你使用 MYSQL ,你需要改变一件事,只需用 DocumentElement替换句子obj2=(SoapObject) obj1.getProperty("NewDataSet");
中的单词 NewDataSet 强>
private static final String NAMESPACE = "http://tempuri.org/";
private static final String URL = "http://localhost/Web_Service.asmx?"; // you can use IP address instead of localhost
private static final String METHOD_NAME = "Function_Name";
private static final String SOAP_ACTION = NAMESPACE + METHOD_NAME;
SoapObject request = new SoapObject(NAMESPACE, METHOD_NAME);
request.addProperty("parm_name", prm_value); // Parameter for Method
SoapSerializationEnvelope envelope = new SoapSerializationEnvelope(SoapEnvelope.VER11);
envelope.dotNet = true;
envelope.setOutputSoapObject(request);
HttpTransportSE androidHttpTransport = new HttpTransportSE(URL);
try {
androidHttpTransport.call(SOAP_ACTION, envelope); //call the eb service Method
} catch (Exception e) {
e.printStackTrace();
} //Next task is to get Response and format that response
SoapObject obj, obj1, obj2, obj3;
obj = (SoapObject) envelope.getResponse();
obj1 = (SoapObject) obj.getProperty("diffgram");
obj2 = (SoapObject) obj1.getProperty("NewDataSet");
for (int i = 0; i < obj2.getPropertyCount(); i++) //the method getPropertyCount() return the number of rows
{
obj3 = (SoapObject) obj2.getProperty(i);
obj3.getProperty(0).toString(); //value of column 1
obj3.getProperty(1).toString(); //value of column 2
//like that you will get value from each column
}
如果您对此有任何疑问,可以写信给我。
答案 4 :(得分:0)
如果预计会有多个结果,那么getResponse()
方法将返回包含各种回复的Vector
。
在这种情况下违规代码变为:
Object result = envelope.getResponse();
// treat result as a vector
String resultText = null;
if (result instanceof Vector)
{
SoapPrimitive element0 = (SoapPrimitive)((Vector) result).elementAt(0);
resultText = element0.toString();
}
tv.setText(resultText);
回答
答案 5 :(得分:0)
我想你不能打电话
androidHttpTransport.call(SOAP_ACTION, envelope);
主线程上的。
Network operations should be done on different Thread.
创建另一个Thread或AsyncTask来调用该方法。