Java 使用Ksoap进行xml解析

我正在使用ksoap2进行xml解析。我已经在我的库中添加了ksoap2.jar文件

import java.io.IOException;
import java.util.Vector;
import javax.microedition.midlet.*;
import javax.microedition.lcdui.*;
import org.ksoap2.SoapEnvelope;
import org.ksoap2.SoapFault;

import org.ksoap2.serialization.SoapObject;
import org.ksoap2.serialization.SoapSerializationEnvelope;
import org.ksoap2.transport.HttpTransport;


public class KsoapMidlet extends MIDlet {

Display display;
String strUserName= "Hello";
String strPassword = "World";
long ParentID;
String ChildIMEINumber;
String ChildName;
String serviceUrl = "http://www.semaphore.co.in/ChildTrackerService/ChildTrackerService.asmx";
String serviceNameSpace = "http://tempuri.org/";
String soapAction ="http://tempuri.org/AddChild";
String methodName = "AddChild";
Vector v=new Vector();



public void startApp() {
display = Display.getDisplay(this);
parsing();
// v.addElement("Hello");
// v.addElement("World");
// Constants d=new Constants();
// d.callSoap(methodName,v);
}

public void pauseApp() {
}

public void destroyApp(boolean unconditional) {
notifyDestroyed();
}
public void parsing(){
SoapObject request = new SoapObject(serviceNameSpace,methodName);
request.addProperty("strUserName", strUserName);
request.addProperty("strPassword", strPassword);
SoapSerializationEnvelope envelope = new SoapSerializationEnvelope(SoapEnvelope.VER11);
envelope.dotNet = true;
envelope.setOutputSoapObject(request);
envelope.bodyOut = request;
envelope.encodingStyle = SoapSerializationEnvelope.ENC;
HttpTransport transport = new HttpTransport("http://www.semaphore.co.in/ChildTrackerService/ChildTrackerService.asmx");
transport.setXmlVersionTag("<?xml version=\"1.0\" encoding=\"UTF-8\"?>");
transport.debug = true;
String result = null;

try {
transport.call("http://tempuri.org/" + methodName, envelope);
result = (envelope.getResponse()).toString();
System.out.println("***********************result********************************");
System.out.println("result=="+result);
System.out.println("Result :" + result.toString());
}
catch (org.xmlpull.v1.XmlPullParserException ex2) {
System.out.println("XmlPullParserException :" + ex2.toString());
System.out.println("Request \n" + transport.requestDump);
System.out.println("Response \n" + transport.responseDump);
} catch (SoapFault sf) {
System.out.println("SoapFault :" + sf.faultstring);
System.out.println("Request \n" + transport.requestDump);
System.out.println("Response \n" + transport.responseDump);
} catch (IOException ioe) {
System.out.println("IOException :" + ioe.toString());
System.out.println("Request \n" + transport.requestDump);
System.out.println("Response \n" + transport.responseDump);
}
}
}

请求 HelloWorld

回应

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我假设HTTP响应是一个带有自定义HTML错误页的HTTP 404。这是我所能想象的唯一一种可能，即如何在web服务中获得答案

可能是传输中的URL。调用（..）错误或服务器端的web服务未激活

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我建议使用WebScarab（Java）或Fiddler（.Net）等本地http代理，并将调用重定向到此本地代理。然后您将看到服务器作为响应发送的内容。

我也遇到了同样的问题。我解决这个问题的方法是更改手机上的个人资料设置。转到GPRS连接设置并将协议更改为HTTP（以前是在WAP上）。

@Robert感谢您的回复我得到了解决方案。这是因为我的设备中设置了代理。

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