Android 访问.NETWebService 遇到的问题，急求教！！

Question

public void onCreate(Bundle savedInstanceState) {
 super.onCreate(savedInstanceState);        setContentView(R.layout.main);
        SoapObject request = new SoapObject("http://tempuri.org/", "HelloWorld");  
        
        SoapSerializationEnvelope envelope = new SoapSerializationEnvelope(SoapEnvelope.VER12);   
      
        envelope.dotNet = true;
        envelope.setOutputSoapObject(request);
        AndroidHttpTransport  ht = new AndroidHttpTransport("http://192.168.1.101/Service/Service1.asmx");
        try {
         //这一步报错java.net.SocketException: Permission denied
         ht.call("http://tempuri.org/HelloWorld",envelope);

SoapObject soapObject = (SoapObject) envelope.getResponse();  

TextView tvResult = (TextView)findViewById(R.string.hello); 

tvResult.setText(soapObject.getProperty(0)+""); 

} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
} catch (XmlPullParserException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}对java.net.SocketException: Permission denied
网上说加权限<uses-permission android:name="android.permission.INTERNET" /> <?xml version="1.0" encoding="utf-8"?>
<manifest xmlns:android="http://schemas.android.com/apk/res/android"
      package="com.test.web"
      android:versionCode="1"
      android:versionName="1.0">    <application android:icon="@drawable/icon" android:label="@string/app_name">
        <activity android:name=".ActivityWeb"
                  android:label="@string/app_name">
            <intent-filter>
                <action android:name="android.intent.action.MAIN" />
                <category android:name="android.intent.category.LAUNCHER" />
            </intent-filter>
        </activity>
<uses-permission android:name="android.permission.INTERNET" /> 
    </application>
</manifest>
最后还是报那个错！有专家分析下

Answer 1

<?xml version="1.0" encoding="utf-8"?>
<manifest xmlns:android="http://schemas.android.com/apk/res/android"
      package="com.test.web"
      android:versionCode="1"
      android:versionName="1.0">    <application android:icon="@drawable/icon" android:label="@string/app_name">
        <activity android:name=".ActivityWeb"
                  android:label="@string/app_name">
            <intent-filter>
                <action android:name="android.intent.action.MAIN" />
                <category android:name="android.intent.category.LAUNCHER" />
            </intent-filter>
        </activity>
        
    </application>    <uses-permission android:name="android.permission.INTERNET" /> 
</manifest>
这样试下，权限语句放在application外面。

Answer 2

对啊，我记得权限应该加在applicaiton的外边的啊

Answer 3

对我忘了，我放下面了,
但现在又报下面错误，帮我看看程序代码 正确否public void onCreate(Bundle savedInstanceState) {
 super.onCreate(savedInstanceState);        setContentView(R.layout.main);
        SoapObject request = new SoapObject("http://tempuri.org/", "HelloWorld");  
        
        SoapSerializationEnvelope envelope = new SoapSerializationEnvelope(SoapEnvelope.VER12);   
      
        envelope.dotNet = true;
        envelope.setOutputSoapObject(request);
        AndroidHttpTransport  ht = new AndroidHttpTransport("http://192.168.1.101/Service/Service1.asmx");
        try {
                      ht.call("http://tempuri.org/HelloWorld",envelope);
            //报java.lang.ClassCastException: org.ksoap2.serialization.SoapPrimitive
            SoapObject soapObject = (SoapObject) envelope.getResponse();  
            
            TextView tvResult = (TextView)findViewById(R.string.hello); 
            
            tvResult.setText(soapObject.getProperty(0)+""); 
            
        } catch (IOException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        } catch (XmlPullParserException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }
}

Answer 4

 SoapObject request = new SoapObject("http://tempuri.org/", "HelloWorld");
引用不对。、
把引用路径全部写上可过、
如：
org.ksoap2.serialization.SoapObject soa=new org.ksoap2.serialization.SoapObject("http://tempuri.org/", "HelloWorld");
检查引用ksoap包时是否有错误！

Answer 5

不是权限的问题，我试过了，在    SoapObject soapObject = (SoapObject) envelope.getResponse(); 这一句有问题我改成这样 Object oj=envelope.getResult();
就通过了，虽然成功了，不知道为什么