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我已经loginform在 android 中创建并检查服务器上的身份验证，但它不会给我输出，我不知道发生了哪个错误！

我的 Android 活动代码：

public class MainActivity extends Activity {
private final String NAMESPACE = "http://ws.webapp.org";
private final String URL = "http://192.168.0.2:8080/WebApp2/services/SearchData?wsdl";
private final String SOAP_ACTION = "http://ws.webapp.org/getSearchData";
private final String METHOD_NAME = "getSearchData";
/** Called when the activity is first created. */
@Override
public void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_main);
    Button login = (Button) findViewById(R.id.button1);
    login.setOnClickListener(new View.OnClickListener() {

        public void onClick(View arg0) {
            loginAction();              
        }
    });
}

private void loginAction(){
    SoapObject request = new SoapObject(NAMESPACE, METHOD_NAME);

    EditText userName = (EditText) findViewById(R.id.editText1);
    String user_Name = userName.getText().toString();
    EditText userPassword = (EditText) findViewById(R.id.editText2);
    String user_Password = userPassword.getText().toString();

  //Pass value for userName variable of the web service
    PropertyInfo unameProp =new PropertyInfo();
    unameProp.setName("userName");//Define the variable name in the web service method
    unameProp.setValue(user_Name);//set value for userName variable
    unameProp.setType(String.class);//Define the type of the variable
    request.addProperty(unameProp);//Pass properties to the variable

  //Pass value for Password variable of the web service
    PropertyInfo passwordProp =new PropertyInfo();
    passwordProp.setName("password");
    passwordProp.setValue(user_Password);
    passwordProp.setType(String.class);
    request.addProperty(passwordProp);

    SoapSerializationEnvelope envelope = new SoapSerializationEnvelope(SoapEnvelope.VER11);
    envelope.setOutputSoapObject(request);
    HttpTransportSE androidHttpTransport = new HttpTransportSE(URL);

    try{

        androidHttpTransport.call(SOAP_ACTION, envelope);
        SoapPrimitive response = (SoapPrimitive)envelope.getResponse();

        TextView result = (TextView) findViewById(R.id.textView1);
        result.setText(response.toString());

    }
    catch(Exception e)
    {
        Toast.makeText(getApplicationContext(), e.getMessage(), Toast.LENGTH_LONG).show();
        e.printStackTrace();          
    }
   }

Androidmanifest.xml文件：我已授予以下权限manifest.xml

请帮忙..！！

4

2 回答 2

2

永远不要在主线程中进行 HTTP 请求/网络调用，始终使用 Asynctask 在后台执行更长的操作，这样它就不会停止 UI 线程。

本教程提供了一个非常好的教程和分步指南。

于 2013-10-18T10:37:15.220 回答
0

在这里，您不使用任何与网络相关的操作的直接调用....

您首先调用后台线程并调用网络相关操作...。

参考这段代码：

块引用

login.setOnClickListener(new View.OnClickListener() {

    public void onClick(View arg0) {

         AddReviewAsyncTask  call  = new AddReviewAsyncTask ();

         call.execute();

    }
});




private class AddReviewAsyncTask extends AsyncTask<String, Void, Void>
{

int responce;
@Override
protected Void doInBackground(String... params) {
    // TODO Auto-generated method stub

     loginAction(); 
    return null;


}

@Override
protected void onPostExecute(Void result) {
    // TODO Auto-generated method stub
    super.onPostExecute(result);
}

@Override
protected void onPreExecute() {
    // TODO Auto-generated method stub
    super.onPreExecute();
}

}
于 2014-01-17T12:41:21.830 回答