Question

我正在使用以下代码来访问Web服务，并显示错误应用程序没有响应。

package com.android.webservice;

import org.ksoap2.SoapEnvelope;
import org.ksoap2.serialization.SoapObject;
import org.ksoap2.serialization.SoapSerializationEnvelope;
import org.ksoap2.transport.HttpTransportSE;

import android.app.Activity;
import android.os.Bundle;
import android.view.View;
import android.view.View.OnClickListener;
import android.widget.Button;
import android.widget.EditText;
import android.widget.TextView;
import android.widget.Toast;

public class WebServiceActivity extends Activity {
    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";


    @Override
    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.main);



        Button getquote = (Button) findViewById(R.id.getquote);  
        getquote.setOnClickListener(new OnClickListener() {  

        public void onClick(View v) { 
            TextView result1;
            result1=(TextView)findViewById(R.id.result1);
        try {


            SoapObject request = new SoapObject(NAMESPACE, METHOD_NAME);
            EditText CompanyName = (EditText) findViewById(R.id.CompanyName); 
            String val1 = (CompanyName.getText().toString());
            request.addProperty("passonString", val1);

            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();

            result1.setText(result.toString());
        } catch (Exception e) {

            result1.setText(e.getMessage());
            }
    }
});
    }
}

可能是因为这是在UI线程中运行，我想在后台运行它使用处理程序我是这个领域的新手，所以有poroblem在后台线程中写它。任何人都可以给我指点，如何在处理程序内编写代码。感谢

Answer 1

试试这个:: AsyncTask可以正确，方便地使用UI线程。该类允许执行后台操作并在UI线程上发布结果，而无需操纵线程和/或处理程序。

异步任务由在后台线程上运行的计算定义，其结果在UI线程上发布。异步任务由3种泛型类型定义，称为Params，Progress和Result，以及4个步骤，称为onPreExecute，doInBackground，onProgressUpdate和onPostExecute

     private class xyz extends AsyncTask<Void, Void, Void> {
    private final ProgressDialog dialog = new ProgressDialog(tranning.this);

    @Override
    protected void onPreExecute() {
        this.dialog.setMessage("Please Wait...");
        this.dialog.show();
        // put your code which preload with processDialog  
    }


    @Override
    protected Void doInBackground(Void... arg0) {
        // put your code here
        return null;
    }

    @Override
    protected void onPostExecute(final Void unused) {
        if (this.dialog.isShowing()) {
          this.dialog.dismiss();

        }   
    }
}

并在main ::

中使用它

new xyz().execute();

For more Elaboration
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Answer 2

您应该使用AsyncTask在后台执行此工作

http://labs.makemachine.net/2010/05/android-asynctask-example/

如何使用处理程序在后台运行任务

2 个答案: