两次点击返回键退出应用

我的APP首页是一个H5登录页面，用的WebView，登录之后进入个人首页，登录后在个人首页按返回键时会返回到WebView的登录页面，不能直接退出应用，我想登录后在个人首页点击返回键后直接退出应用。个人分析是Activity栈的原因。WebViewActivity在下面，IndexPageActivity在上面。

转载自：http://blog.youkuaiyun.com/sunnyfans/article/details/7688092

上面链接中的方法完美解决了我的问题，感谢！

1. 在项目中键一个SysApplication类，内容如下

import java.util.LinkedList;
import java.util.List;
import android.app.Activity;
import android.app.AlertDialog;
import android.app.Application;
import android.content.DialogInterface;
import android.content.Intent;

public class SysApplication extends Application {
    private List<Activity> mList = new LinkedList<Activity>();
    private static SysApplication instance;

    private SysApplication() {
    }
    public synchronized static SysApplication getInstance() {
        if (null == instance) {
            instance = new SysApplication();
        }
        return instance;
    }
    // add Activity
    public void addActivity(Activity activity) {
        mList.add(activity);
    }

    public void exit() {
        try {
            for (Activity activity : mList) {
                if (activity != null)
                    activity.finish();
            }
        } catch (Exception e) {
            e.printStackTrace();
        } finally {
            System.exit(0);
        }
    }
    public void onLowMemory() {
        super.onLowMemory();
        System.gc();
    }
}

2. 然后在每个Activity的onCreate()方法中添加

SysApplication.getInstance().addActivity(this);

具体如下：

    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        SysApplication.getInstance().addActivity(this);
        requestWindowFeature(Window.FEATURE_NO_TITLE);
        setContentView(R.layout.activity_webview);
        webView = (WebView) findViewById(R.id.webView);
        ......
        Intent intent = getIntent();
        String url = intent.getStringExtra("url");
        webView.loadUrl(url);

    }

    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        SysApplication.getInstance().addActivity(this);
        requestWindowFeature(Window.FEATURE_NO_TITLE);
        setContentView(R.layout.activity_index_page);
        initView();
        updateView();
    }

        在需要退出的地方调用

SysApplication.getInstance().exit();

我在应用中还使用了点击两次退出，

转载自：http://www.open-open.com/lib/view/open1392186781082.html

具体实现如下：

    // 定义一个变量，来标识是否退出
    private static boolean isExit = false;

    private static Handler mHandler = new Handler() {

        @Override
        public void handleMessage(Message msg) {
            super.handleMessage(msg);
            isExit = false;
        }
    };
    @Override
    public boolean onKeyDown(int keyCode, KeyEvent event) {
        if (keyCode == KeyEvent.KEYCODE_BACK) {
            exit();
            return true;
        }
        return super.onKeyDown(keyCode, event);
    }

    private void exit() {
        if (!isExit) {
            isExit = true;
            Toast.makeText(getApplicationContext(), "再按一次后退键退出程序",
                    Toast.LENGTH_SHORT).show();
            // 利用handler延迟发送更改状态信息
            mHandler.sendEmptyMessageDelayed(0, 2000);
        } else {
            Log.e(TAG, "exit application");
            SysApplication.getInstance().exit();

        }
    }

把另一种退出方式也摘录上，方便查找

package com.example.clickexittest; 
   
import android.app.Activity; 
import android.os.Bundle; 
import android.util.Log; 
import android.view.KeyEvent; 
import android.widget.Toast; 
   
public class MainActivity extends Activity { 
   
    private static final String TAG = MainActivity.class.getSimpleName(); 
   
    private long clickTime = 0; //记录第一次点击的时间 
   
    @Override 
    protected void onCreate(Bundle savedInstanceState) { 
        super.onCreate(savedInstanceState); 
        setContentView(R.layout.activity_main); 
   
    } 
   
    @Override 
    public boolean onKeyDown(int keyCode, KeyEvent event) { 
        if (keyCode == KeyEvent.KEYCODE_BACK) { 
            exit(); 
            return true; 
        } 
        return super.onKeyDown(keyCode, event); 
    } 
   
    private void exit() { 
        if ((System.currentTimeMillis() - clickTime) > 2000) { 
            Toast.makeText(getApplicationContext(), "再按一次后退键退出程序", 
                    Toast.LENGTH_SHORT).show(); 
            clickTime = System.currentTimeMillis(); 
        } else { 
            Log.e(TAG, "exit application"); 
            this.finish(); 
//          System.exit(0); 
        } 
    } 
}