Android初级教程_按两次返回键退出程序和长按返回键退出程序

本文详细介绍了如何通过双击返回键与长按返回键实现快速退出应用的功能，包括代码实现与原理解析。

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以上两个功能主要是参考了一下博客的:

http://blog.youkuaiyun.com/chenshijun0101/article/details/7682210

http://blog.youkuaiyun.com/winlinking/article/details/6596455

方法一：

按两次返回键退出程序:

public class Exit {

    private boolean isExit = false;
   private Runnable task = new Runnable() {
       @Override
       public void run() {
           isExit = false;
       }
   };

   public void doExitInOneSecond() {
       isExit = true;
       HandlerThread thread = new HandlerThread("doTask");
       thread.start();
       new Handler(thread.getLooper()).postDelayed(task, 1000);
   }

   public boolean isExit() {
       return isExit;
   }

   public void setExit(boolean isExit) {
       this.isExit = isExit;
   }
}

Activity:

public class Back2FinishAppActivity extends Activity {
   Exit exit = new Exit();

    /** Called when the activity is first created. */
    @Override
    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.main);
    }



    @Override
      public boolean onKeyDown(int keyCode, KeyEvent event) {
        if (keyCode == KeyEvent.KEYCODE_BACK) {
                pressAgainExit();
                return true;
                }

          return super.onKeyDown(keyCode, event);
      }

    private void pressAgainExit() {
            if (exit.isExit()) {
                finish();
            } else {
                Toast.makeText(getApplicationContext(), "再按一次退出程序",
                        Toast.LENGTH_SHORT).show();
                exit.doExitInOneSecond();
            }
        }
}

长按返回键退出程序:

public boolean dispatchKeyEvent(KeyEvent event)
    {
        int keyCode=event.getKeyCode();
        switch(keyCode)
        {
            case KeyEvent.KEYCODE_BACK: {
                 if(event.isLongPress())
                 {
                     this.stopService(intent);
                     System.exit(0);
                     return true;
                 }else
                 {
                     return false;

                 }
            }
        }
        return super.dispatchKeyEvent(event);

    }

方法二：

private long exitTime = 0;
@Override
public boolean onKeyDown(int keyCode, KeyEvent event) {
if(keyCode == KeyEvent.KEYCODE_BACK && event.getAction() == KeyEvent.ACTION_DOWN){
if((System.currentTimeMillis()-exitTime) > 2000){
Toast.makeText(getApplicationContext(), "再按一次退出程序", Toast.LENGTH_SHORT).show();
exitTime = System.currentTimeMillis();
} else {
finish();
System.exit(0);
}
return true;
}
return super.onKeyDown(keyCode, event);
}