Android_按两次返回键退出程序和长按返回键退出程序

以上两个功能主要是参考了一下博客的:

http://blog.csdn.net/chenshijun0101/article/details/7682210

http://blog.csdn.net/winlinking/article/details/6596455

方法一：

按两次返回键退出程序:

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();
            }
        }
}  

辅助类Exit.java

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;
}  

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); 

}