Android_按两次返回键退出程序和长按返回键退出程序

最新推荐文章于 2021-05-27 17:05:49 发布

原创最新推荐文章于 2021-05-27 17:05:49 发布 · 687 阅读

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本文介绍了如何在Android应用中实现双击返回键退出程序的功能，包括使用辅助类来记录点击次数和延迟时间的方法，以及通过长按返回键直接退出程序的实现方式。

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:

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); 

}