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  • Android 按二次后退键退出应用程序

    前言
             欢迎大家我分享和推荐好用的代码段~~
    声明
             欢迎转载,但请保留文章原始出处:
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    http://www.csdn.net
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    正文


    <span style="font-size:10px;">private static Boolean isExit = false;
        private static Boolean hasTask = false;
        Timer tExit = new Timer();
        TimerTask task = new TimerTask() {
             
            @Override
            public void run() {
                isExit = false;
                hasTask = true;
            }
        };
         
        @Override
        public boolean onKeyDown(int keyCode, KeyEvent event) {
            System.out.println("TabHost_Index.java onKeyDown");
            if (keyCode == KeyEvent.KEYCODE_BACK) {
                if(isExit == false ) {
                    isExit = true;
                    Toast.makeText(this, "再按一次后退键退出应用程序", Toast.LENGTH_SHORT).show();
                    if(!hasTask) {
                        tExit.schedule(task, 2000);
                    }
                } else {
                    finish();
                    System.exit(0);
                }
            }
            return false;
        }
    </span>
    <span style="font-size:10px;">private long waitTime = 2000;  
    	private long touchTime = 0; 
    	@Override
    	public boolean onKeyDown(int keyCode, KeyEvent event) {
    		if (event.getAction() == KeyEvent.ACTION_DOWN && KeyEvent.KEYCODE_BACK == keyCode) {
    			long currentTime = System.currentTimeMillis();
    			if ((currentTime - touchTime) >= waitTime) {
    				Toast.makeText(context, "再按一次退出程序", Toast.LENGTH_SHORT).show();
    				touchTime = currentTime;
    			} else {
    				finish();
    				System.exit(0);
    			}
    			return true;
    		}
        	
    		return super.onKeyDown(keyCode, event);
    	}
    </span>


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  • 原文地址:https://www.cnblogs.com/mengfanrong/p/5237519.html
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