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~~
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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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