Android点击两次返回键,退出应用程序
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(this, "在点击一次返回键将退出", 1000).show(); exitTime = System.currentTimeMillis(); } else { finish(); System.exit(0); } return true; } return super.onKeyDown(keyCode, event); }