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) { ToastUtils.showMessage("再按一次退出应用程序"); exitTime = System.currentTimeMillis(); } else { Intent intent = new Intent(Intent.ACTION_MAIN); intent.setFlags(Intent.FLAG_ACTIVITY_NEW_TASK); intent.addCategory(Intent.CATEGORY_HOME); startActivity(intent); return true; } return true; } return super.onKeyDown(keyCode, event); }
重写返回键 小于2秒退出
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转载自blog.csdn.net/m0_37358427/article/details/80900102
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