longfirstTime=0;@OverridepublicbooleanonKeyDown(intkeyCode,KeyEventevent){if(keyCode==KeyEvent.KEYCODE_BACK&&event.getRepeatCount()==0){longsecondTime=System.currentTimeMillis();if(secondTime-firstTime>2000){//如果两次按键时间间隔大于2000毫秒，则不退出Toast.makeText(FragmentMainActivity.this,"再按一次退出",Toast.LENGTH_SHORT).show();firstTime=secondTime;//更新firstTimereturntrue;}else{System.exit(0);}}returnfalse;}