二次判断退出

 1 private long exitTime = 0;
 2 
 3 @Override
 4 public boolean onKeyDown(int keyCode, KeyEvent event) {
 5 if(keyCode == KeyEvent.KEYCODE_BACK && event.getAction() == KeyEvent.ACTION_DOWN){
 6 if((System.currentTimeMillis()-exitTime) > 2000){ 
 7 Toast.makeText(getApplicationContext(), "再按一次退出程序", Toast.LENGTH_SHORT).show(); 
 8 exitTime = System.currentTimeMillis(); 
 9 } else {
10 finish();
11 System.exit(0);
12 }
13 return true; 
14 }
15 return super.onKeyDown(keyCode, event);
16 }