android程序点击两次返回键程序退出（方法总结）

来源：互联网发布：仙剑奇侠传98 boss数据编辑：程序博客网时间：2024/05/16 07:51

出处:http://tjlibaoh.blog.163.com/blog/static/2112264132012984518743/

方法一、

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;          }      };  public boolean onKeyDown(int keyCode, KeyEvent event) {                  // TODO Auto-generated method stub                  if(keyCode == KeyEvent.KEYCODE_BACK){  //                        System.out.println("user back down");                          if(isExit == false ) {                                  isExit = true;                                  Toast.makeText(this, "再按一次退出程序", Toast.LENGTH_SHORT).show();                                  if(!hasTask) {                                          tExit.schedule(task, 2000);                                  }} else {

      }                                  finish();                                  System.exit(0);                          }                  }                                          return false;          }

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方法二、

boolean isExit=false;
    Handler mHandler = new Handler(){
        @Override
        public void handleMessage(Message msg) {
            super.handleMessage(msg);
            isExit=false;
        }

    };

    @Override
    public boolean onKeyDown(int keyCode, KeyEvent event) {
        // TODO Auto-generated method stub
        if(keyCode == KeyEvent.KEYCODE_BACK){
            if(!isExit){
                isExit=true;
                Toast.makeText(getApplicationContext(), "再按一次退出程序", Toast.LENGTH_SHORT).show();
                //利用handler延迟发送更改状态信息
                mHandler.sendEmptyMessageDelayed(0, 2000);
            }
            else{
                finish();
                System.exit(0);
            }
        }
        return false;
    }

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方法三、

private long mExitTime = 0;
        public boolean onKeyDown(int keyCode, KeyEvent event) {
                if (keyCode == KeyEvent.KEYCODE_BACK) {
                        if ((System.currentTimeMillis() - mExitTime) > 2000) {
                                Object mHelperUtils;
                                Toast.makeText(this, "再按一次退出程序", Toast.LENGTH_SHORT).show();
                                mExitTime = System.currentTimeMillis();

                        } else {
                                finish();
                        }
                        return true;
                }
                return super.onKeyDown(keyCode, event);
        }
}

android程序 点击两次返回键程序退出（方法总结）

android程序点击两次返回键程序退出（方法总结）