无法在Handler()

时间:2016-02-18 08:25:25

标签: android multithreading handler

我从后台线程调用一个Handler类。在Handler类中,我试图显示一个toast。理论上它应该完美地工作,因为Handler是将UI任务转发到主UI线程的队列。但是,在我的情况下,我得到例外。

private void firstTimeLogin() {

        final LoginUiThreadHandler loginHandler = new LoginUiThreadHandler();

        new Thread(new Runnable() {
            @Override
            public void run() {
                Message m = loginHandler.obtainMessage();

                Bundle bund = new Bundle();
                bund.putInt("loginResult", 1);
                m.setData(bund);

                loginHandler.handleMessage(m);
            }
        }).start();
    }

    private class LoginUiThreadHandler extends Handler {

        @Override
        public void handleMessage(Message msg) {
            super.handleMessage(msg);

            int loginResult = msg.getData().getInt("loginResult");
            if(loginResult == 0) 
                Toast.makeText(getActivity().getApplicationContext(), "Login success", Toast.LENGTH_SHORT).show();

        }
    }

我做错了什么?

1 个答案:

答案 0 :(得分:-1)

替换为 -

LoginUiThreadHandler loginHandler = new LoginUiThreadHandler(Looper.getMainLooper());

而不是 -

LoginUiThreadHandler loginHandler = new LoginUiThreadHandler();