Android从匿名类中获取活动

Question

我是Android开发和Java的新手.这是基本设置:我有一个启动画面,AsyncTask用于检查服务器可用性.在这个帖子之后,我在我的活动中做了一个回调.这比做工作更有意义,OnPostExecute()因为我想在不同的活动中重用这个任务.

但是,在我的回调中,我会检查状态是否正常.如果是,它应该启动下一个活动.但是从我的回调的上下文中,我不知道如何获得我的Activity引用,我需要它作为Intent的参数.

这是OnCreate下我的Activity中的代码:

    //Check server status
    CheckServiceTask t = new CheckServiceTask(new OnTaskCompleted<ShaggyServiceStatus>() {
        @Override
        public void onTaskCompleted(ShaggyServiceStatus result) {
            Log.i(TAG, "Callback. Result: " + result.getStatus());
            ProgressBar pb = (ProgressBar) findViewById(R.id.splash_progress);
            pb.setVisibility(View.INVISIBLE);

            if (result.getStatusCode() == 999){
                TextView t = (TextView) findViewById(R.id.splash_status_text);
                t.setText(result.getStatus());
                return;
            }

            Intent i = new Intent(getActivity(), LoginActivity.class);
            startActivity(i);
            finish();

        }
    });

    t.execute();

失败的部分是getActivity().那个电话不可用.使用this抛出错误(我理解,因为我在上下文中OnTaskCompleted).

为了完整性,这是OnTaskCompleted的接口:

public interface OnTaskCompleted<T> {
    public void onTaskCompleted(T result);
}

这是CheckServiceTask类:

public class CheckServiceTask extends AsyncTask<Void, Void, ShaggyServiceStatus>{
    private static final String TAG = "coo";

    public OnTaskCompleted<ShaggyServiceStatus> listener;

    public CheckServiceTask (OnTaskCompleted<ShaggyServiceStatus> l){
        this.listener = l;
    }

    @Override
    protected ShaggyServiceStatus doInBackground(Void... params) {
        try {
            Log.i(TAG, "Connecting to server...");
            //TODO: Make this a setting
            final String url = "https://someplace.com/status";
            RestTemplate restTemplate = new RestTemplate();
            restTemplate.getMessageConverters().add(new MappingJackson2HttpMessageConverter());
            //restTemplate.postForObject("url", bodyObject, ShaggySer.class);
            ShaggyServiceStatus sss = restTemplate.getForObject(url, ShaggyServiceStatus.class);
            Log.d(TAG, "Got the status.");
            return sss;
        } catch (Exception e) {
            //TODO: Exception handling
            Log.e(TAG, e.getMessage(), e);
        }

        //If we're here, it's not okay.
        ShaggyServiceStatus r = new ShaggyServiceStatus("Cannot connect to server", 999, "none");

        return r;
    }

    @Override
    protected void onPostExecute(ShaggyServiceStatus result) {
        super.onPostExecute(result);
        listener.onTaskCompleted(result);
    }
}

Answer 1

使用 CurrentClassName.this

Intent i = new Intent(CurrentClassName.this, LoginActivity.class);

getActivity():与Fragments一起使用

返回此片段当前与之关联的Activity.

class.this与this嵌套类时不同.

Answer 2

如果您使用的活动名为" MyActivity",则可以执行以下操作:

MyActivity.this

这段代码将返回外部类的这个"当前"对象