java - OnPostExecute 被跳过

Question

当我使用 new GiveMeJson2.execute(ref_id).get() 调用异步类时，

我的应用程序进入 doInBackground，创建一个 JSON 字符串并返回它。

此时，我认为应用程序必须进入 onPostExecute 并解析该 JSON。

public class LiveArrivalActivity extends AppCompatActivity {
ListViewLiveArrival adapter;
ListView listView;
public static ArrayList<LiveArrivalHelper> list = new ArrayList<LiveArrivalHelper>();

@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_live_arrival);

    listView = findViewById(R.id.listView);

    FloatingActionButton fab = (FloatingActionButton) findViewById(R.id.fab);
    fab.setOnClickListener(new View.OnClickListener() {
        @Override
        public void onClick(View view) {
            Snackbar.make(view, "Replace with your own action", Snackbar.LENGTH_LONG)
                    .setAction("Action", null).show();
        }
    });

    Bundle b = getIntent().getExtras();
    Station station = (Station) b.get("Station");
    String ref_id = station.getRef_id();
    try {
        new GiveMeJson2().execute(ref_id).get();
    } catch (InterruptedException e) {
        e.printStackTrace();
    } catch (ExecutionException e) {
        e.printStackTrace();
    }
    Log.i("OUT", list.toString());

    adapter = new ListViewLiveArrival(this, list);
    listView.setAdapter(adapter);
}    }

class GiveMeJson2 extends AsyncTask<String, Void, String> {
String jsonText = "";
@Override
protected String doInBackground(String... stations) {

    try {
        URL url = new URL("https://www.trola.si/" + stations[0]);
        HttpURLConnection httpURLConnection = (HttpURLConnection) url.openConnection();
        httpURLConnection.setRequestProperty("Accept", "application/json");
        InputStream inputStream = httpURLConnection.getInputStream();
        BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(inputStream));
        jsonText = readAll(bufferedReader);
    } catch (MalformedURLException e) {
        e.printStackTrace();
    } catch (IOException e) {
        e.printStackTrace();
    }
    return jsonText;
}

@Override
protected void onPostExecute(String s) {
    super.onPostExecute(s);
    try {
        JSONArray jsonArray = new JSONObject(s).getJSONArray("stations");
        JSONArray buses = jsonArray.getJSONObject(0).getJSONArray("buses");
        for (int i = 0; i < buses.length(); i++) {
            JSONObject jsonObject = buses.getJSONObject(i);
            String direction = jsonObject.getString("direction");
            String number = jsonObject.getString("number");
            JSONArray arrivalsArray = jsonObject.getJSONArray("arrivals");


            if (arrivalsArray.length() > 0){
                String arrivals = "";
                for (int j = 0; j < arrivalsArray.length(); j++) {
                    arrivals += arrivalsArray.get(j) + " min ";
                }
                Log.i("OUT", number+direction+arrivals.toString());
                LiveArrivalHelper liveArrivalHelper = new LiveArrivalHelper(direction,number,arrivals);
                LiveArrivalActivity.list.add(liveArrivalHelper);
            }
        }

    } catch (JSONException e) {
        e.printStackTrace();
    }
}

private String readAll(Reader rd) throws IOException {
    StringBuilder sb = new StringBuilder();
    int cp;
    while ((cp = rd.read()) != -1) {
        sb.append((char) cp);
    }
    return sb.toString();
} }

关于为什么这样做有什么建议吗？

Answer 1

当您在 AsyncTask 上调用 .get() 时，它会强制将其称为同步。在本例中是在主线程上。

这是不允许的。
所以AsyncTask进入doInBackground，那么Android系统立即抛出错误。

如果您查看 LogCat，错误将打印在那里。

您需要从 AsyncTask 中删除 .get() 并实际异步处理它。