如何仅在应用程序打开时每隔几秒钟运行一次 Android 代码



目前,我的应用需要每 5 秒执行一次任务,其中包括更新使用 Google 地图 API 的地图中标记的位置。但是,当应用程序打开并且人们使用它时,它只需要每 5 秒执行一次。

据互联网上的人称,似乎 AsyncTask 即使在应用程序关闭时也能运行,导致它耗尽了大量的电池寿命。

如何使任务在且仅当应用程序打开或刚刚启动时运行?

private class MyAsyncTask extends AsyncTask<String, String, Void> {
    InputStream inputStream = null;
    String result = "";
    @Override
    protected Void doInBackground(String... params) {
        String url_select = "https://www.example.com/"; // generic
        try {
            URL url = new URL(url_select);
            HttpURLConnection connectTo = (HttpURLConnection) url.openConnection();
            connectTo.setRequestProperty("User-Agent", "");
            connectTo.setRequestMethod("POST");
            connectTo.setDoInput(true);
            connectTo.connect();
            // Read content & Log
            inputStream = connectTo.getInputStream();
        } catch (IOException e4) {
            Log.e("IOException", e4.toString());
            e4.printStackTrace();
        }
        // Convert response to string using String Builder
        try {
            BufferedReader bReader = new BufferedReader(new InputStreamReader(inputStream, "utf-8"), 8);
            StringBuilder sBuilder = new StringBuilder();
            String line = null;
            while ((line = bReader.readLine()) != null) {
                sBuilder.append(line + "n");
            }
            inputStream.close();
            result = sBuilder.toString();
        } catch (Exception e) {
            Log.e("St.Build,BuffRead", "Error converting result " + e.toString());
        }
        return null;
    } // protected Void doInBackground(String... params)
    @Override
    protected void onPostExecute(Void v) {
        //parse JSON data
        try {
            JSONObject busData = new JSONObject(result);
            if (busData.getString("title").equals("Bus Positions")) {
                for (Marker mark: allBuses) {
                    mark.remove(); // Remove marker
                }
                allBuses.clear();
                JSONObject resultSet = busData.getJSONObject("ResultSet");
                JSONArray results = resultSet.getJSONArray("Result");
                for (int i = 0; i < results.length(); i++) {
                    JSONObject bus = results.getJSONObject(i);
                    double lat = bus.getDouble("lat");
                    double lng = bus.getDouble("lng");
                    int call_name = bus.getInt("call_name");
                    String bus_type = "bus";
                    LatLng busLocation = new LatLng(lat,lng);
                    Marker busMark = map.addMarker(new MarkerOptions()
                            .position(busLocation)
                            .title(bus_type)
                            .icon(BitmapDescriptorFactory.fromResource(R.id.action_search)));
                    allBuses.add(busMark);
                }
            }
        } catch (JSONException e) {
            Log.e("JSONException", "Error: " + e.toString());
        } // catch (JSONException e)
    } // protected void onPostExecute(Void v)
}

您可以使用一些解决方法来创建长时间运行的异步任务,并根据活动的生命周期管理其生命周期。

You can cancel the AsyncTask in the onStop method of you Activity.

在我在网上的搜索中,我看到了几种不同的方法,并想知道这样做的最佳方法是什么。

安排服务器呼叫的最佳方式?

我看到的选项是:

1. 定时器
2. 计划线程池执行器

                  ***Timer***



int delay = 5000; // delay for 5 sec
int period = 1000; // repeat every sec
Timer timer = new Timer();
timer.scheduleAtFixedRate(new TimerTask() {
public void run() {
// // call your MyAsyncTask class
          }
     }, delay, period);
}

*计时器有一些缺点,可以通过ScheduledThreadPoolExecutor解决。*

           ***ScheduledThreadPoolExecutor.***

ScheduledExecutorService   executorService = Executors.newScheduledThreadPool(1);
    executorService.scheduleAtFixedRate(new Runnable() {
        @Override
        public void run() {
            runOnUiThread(new Runnable() {
                @Override
                public void run() {
                    Log.d("EXESER", "CH EXES");
                    // Your code

                }
            });
        }
    }, 1

0, 40, 时间单位.秒);

你有什么意见?

最新更新