安卓FusedLocationProviderApi:传入的intent没有LocationResult或者LocationAvailability

Question

我正在尝试通过谷歌的FusedLocationProviderApi订阅位置更新。我希望在后台接收更新，这样即使应用程序被终止，我也会收到更新。根据在线文档，我尽可能地编写了以下代码。注意:这是在意图服务中完成的，而不是在UI线程上完成的，这就是我使用阻塞connect/result方法的原因。

private void startLocationServices(String deviceId, int pollingInterval) {
    Log.i(TAG, "Starting location services with interval: " + pollingInterval + "ms");
    PowerManager powerManager = (PowerManager) getSystemService(POWER_SERVICE);
    final PowerManager.WakeLock wakeLock = powerManager.newWakeLock(PowerManager.PARTIAL_WAKE_LOCK, TAG);
    wakeLock.acquire();

    final GoogleApiClient googleApiClient =
            new GoogleApiClient.Builder(this)
                    .addApi(LocationServices.API)
                    .build();

    ConnectionResult result = googleApiClient.blockingConnect();

    if (!result.isSuccess() || !googleApiClient.isConnected()) {
        Log.e(TAG, "Failed to connect to Google Api");
        wakeLock.release();
        return;
    }

    LocationRequest locationRequest = new LocationRequest();
    locationRequest.setInterval(pollingInterval);
    locationRequest.setFastestInterval(10000);
    locationRequest.setPriority(LocationRequest.PRIORITY_BALANCED_POWER_ACCURACY);

    Intent locationIntent = new Intent(this, GeoBroadcastReceiver.class);
    locationIntent.putExtra(EXTRA_LOCATION_UPDATE_DEVICE_ID, deviceId);
    locationIntent.setAction(GeoBroadcastReceiver.ACTION_LOCATION_UPDATE);
    PendingIntent locationPendingIntent = PendingIntent.getBroadcast(
            this, 0, locationIntent, PendingIntent.FLAG_UPDATE_CURRENT);

    PendingResult pendingResult = LocationServices.FusedLocationApi
            .requestLocationUpdates(googleApiClient, locationRequest, locationPendingIntent);

    Result requestResult = pendingResult.await();
    Status requestStatus = requestResult.getStatus();

    if (requestStatus.isSuccess()) {
        Log.i(TAG, "Successfully subscribed to location updates.");
    } else {
        Log.e(TAG, String.format(
                        "Failed subscribe to location updates. Error code: %d, Message: %s.",
                        requestStatus.getStatusCode(),
                        requestStatus.getStatusMessage()));
    }

    googleApiClient.disconnect();
    wakeLock.release();
}

当我运行这段代码时，我看到requestStatus.isSuccess()返回true，这表明我已经成功订阅了位置更新。此外，扩展WakefulBroadcastReceiver的GeoBroadcastReciever以正确的轮询间隔和正确的操作接收意图。到目前为止，看起来还不错。下面是我在GeoBroadcastReceiver的onReceive方法中所做的事情

    if (LocationResult.hasResult(intent)) {
        LocationResult locationResult = LocationResult.extractResult(intent);
        Location location = locationResult.getLastLocation();
        if (location != null) {
            GeoMonitoringService.wakefulLocationUpdate(context, location);
        } else {
            Log.e(TAG, "LocationResult does not contain a LastLocation.");
        }
    } else {
        Log.e(TAG, "Intent does not contain a LocationResult.");
    }

问题是，每当意图进入时，它既不包含LocationResult，也不包含LocationAvailabilityResult。我在调试器中检查了传入的intent，intent的附加内容中唯一的一项是我在设置intent时添加的额外内容(设备id)。因此，LocationResult.hasResult()返回false。每一次。

我在运行4.0.1的Galaxy Note 4和运行5.1.1的Nexus4上进行了测试，结果相同。

如果我禁用电话上的定位，我将完全停止接收意向，正如预期的那样。

Answer 1

从挂起的intent中删除多余的内容，否则定位结果不会下发。我在文档中找不到这一点的解释，但我在经过多次试验和错误后找到了答案。

Answer 2

一个变通方法(Christophe Beyls建议只使用Intent数据)所以，因为我只需要发送几个参数，所以我这样做:在requestLocationUpdates之前构建Intent时:intent.setData(“http://a.com/a？”+ Param1+ "?“+ Param2+ "?”+ Param3)；在BroadcastReceiver中: String[] parameters = intent.getDataString().split("")；这很好用，并且Intent确实返回位置。

Answer 3

您可以使用：

int id = 7;
String name = "myName";
uriBuilder.scheme("http")
            .authority("workaround.com")
            .appendPath("extra")
            .appendQueryParameter("id", String.valueOf(id))
            .appendQueryParameter("name", name);
intent.setData(uriBuilder.build());

和

@Override
protected void onHandleIntent(Intent intent) {
    if (LocationResult.hasResult(intent)) {
        int id = Integer.valueOf(uri.getQueryParameter("id"));
        String name = uri.getQueryParameter("name");
        ....
    }
}