获取用户在Android中的位置的好方法

Question

问题：

尽快将用户的当前位置置于阈值内，同时节省电池电量。

为什么问题是一个问题：

首先，android有两个提供商;网络和GPS。有时网络更好，有时GPS更好。

“更好”是指速度与准确度之比。如果我几乎可以立即获得位置并且不打开GPS，我愿意牺牲几米精度。

其次，如果您要求更新位置更改，则在当前位置稳定时不会发送任何内容。

谷歌有一个确定“最佳”位置的例子：http://developer.android.com/guide/topics/location/obtaining-user-location.html#BestEstimate 但我认为它不应该接近它应该/可能的好。

我有点困惑为什么谷歌没有标准化的位置API，开发者不应该关心位置来自哪里，你应该只指定你想要的东西，手机应该为你选择。

我需要帮助的是：

我需要找到一个很好的方法来确定“最佳”位置，也许是一些启发式或者可能通过某些第三方库。

这并不意味着确定最好的提供商！我可能会使用所有提供商并选择其中最好的。

应用程序的背景：

该应用程序将以固定间隔收集用户的位置（假设每10分钟左右）并将其发送到服务器。该应用程序应尽可能节省电池，并且位置应具有X（50-100？）米的精度。

目标是以后能够在地图上绘制白天用户的路径，因此我需要足够的准确性。

其他：

您认为对于期望和接受的准确度的合理价值是什么？我一直在使用100米接受，并且根据需要使用30米，这要问多少？我希望以后能够在地图上绘制用户的路径。期望100米，接受500米更好吗？

此外，现在我每次更新GPS的时间最长为60秒，如果你在室内的准确度可能超过200米，这个位置太短了吗？

这是我目前的代码，感谢任何反馈（除了缺少错误检查，这是TODO）：

protected void runTask() {
    final LocationManager locationManager = (LocationManager) context
            .getSystemService(Context.LOCATION_SERVICE);
    updateBestLocation(locationManager
            .getLastKnownLocation(LocationManager.GPS_PROVIDER));
    updateBestLocation(locationManager
            .getLastKnownLocation(LocationManager.NETWORK_PROVIDER));
    if (getLocationQuality(bestLocation) != LocationQuality.GOOD) {
        Looper.prepare();
        setLooper(Looper.myLooper());
        // Define a listener that responds to location updates
        LocationListener locationListener = new LocationListener() {

            public void onLocationChanged(Location location) {
                updateBestLocation(location);
                if (getLocationQuality(bestLocation) != LocationQuality.GOOD)
                    return;
                // We're done
                Looper l = getLooper();
                if (l != null) l.quit();
            }

            public void onProviderEnabled(String provider) {}

            public void onProviderDisabled(String provider) {}

            public void onStatusChanged(String provider, int status,
                    Bundle extras) {
                // TODO Auto-generated method stub
                Log.i("LocationCollector", "Fail");
                Looper l = getLooper();
                if (l != null) l.quit();
            }
        };
        // Register the listener with the Location Manager to receive
        // location updates
        locationManager.requestLocationUpdates(
                LocationManager.GPS_PROVIDER, 1000, 1, locationListener,
                Looper.myLooper());
        locationManager.requestLocationUpdates(
                LocationManager.NETWORK_PROVIDER, 1000, 1,
                locationListener, Looper.myLooper());
        Timer t = new Timer();
        t.schedule(new TimerTask() {

            @Override
            public void run() {
                Looper l = getLooper();
                if (l != null) l.quit();
                // Log.i("LocationCollector",
                // "Stopping collector due to timeout");
            }
        }, MAX_POLLING_TIME);
        Looper.loop();
        t.cancel();
        locationManager.removeUpdates(locationListener);
        setLooper(null);
    }
    if (getLocationQuality(bestLocation) != LocationQuality.BAD) 
        sendUpdate(locationToString(bestLocation));
    else Log.w("LocationCollector", "Failed to get a location");
}

private enum LocationQuality {
    BAD, ACCEPTED, GOOD;

    public String toString() {
        if (this == GOOD) return "Good";
        else if (this == ACCEPTED) return "Accepted";
        else return "Bad";
    }
}

private LocationQuality getLocationQuality(Location location) {
    if (location == null) return LocationQuality.BAD;
    if (!location.hasAccuracy()) return LocationQuality.BAD;
    long currentTime = System.currentTimeMillis();
    if (currentTime - location.getTime() < MAX_AGE
            && location.getAccuracy() <= GOOD_ACCURACY)
        return LocationQuality.GOOD;
    if (location.getAccuracy() <= ACCEPTED_ACCURACY)
        return LocationQuality.ACCEPTED;
    return LocationQuality.BAD;
}

private synchronized void updateBestLocation(Location location) {
    bestLocation = getBestLocation(location, bestLocation);
}

// Pretty much an unmodified version of googles example
protected Location getBestLocation(Location location,
        Location currentBestLocation) {
    if (currentBestLocation == null) {
        // A new location is always better than no location
        return location;
    }
    if (location == null) return currentBestLocation;
    // Check whether the new location fix is newer or older
    long timeDelta = location.getTime() - currentBestLocation.getTime();
    boolean isSignificantlyNewer = timeDelta > TWO_MINUTES;
    boolean isSignificantlyOlder = timeDelta < -TWO_MINUTES;
    boolean isNewer = timeDelta > 0;
    // If it's been more than two minutes since the current location, use
    // the new location
    // because the user has likely moved
    if (isSignificantlyNewer) {
        return location;
        // If the new location is more than two minutes older, it must be
        // worse
    } else if (isSignificantlyOlder) {
        return currentBestLocation;
    }
    // Check whether the new location fix is more or less accurate
    int accuracyDelta = (int) (location.getAccuracy() - currentBestLocation
            .getAccuracy());
    boolean isLessAccurate = accuracyDelta > 0;
    boolean isMoreAccurate = accuracyDelta < 0;
    boolean isSignificantlyLessAccurate = accuracyDelta > 200;
    // Check if the old and new location are from the same provider
    boolean isFromSameProvider = isSameProvider(location.getProvider(),
            currentBestLocation.getProvider());
    // Determine location quality using a combination of timeliness and
    // accuracy
    if (isMoreAccurate) {
        return location;
    } else if (isNewer && !isLessAccurate) {
        return location;
    } else if (isNewer && !isSignificantlyLessAccurate
            && isFromSameProvider) {
        return location;
    }
    return bestLocation;
}

/** Checks whether two providers are the same */
private boolean isSameProvider(String provider1, String provider2) {
    if (provider1 == null) {
        return provider2 == null;
    }
    return provider1.equals(provider2);
}

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获取用户在Android中的位置的好方法

问题：

为什么问题是一个问题：

我需要帮助的是：

应用程序的背景：

其他：

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