Android上的LocationManager API似乎对一个只需要偶尔粗略估计用户位置的应用程序来说有点麻烦。

我正在开发的应用程序本身并不是一个定位应用程序,但它确实需要获取用户的位置,以便显示附近企业的列表。它不需要担心用户是否在移动或类似的事情。

以下是我想做的:

向用户显示附近位置的列表。预加载用户的位置,以便在“活动X”中需要它时,它将可用。我并不特别关心更新的准确性或频率。只要不太远,只要抓住一个位置就足够了。也许如果我想变得更漂亮,我会每隔几分钟左右更新一次位置,但这不是一个很大的优先事项。适用于任何具有GPS或网络位置提供商的设备。

这似乎并不难,但在我看来,我必须组建两个不同的位置提供商(GPS和NETWORK),并管理每个提供商的生命周期。不仅如此,我还必须在多个活动中复制相同的代码以满足#2。过去,我曾尝试使用getBestProvider()将解决方案简化为仅使用一个位置提供程序,但这似乎只提供了最好的“理论”提供程序,而不是实际会为您提供最佳结果的提供程序。

有没有更简单的方法来实现这一点?


当前回答

使用以下代码,它将提供可用的最佳提供商:

String locCtx = Context.LOCATION_SERVICE; 

LocationManager locationMgr = (LocationManager) ctx.getSystemService(locCtx);

Criteria criteria  = new Criteria();
criteria.setAccuracy(Criteria.ACCURACY_FINE);
criteria.setAltitudeRequired(false);
criteria.setBearingRequired(false);
criteria.setCostAllowed(true);
criteria.setPowerRequirement(Criteria.POWER_LOW);

String provider = locationMgr.getBestProvider(criteria, true);

System.out.println("Best Available provider::::"+provider);

其他回答

使用Fedor的解决方案,我经历了多次执行回调gotLocation。当gotLocation方法“足够长”时,这似乎是由于重写的LocationListener.onLocationChanged方法中的竞争条件所致。我不确定,但我猜removeUpdates会阻止Looper队列中新消息的入队,但它不会删除那些已经入队但尚未使用的消息。这就是比赛条件。

为了减少这种错误行为的可能性,可以在激发onLocationChanged事件之前调用removeUpdates,但我们仍然有竞争条件。

我找到的最佳解决方案是用requestSingleUpdate替换requestLocationUpdates。

这是我的版本,基于Fedor的解决方案,使用Handler向looper线程发送消息:

public class LocationResolver {
    private Timer timer;
    private LocationManager locationManager;
    private LocationResult locationResult;
    private boolean gpsEnabled = false;
    private boolean networkEnabled = false;
    private Handler locationTimeoutHandler;

    private final Callback locationTimeoutCallback = new Callback() {
        public boolean handleMessage(Message msg) {
            locationTimeoutFunc();
            return true;
        }

        private void locationTimeoutFunc() {   
            locationManager.removeUpdates(locationListenerGps);
            locationManager.removeUpdates(locationListenerNetwork);

            Location networkLocation = null, gpsLocation = null;
            if (gpsEnabled)
                gpsLocation = locationManager.getLastKnownLocation(LocationManager.GPS_PROVIDER);
            if (networkEnabled)
                networkLocation = locationManager.getLastKnownLocation(LocationManager.NETWORK_PROVIDER);

            // if there are both values use the latest one
            if (gpsLocation != null && networkLocation != null) {
                if (gpsLocation.getTime() > networkLocation.getTime())
                    locationResult.gotLocation(gpsLocation);
                else
                    locationResult.gotLocation(networkLocation);
                return;
            }

            if (gpsLocation != null) {
                locationResult.gotLocation(gpsLocation);
                return;
            }
            if (networkLocation != null) {
                locationResult.gotLocation(networkLocation);
                return;
            }
            locationResult.gotLocation(null);           
        }
    };
    private final LocationListener locationListenerGps = new LocationListener() {
        public void onLocationChanged(Location location) {              
            timer.cancel();
            locationResult.gotLocation(location);
            locationManager.removeUpdates(this);
            locationManager.removeUpdates(locationListenerNetwork);
        }

        public void onProviderDisabled(String provider) {
        }

        public void onProviderEnabled(String provider) {
        }

        public void onStatusChanged(String provider, int status, Bundle extras) {
        }
    };
    private final LocationListener locationListenerNetwork = new LocationListener() {
        public void onLocationChanged(Location location) {    
            timer.cancel(); 
            locationResult.gotLocation(location);
            locationManager.removeUpdates(this);
            locationManager.removeUpdates(locationListenerGps);
        }

        public void onProviderDisabled(String provider) {
        }

        public void onProviderEnabled(String provider) {
        }

        public void onStatusChanged(String provider, int status, Bundle extras) {
        }
    };

    public void prepare() {
        locationTimeoutHandler = new Handler(locationTimeoutCallback);
    }

    public synchronized boolean getLocation(Context context, LocationResult result, int maxMillisToWait) {
        locationResult = result;
        if (locationManager == null)
            locationManager = (LocationManager) context.getSystemService(Context.LOCATION_SERVICE);

        // exceptions will be thrown if provider is not permitted.
        try {
            gpsEnabled = locationManager.isProviderEnabled(LocationManager.GPS_PROVIDER);
        } catch (Exception ex) {
        }
        try {
            networkEnabled = locationManager.isProviderEnabled(LocationManager.NETWORK_PROVIDER);
        } catch (Exception ex) {
        }

        // don't start listeners if no provider is enabled
        if (!gpsEnabled && !networkEnabled)
            return false;

        if (gpsEnabled)
            locationManager.requestSingleUpdate(LocationManager.GPS_PROVIDER, locationListenerGps, Looper.myLooper());
            //locationManager.requestLocationUpdates(LocationManager.GPS_PROVIDER, 0, 0, locationListenerGps);
        if (networkEnabled)
            locationManager.requestSingleUpdate(LocationManager.NETWORK_PROVIDER, locationListenerNetwork, Looper.myLooper());
            //locationManager.requestLocationUpdates(LocationManager.NETWORK_PROVIDER, 0, 0, locationListenerNetwork);

        timer = new Timer();
        timer.schedule(new GetLastLocationTask(), maxMillisToWait);
        return true;
    }

    private class GetLastLocationTask extends TimerTask {
        @Override
        public void run() { 
            locationTimeoutHandler.sendEmptyMessage(0);
        }
    }

    public static abstract class LocationResult {
        public abstract void gotLocation(Location location);
    }
}

我从定制的looper线程中使用这个类,如下所示:

public class LocationGetter {
    private final Context context;
    private Location location = null;
    private final Object gotLocationLock = new Object();
    private final LocationResult locationResult = new LocationResult() {            
        @Override
        public void gotLocation(Location location) {
            synchronized (gotLocationLock) {
                LocationGetter.this.location = location;
                gotLocationLock.notifyAll();
                Looper.myLooper().quit();
            }
        }
    };

    public LocationGetter(Context context) {
        if (context == null)
            throw new IllegalArgumentException("context == null");

        this.context = context;
    }

    public synchronized Coordinates getLocation(int maxWaitingTime, int updateTimeout) {
        try {
            final int updateTimeoutPar = updateTimeout;
            synchronized (gotLocationLock) {            
                new Thread() {
                    public void run() {
                        Looper.prepare();
                        LocationResolver locationResolver = new LocationResolver();
                        locationResolver.prepare();
                        locationResolver.getLocation(context, locationResult, updateTimeoutPar);
                        Looper.loop();
                    }
                }.start();

                gotLocationLock.wait(maxWaitingTime);
            }
        } catch (InterruptedException e1) {
            e1.printStackTrace();
        }

        if (location != null)
            coordinates = new Coordinates(location.getLatitude(), location.getLongitude());
        else
            coordinates = Coordinates.UNDEFINED;
        return coordinates; 
    }
}

其中Coordinates是一个简单的类,具有两个财产:纬度和经度。

你可以一直使用LocationManager.getLastKnownLocation(),但就像它说的那样,它可能已经过时了。

获取一般位置的一个简单方法是注册网络(通常很快)。

LocationManager locationManager = (LocationManager) this.getSystemService(Context.LOCATION_SERVICE);
locationManager.requestLocationUpdates(
     LocationManager.NETWORK_PROVIDER, 1000, 1000, this);

然后做

locationManager.removeUpdates(this);

在侦听器的onLocationChanged()方法中。

public static Location getBestLocation(Context ctxt) {
    Location gpslocation = getLocationByProvider(
        LocationManager.GPS_PROVIDER, ctxt);
    Location networkLocation = getLocationByProvider(
        LocationManager.NETWORK_PROVIDER, ctxt);
    Location fetchedlocation = null;
    // if we have only one location available, the choice is easy
    if (gpslocation != null) {
        Log.i("New Location Receiver", "GPS Location available.");
        fetchedlocation = gpslocation;
    } else {
        Log.i("New Location Receiver",
            "No GPS Location available. Fetching Network location lat="
                + networkLocation.getLatitude() + " lon ="
                + networkLocation.getLongitude());
        fetchedlocation = networkLocation;
    }
    return fetchedlocation;
}

/**
 * get the last known location from a specific provider (network/gps)
 */
private static Location getLocationByProvider(String provider, Context ctxt) {
    Location location = null;
    // if (!isProviderSupported(provider)) {
    // return null;
    // }
    LocationManager locationManager = (LocationManager) ctxt
            .getSystemService(Context.LOCATION_SERVICE);
    try {
        if (locationManager.isProviderEnabled(provider)) {
            location = locationManager.getLastKnownLocation(provider);
        }
    } catch (IllegalArgumentException e) {
        Log.i("New Location Receiver", "Cannot access Provider " + provider);
    }
    return location;
}

这里有点晚了,但在这种情况下,我会使用谷歌地图API,并使用谷歌地图的纬度和经度API标记附近的位置。另外,如果你能在地图上显示他/她的位置,用户体验会更好。无需担心用户位置的更新或使用android api搜索。让谷歌地图为您处理内部内容。

@emmby可能已经在他的应用程序中解决了这个问题,但为了将来的参考,我向其他开发人员推荐的是查看谷歌地图API中的位置特定内容。

编辑:在谷歌地图中显示用户位置的链接

实际上,我们可以使用两个提供商(GPS和网络)。他们只是分享一个公共听众:

locationManager.requestLocationUpdates(LocationManager.NETWORK_PROVIDER, 10 * 1000, (float) 10.0, listener);
locationManager.requestLocationUpdates(LocationManager.GPS_PROVIDER, 90 * 1000, (float) 10.0, listener);

这是必要的,因为总是需要及时调用OnLocationChanged()方法。