Android GPS 查询位置数据不正确

我没有为此使用模拟位置...事实上，代码上周运行良好。

我有一个应用程序，它收集 GPS 数据并使用应用程序本身生成的 X、Y 坐标输出谷歌地图链接。我不是 100% 确定为什么它没有按应有的方式工作，但是当我请求应用程序根据手机提供的 GPS 位置创建谷歌地图链接时，它告诉我我距离我的点有 5 - 6 个街区原点（我当时实际所在的位置）不完全是我想要的

Knowns:

• 我已设置适当的权限
• 上周所有代码都运行良好

以下是收集 GPS 信息的代码：

        Toast.makeText(context, "Gathering GPS Data...", Toast.LENGTH_SHORT).show();
    gps = new gpsTracker(Settings.this);
    if(gps.canGetLocation()){
        try{
            gps.getLocation();
            lon = gps.getLongitude();
            lat = gps.getLatitude();
            Toast.makeText(getApplicationContext(), "Your location is - \nlat: " + lat + "\nlon: " + lon, Toast.LENGTH_LONG).show();
        }
        catch(Exception  e){}
    }
    else{
        gps.showSettingsAlert();
    }

上述所有类均来自：

import android.app.AlertDialog;
import android.app.Service;
import android.content.Context;
import android.content.DialogInterface;
import android.content.Intent;
import android.location.Location;
import android.location.LocationListener;
import android.location.LocationManager;
import android.os.Bundle;
import android.os.IBinder;
import android.provider.Settings;
import android.util.Log;

public class gpsTracker extends Service implements LocationListener {

private final Context mContext;

// flag for GPS status
boolean isGPSEnabled = false;

// flag for network status
boolean isNetworkEnabled = false;

// flag for GPS status
boolean canGetLocation = false;

Location location; // location
double latitude; // latitude
double longitude; // longitude

// The minimum distance to change Updates in meters
private static final long MIN_DISTANCE_CHANGE_FOR_UPDATES = 10; // 10 meters

// The minimum time between updates in milliseconds
private static final long MIN_TIME_BW_UPDATES = 1000 * 60 * 1; // 1 minute

// Declaring a Location Manager
protected LocationManager locationManager;

public gpsTracker(Context context) {
    this.mContext = context;
    getLocation();
}

public Location getLocation() {
    try {
        locationManager = (LocationManager) mContext
                .getSystemService(LOCATION_SERVICE);

        // getting GPS status
        isGPSEnabled = locationManager
                .isProviderEnabled(LocationManager.GPS_PROVIDER);

        // getting network status
        isNetworkEnabled = locationManager
                .isProviderEnabled(LocationManager.NETWORK_PROVIDER);

        if (!isGPSEnabled && !isNetworkEnabled) {
            // no network provider is enabled
        } else {
            this.canGetLocation = true;
            // First get location from Network Provider
            if (isNetworkEnabled) {
                locationManager.requestLocationUpdates(
                        LocationManager.NETWORK_PROVIDER,
                        MIN_TIME_BW_UPDATES,
                        MIN_DISTANCE_CHANGE_FOR_UPDATES, this);
                Log.d("Network", "Network");
                if (locationManager != null) {
                    location = locationManager
                            .getLastKnownLocation(LocationManager.NETWORK_PROVIDER);
                    if (location != null) {
                        latitude = location.getLatitude();
                        longitude = location.getLongitude();
                    }
                }
            }
            // if GPS Enabled get lat/long using GPS Services
            if (isGPSEnabled) {
                if (location == null) {
                    locationManager.requestLocationUpdates(
                            LocationManager.GPS_PROVIDER,
                            MIN_TIME_BW_UPDATES,
                            MIN_DISTANCE_CHANGE_FOR_UPDATES, this);
                    Log.d("GPS Enabled", "GPS Enabled");
                    if (locationManager != null) {
                        location = locationManager
                                .getLastKnownLocation(LocationManager.GPS_PROVIDER);
                        if (location != null) {
                            latitude = location.getLatitude();
                            longitude = location.getLongitude();
                        }
                    }
                }
            }
        }

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

    return location;
}

/**
 * Stop using GPS listener
 * Calling this function will stop using GPS in your app
 * */
public void stopUsingGPS(){
    if(locationManager != null){
        locationManager.removeUpdates(gpsTracker.this);
    }
}

/**
 * Function to get latitude
 * */
public double getLatitude(){
    if(location != null){
        latitude = location.getLatitude();
    }

    // return latitude
    return latitude;
}

/**
 * Function to get longitude
 * */
public double getLongitude(){
    if(location != null){
        longitude = location.getLongitude();
    }

    // return longitude
    return longitude;
}

/**
 * Function to check GPS/wifi enabled
 * @return boolean
 * */
public boolean canGetLocation() {
    return this.canGetLocation;
}

/**
 * Function to show settings alert dialog
 * On pressing Settings button will lauch Settings Options
 * */
public void showSettingsAlert(){
    AlertDialog.Builder alertDialog = new AlertDialog.Builder(mContext);

    // Setting Dialog Title
    alertDialog.setTitle("GPS is settings");

    // Setting Dialog Message
    alertDialog.setMessage("GPS is not enabled. Do you want to go to settings menu?");

    // On pressing Settings button
    alertDialog.setPositiveButton("Settings", new DialogInterface.OnClickListener() {
        public void onClick(DialogInterface dialog,int which) {
            Intent intent = new Intent(Settings.ACTION_LOCATION_SOURCE_SETTINGS);
            mContext.startActivity(intent);
        }
    });

    // on pressing cancel button
    alertDialog.setNegativeButton("Cancel", new DialogInterface.OnClickListener() {
        public void onClick(DialogInterface dialog, int which) {
            dialog.cancel();
        }
    });

    // Showing Alert Message
    alertDialog.show();
}

@Override
public void onLocationChanged(Location location) {
}

@Override
public void onProviderDisabled(String provider) {
}

@Override
public void onProviderEnabled(String provider) {
}

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

@Override
public IBinder onBind(Intent arg0) {
    return null;
}

根据指定的内容[4-5块的差异]，您可能会获得location from networkProvider only.

有了这个gpsTracker问题中提到的代码，
需要进行一些修改，而不是按原样使用代码：

1.有2个if验证来源的循环location是否启用且否 'else':

 if (isNetworkEnabled) {
                locationManager.requestLocationUpdates(
                        LocationManager.NETWORK_PROVIDER,
                        MIN_TIME_BW_UPDATES,
                        MIN_DISTANCE_CHANGE_FOR_UPDATES, this);
                Log.d("Network", "Network");
                if (locationManager != null) {
                    location = locationManager
                            .getLastKnownLocation(LocationManager.NETWORK_PROVIDER);
                    if (location != null) {
                        latitude = location.getLatitude();
                        longitude = location.getLongitude();
                    }
                }
            }
            // if GPS Enabled get lat/long using GPS Services
            if (isGPSEnabled) {
                if (location == null) {
                    locationManager.requestLocationUpdates(
                            LocationManager.GPS_PROVIDER,
                            MIN_TIME_BW_UPDATES,
                            MIN_DISTANCE_CHANGE_FOR_UPDATES, this);
                    Log.d("GPS Enabled", "GPS Enabled");
                    if (locationManager != null) {
                        location = locationManager
                                .getLastKnownLocation(LocationManager.GPS_PROVIDER);
                        if (location != null) {
                            latitude = location.getLatitude();
                            longitude = location.getLongitude();
                        }
                    }
                }
            }  

这意味着当您可以获得该应用程序时，该应用程序将完成两倍的工作location来自两个来源。此外，来源location获得的总是含糊不清的。

当您只需要近似值时，这很好location主要不应该为空。

如果你只想使用这个类来获取location，尝试构建if-else根据要求并确保不会重复location获得。
Eg. if GPS具有更高的偏好，适用于您的情况，请改变if上面并把网络状况加上else like:

if (isGPSEnabled) {
                if (location == null) {
                    locationManager.requestLocationUpdates(
                            LocationManager.GPS_PROVIDER,
                            MIN_TIME_BW_UPDATES,
                            MIN_DISTANCE_CHANGE_FOR_UPDATES, this);
                    Log.d("GPS Enabled", "GPS Enabled");
                    if (locationManager != null) {
                        location = locationManager
                                .getLastKnownLocation(LocationManager.GPS_PROVIDER);
                        if (location != null) {
                            latitude = location.getLatitude();
                            longitude = location.getLongitude();
                        }
                    }
                }
            } else if (isNetworkEnabled) {
            locationManager.requestLocationUpdates(
                    LocationManager.NETWORK_PROVIDER,
                    MIN_TIME_BW_UPDATES,
                    MIN_DISTANCE_CHANGE_FOR_UPDATES, this);
            Log.d("Network", "Network");
            if (locationManager != null) {
                location = locationManager
                        .getLastKnownLocation(LocationManager.NETWORK_PROVIDER);
                if (location != null) {
                    latitude = location.getLatitude();
                    longitude = location.getLongitude();
                }
            }
        }

根据您的要求，我建议删除网络提供商部分并获取location仅来自 GPS，尤其是在视线不成问题的情况下。

当您的代码工作正常时，它必须从 GPS 获取位置并将其设置在对象中。但因为这两个“if“ 和不 ”else“，你永远不会知道是否location获得的是从网络或GPS - 你可以检查位置.getProvider() http://developer.android.com/reference/android/location/Location.html#getProvider()的条件内canGetLocation()

2.此外，您还可以记录消息或提示针对某个特定来源的某些操作...例如：
在这一部分中：

 if (!isGPSEnabled && !isNetworkEnabled) {
            // no network provider is enabled
        }    

只需将其分成两个不同的if(s) 和相应的代码。截至目前，这里什么也没有发生，所以除非您从外部检查，否则您不知道两者是否都被禁用。

建议：尝试一下使用 GooglePlayServices 的 LocationClient http://developer.android.com/reference/com/google/android/gms/location/LocationClient.html for 检索当前位置 http://developer.android.com/training/location/retrieve-current.html。我发现它更可靠、更有用。检查这个融合位置提供商示例 http://www.kpbird.com/2013/06/fused-location-provider-example.html， 环境位置请求对象 http://developer.android.com/reference/com/google/android/gms/location/LocationRequest.html根据你的要求是关键。

另一个更新：刚刚发现：关于堆栈溢出的有用问题 - 获取用户位置的好方法 https://stackoverflow.com/questions/6181704/good-way-of-getting-the-users-location-in-android?rq=1

为查找此问题/答案的任何人提供更新
关于LocationClient的建议；
下不再找到 LocationClientcom.google.android.gms.location https://developer.android.com/reference/com/google/android/gms/location/package-summary.html， 参考：
Android play services 6.5：LocationClient 丢失 https://stackoverflow.com/questions/27372638/android-play-services-6-5-locationclient-is-missing