我可以使用 fusedlocationproviderapi 并强制仅使用 GPS 吗?
Can I use fusedlocationproviderapi and force using GPS only?
我在 Android 应用程序中使用 fusedlocationproviderapi 来传输用户位置。这工作正常,除了有时位置会跳到镇上的某个位置,然后显示相同的位置。我相信该应用程序会从 GPS 切换到手机信号塔试用,而在我们地区这并不适用。(MN) 我使用
<uses-permission android:name="android.permission.ACCESS_FINE_LOCATION" />
在清单文件中,我会很好地忽略错误位置或检测这些错误位置而不传输它们。
mLocationRequest = LocationRequest.create();
mLocationRequest.setInterval(UpdateSeconds * 1000);
mLocationRequest.setFastestInterval(UpdateSeconds * 1000);
mLocationRequest.setPriority(LocationRequest.PRIORITY_HIGH_ACCURACY);
我使用上面的代码,当我获得 GPS 位置时效果很好。
如果您只对来自 GPS 的数据感兴趣,可以使用 LocationManager with GPS_PROVIDER。如果您只对 GPS 感兴趣,我认为使用 FusedLocationProviderApi 没有任何好处。
如果新位置比当前最佳位置 "better",您可以决定每次位置更新。这样可以避免超出您要求的准确度:
protected boolean isBetterLocation(Location location,
Location currentBestLocation) {
final int TWO_MINUTES = 1000 * 60 * 2;
if (currentBestLocation == null) {
// A new location is always better than no location
return true;
}
// 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 true;
// If the new location is more than two minutes older, it must be
// worse
} else if (isSignificantlyOlder) {
return false;
}
// 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;
// Determine location quality using a combination of timeliness and
// accuracy
if (isMoreAccurate) {
return true;
} else if (isNewer && !isLessAccurate) {
return true;
} else if (isNewer && !isSignificantlyLessAccurate) {
return true;
}
return false;
}
我在 Android 应用程序中使用 fusedlocationproviderapi 来传输用户位置。这工作正常,除了有时位置会跳到镇上的某个位置,然后显示相同的位置。我相信该应用程序会从 GPS 切换到手机信号塔试用,而在我们地区这并不适用。(MN) 我使用
<uses-permission android:name="android.permission.ACCESS_FINE_LOCATION" />
在清单文件中,我会很好地忽略错误位置或检测这些错误位置而不传输它们。
mLocationRequest = LocationRequest.create();
mLocationRequest.setInterval(UpdateSeconds * 1000);
mLocationRequest.setFastestInterval(UpdateSeconds * 1000);
mLocationRequest.setPriority(LocationRequest.PRIORITY_HIGH_ACCURACY);
我使用上面的代码,当我获得 GPS 位置时效果很好。
如果您只对来自 GPS 的数据感兴趣,可以使用 LocationManager with GPS_PROVIDER。如果您只对 GPS 感兴趣,我认为使用 FusedLocationProviderApi 没有任何好处。
如果新位置比当前最佳位置 "better",您可以决定每次位置更新。这样可以避免超出您要求的准确度:
protected boolean isBetterLocation(Location location,
Location currentBestLocation) {
final int TWO_MINUTES = 1000 * 60 * 2;
if (currentBestLocation == null) {
// A new location is always better than no location
return true;
}
// 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 true;
// If the new location is more than two minutes older, it must be
// worse
} else if (isSignificantlyOlder) {
return false;
}
// 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;
// Determine location quality using a combination of timeliness and
// accuracy
if (isMoreAccurate) {
return true;
} else if (isNewer && !isLessAccurate) {
return true;
} else if (isNewer && !isSignificantlyLessAccurate) {
return true;
}
return false;
}