问题描述
我想存储一个 Location 对象并且我正在尝试选择一种好的方法来做到这一点.请给我建议如何做到这一点
I would like to store a Location object and i'm trying to pick a good way to do it.Please give me advice that how to do this
我正在使用此代码,但是当我从首选项中获取位置时,位置会像这样返回...
I'm using this code but when i get the location from Preferences then Location return like this...
位置[mProvider=STORAGE,mTime=0,mLatitude=30.0,mLongitude=76.0,mHasAltitude=false,mAltitude=0.0,mHasSpeed=false,mSpeed=0.0,mHasBearing=false,mBearing=0.0,mHasAccuracy=false,mAccuracy=0.0,mExtras=null]
Location[mProvider=STORAGE,mTime=0,mLatitude=30.0,mLongitude=76.0,mHasAltitude=false,mAltitude=0.0,mHasSpeed=false,mSpeed=0.0,mHasBearing=false,mBearing=0.0,mHasAccuracy=false,mAccuracy=0.0,mExtras=null]
/** Store Location object in SharedPreferences */
public void storeLocation(Context context, Location location) {
SharedPreferences settings;
Editor editor;
try {
JSONObject locationJson = new JSONObject();
locationJson.put(LATITUDE, location.getLatitude());
locationJson.put(LONGITUDE, location.getLongitude());
settings = context.getSharedPreferences(PREFS_NAME,Context.MODE_PRIVATE);
editor = settings.edit();
editor.putString(KEY_LOCATION, locationJson.toString());
editor.commit();
Log.i("location_util_store", "Location" + location);
} catch (Exception e) {
e.printStackTrace();
}
}
/** Retrieve Location object from SharedPreferences
* @return */
public Location getPrevLocation(Context context) {
SharedPreferences settings;
try {
settings = context.getSharedPreferences(PREFS_NAME,Context.MODE_PRIVATE);
String jsonLocation = settings.getString(KEY_LOCATION, null);
if (jsonLocation != null) {
JSONObject locationJson = new JSONObject(jsonLocation);
Location location = new Location("STORAGE");
location.setLatitude(locationJson.getInt(LATITUDE));
location.setLongitude(locationJson.getInt(LONGITUDE));
Log.i("location_util_get", "Location" + location);
return location;
}
} catch (Exception e) {
e.printStackTrace();
}
return null;
}
推荐答案
最佳解决方案:提取位置纬度 &长参数作为字符串并将它们保存在首选项中.您还可以根据您的用例从位置提取其他信息.
Best Solution: Extract location lat & long parameters as string and save them in prefs. You can extract other information as well from location depending upon your usecase.
保存位置:-
if (location == null) {
sharedPreferences.edit().removeKey("LOCATION_LAT").apply();
sharedPreferences.edit().removeKey("LOCATION_LON").apply();
sharedPreferences.edit().removeKey("LOCATION_PROVIDER").apply();
} else {
sharedPreferences.edit().putString("LOCATION_LAT", String.valueOf(location.getLatitude())).apply();
sharedPreferences.edit().putString("LOCATION_LON", String.valueOf(location.getLongitude())).apply();
sharedPreferences.edit().putString("LOCATION_PROVIDER", location.getProvider()).apply();
}
检索位置:-
String lat = sharedPreferences.getString("LOCATION_LAT", null);
String lon = sharedPreferences.getString("LOCATION_LON", null);
Location location = null;
if (lat != null && lon != null) {
String provider = sharedPreferences.getString("LOCATION_PROVIDER", null);
location = new Location(provider);
location.setLatitude(Double.parseDouble(lat));
location.setLongitude(Double.parseDouble(lon));
}
return location;
糟糕的解决方案:改用 Gson 来保存您的位置:
Bad Solution: Use Gson to persist your Location instead:
保存位置:
String json = location == null ? null : new Gson().toJson(location);
sharedPreferences.edit().putString("Location", json).apply();
检索位置:
String json = sharedPreferences.getString("location", null);
return json == null ? null : new Gson().fromJson(json, Location.class);
如此答案中所述.
这不应该崩溃,但它会在某些设备上崩溃.如此处所述.
This shouldn't crash but it crashes on some devices. As mentioned here.
这篇关于如何在 SharedPreferences 中存储和检索 Location 对象?的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!