如何在 SharedPreferences 中存储和检索 Location 对象? [英] How to store and retrieve Location object in SharedPreferences?

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.

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