Question

我想存储一个Location对象，并试图选择一个好方法来做到这一点。我只有一个小对象，我需要它是私有的，所以到目前为止，SharedPreferences或内部存储对我来说是最有意义的。

我看到一个对象可以是written to a bytearray, stored as a String in SharedPreferences, then written back to an object as described here。

这是一种合理的方法吗？还有更好的方法吗？

感谢您的任何建议。

Answer 1

在保存简单数据时，我更喜欢使用JSON对象，因为代码对于将来的维护者来说比操作字节数组更简单，更容易理解。由于对象很小，因此使用字符串而不是字节数组的大小代价并不重要。

private static final String LATITUDE = "com.somepackage.name.LATITUDE";
private static final String LONGITUDE = "com.somepackage.name.LONGITUDE";

/**
 * Save a location/key pair.
 * 
 * @param key the key associated with the location
 * @param location the location for the key
 * @return true if saved successfully false otherwise
 */
public boolean saveLocation(String key, Location location) {
    LOG.info("Saving location");
    try {
        JSONObject locationJson = new JSONObject();

        locationJson.put(LATITUDE, location.getLatitude());
        locationJson.put(LONGITUDE, location.getLongitude());
        //other location data
        SharedPreferences.Editor edit = preferences.edit();
        edit.putString(key, locationJson.toString());
        edit.commit();
    } catch (JSONException e) {
        LOG.error("JSON Exception", e);
        return false;
    }

    LOG.info("Location {}  saved successfully at key: {}", preferences.getString(key, null),key);
    return true;
}

/**
 * Gets location data for a key.
 * 
 * @param key the key for the saved location
 * @return a {@link Location} object or null if there is no entry for the key
 */
public Location getLocation(String key) {
    LOG.info("Retrieving location at key {} ", key);
    try {
        String json = preferences.getString(key, null);

        if (json != null) {
            JSONObject locationJson = new JSONObject(json);
            Location location = new Location(STORAGE);
            location.setLatitude(locationJson.getInt(LATITUDE));
            location.setLongitude(locationJson.getInt(LONGITUDE));
            LOG.info("Returning location: {}" , location);
            return location;
        }
    } catch (JSONException e) {
        LOG.error("JSON Exception", e);
    }

    LOG.warn("No location found at key {}",  key);
    //or throw exception depending on your logic
    return null;
}

存储Location对象的好方法是什么？

1 个答案: