android - rxjava merge 傳回Object物件資料如何快取

使用 rxjava 的 merge 方法將兩個 api 傳回的資料物件結合得到 Object，然後我想使用 SharedPreferences 方法快取Object。

我嘗試按照網路上的方法保存Objcet，卻沒有效果。請大夥幫忙看看。

Bean：

public class LifeSuggestionResult implements Serializable{ ... }
public class WeatherFuture implements Serializable { ... }

ObjectUtil( 使用 SharedPreferences 儲存並取得 Object)：

public class ObjectUtil {
    public static void setObject(String key, Object object, Context context) {

        SharedPreferences.Editor editor = PreferenceManager.getDefaultSharedPreferences(context).edit();

        ByteArrayOutputStream baos = new ByteArrayOutputStream();
        ObjectOutputStream out = null;
        try {

            out = new ObjectOutputStream(baos);
            out.writeObject(object);
            String objectVal = new String(Base64.encode(baos.toByteArray(), Base64.DEFAULT));
            editor.putString(key, objectVal);
            editor.commit();

        } catch (IOException e) {
            e.printStackTrace();
        } finally {
            try {
                if (baos != null) {
                    baos.close();
                }
                if (out != null) {
                    out.close();
                }
            } catch (IOException e) {
                e.printStackTrace();
            }
        }
    }

    public static <T> T getObject(String key, Context context) {
        SharedPreferences sp = PreferenceManager.getDefaultSharedPreferences(context);
        if (sp.contains(key)) {
            String objectVal = sp.getString(key, null);
            byte[] buffer = Base64.decode(objectVal, Base64.DEFAULT);
            ByteArrayInputStream bais = new ByteArrayInputStream(buffer);
            ObjectInputStream ois = null;
            try {
                ois = new ObjectInputStream(bais);
                T t = (T) ois.readObject();
                return t;
            } catch (StreamCorruptedException e) {
                e.printStackTrace();
            } catch (IOException e) {
                e.printStackTrace();
            } catch (ClassNotFoundException e) {
                e.printStackTrace();
            } finally {
                try {
                    if (bais != null) {
                        bais.close();
                    }
                    if (ois != null) {
                        ois.close();
                    }
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
        }
        return null;
    }
}

rxjava部分：

public void getCurrentWeather(final String city) {
        Observable<WeatherFuture> weatherFutureObservable = new WeatherService().getFutureWeather(city, "zh-Hans", "c");
        Observable<LifeSuggestionResult> lifeSuggestionResultObservable = new WeatherService().getAirQuality(city, "zh-Hans", "city");
        Observable.merge(weatherFutureObservable, lifeSuggestionResultObservable)
                .subscribeOn(Schedulers.io())
                .observeOn(AndroidSchedulers.mainThread())
                .subscribe(new Subscriber<Object>() {
                    @Override
                    public void onCompleted() {
                                       
                    }

                    @Override
                    public void onNext(Object o) {
                        setWeatherInfo(o);
                    }

                    @Override
                    public void onError(Throwable e) {

                    }
                });


    }

    public void setWeatherInfo(Object o) {
 
        if (o instanceof WeatherFuture) {
            ObjectUtil.setObject("WeatherFuture", o, MainActivity.this);
            ...
  }

        } else if (o instanceof LifeSuggestionResult) {
            ObjectUtil.setObject("LifeSuggestion", o, MainActivity.this);
            ...
    }

世界只因有你2756 天前665

android - rxjava merge 傳回Object物件資料如何快取

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