android - rxjava merge 返回Object对象数据如何缓存

问题描述

使用 rxjava 的 merge 方法将两个 api 返回的数据对象结合得到 Object，然后我想使用 SharedPreferences 方法缓存Object。

我尝试按照网上的方法保存Objcet，却没有效果。请大伙帮忙看看。

Bean：

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

ObjectUtil( 使用 SharedPreferences 保存和获取 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); ... }

问题解答

给你一个思路，使用排除法1、在其他地方单独使用ObjectUtil，看是否可以存储和取出一个假数据Object2、在onNext中打印出Object的内容，看是否是预期的Object3、假如两者都没问题，那就看你的setWeatherInfo方法是否有问题