我想在后台线程上保存对 Realm 的改造响应,然后将其传递给 UI 线程,但这有点棘手,因为 Realm 对线程非常敏感。所以代码看起来像这样,请将您的编辑提交给所有更好的解决方案:)
restApi.userRealmList()
.doOnNext(userRealmModels -> {
if (userRealmModels != null){
mRealm = Realm.getInstance(mContext);
mRealm.asObservable()
.map(realm -> mRealm.copyToRealmOrUpdate(userEntity))
.subscribe(new Subscriber<Object>() {
@Override
public void onCompleted() {
}
@Override
public void onError(Throwable e) {
e.printStackTrace();
}
@Override
public void onNext(Object o) {
Log.d("RealmManager", "user added!");
}
});
}})
.map(userEntityDataMapper::transformAll)
.subscribeOn(Schedulers.io())
.observeOn(AndroidSchedulers.mainThread())
.subscribe(new Subscriber<List<User>>() {
@Override
public void onCompleted() {
hideViewLoading();
}
@Override
public void onError(Throwable e) {
hideViewLoading();
showErrorMessage(new DefaultErrorBundle((Exception) e));
showViewRetry();
}
@Override
public void onNext(List<User> users) {
showUsersCollectionInView(users);
}
});
最佳答案
您的代码看起来无法编译?例如。什么是userEntity
。另外,您的 copyToRealmOrUpdate 不在事务内,因此也会崩溃,但它与线程无关。
如果您想在将某些数据发送到 UI 之前将其保存为副作用,您应该能够执行以下操作:
restApi.userRealmList()
.doOnNext(userRealmModels -> {
if (userRealmModels != null) {
Realm realm = Realm.getInstance(mContext);
realm.beginTransaction();
realm.copyToRealmOrUpdate(userRealmModels);
realm.commitTransaction();
realm.close();
}})
.map(userEntityDataMapper::transformAll)
.subscribeOn(Schedulers.io())
.observeOn(AndroidSchedulers.mainThread())
.subscribe(new Subscriber<List<User>>() {
@Override
public void onCompleted() {
hideViewLoading();
}
@Override
public void onError(Throwable e) {
hideViewLoading();
showErrorMessage(new DefaultErrorBundle((Exception) e));
showViewRetry();
}
@Override
public void onNext(List<User> users) {
showUsersCollectionInView(users);
}
});
关于multithreading - 如何将 Retrofit 2 与 Realm 和 RxJava 结合起来,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/35897737/