我想将对领域的改造响应保存在后台线程中,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。因此代码看起来像这样,请对所有更好的解决方案提交修改:)
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);
}
});
您的代码看起来不像可以编译吗?例如什么userEntity
。另外,您的copyToRealmOrUpdate不在事务内,因此也会崩溃,但与线程无关。
如果要在将某些数据发送到UI之前将其另存为副作用,则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);
}
});
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