RxJava2 toList() 从不发出

所以我有以下 Disposable 不起作用。我使用 Room 从表中获取所有行作为列表，将它们映射到某个内容并创建一个列表，然后它不会从那里继续。

storedSuggestionDao
    .getSuggestionsOrderByType() //Flowable
    .doOnNext(storedSuggestions -> Timber.e("storedSuggestions: " + storedSuggestions)) //this work
    .flatMapIterable(storedSuggestions -> storedSuggestions)
    .map(Selection::create) ))
    .doOnNext(selection -> Timber.e("Selection: " + selection)) // works
    .toList()
    .toObservable() // nothing works after this...
    .doOnNext(selections -> Timber.d("selections: " + selections))
    .map(SuggestionUiModel::create)
    .doOnNext(suggestionUiModel -> Timber.d("suggestionUiModel: " + suggestionUiModel))
    .subscribe();

来自第三方的这些类型的数据源通常是无限源，但是toList()需要有限源。我猜你想处理这个集合storedSuggestions并把它放在一起。你可以通过内部转变来实现这一点：

storedSuggestionDao
.getSuggestionsOrderByType() //Flowable
.doOnNext(storedSuggestions -> Timber.e("storedSuggestions: " + storedSuggestions)) //this work
// -------------------------------------
.flatMapSingle(storedSuggestions -> 
    Flowable.fromIterable(storedSuggestions)
    .map(Selection::create)
    .doOnNext(selection -> Timber.e("Selection: " + selection))
    .toList()
)
// -------------------------------------
.doOnNext(selections -> Timber.d("selections: " + selections))
.map(SuggestionUiModel::create)
.doOnNext(suggestionUiModel -> Timber.d("suggestionUiModel: " + suggestionUiModel))
.subscribe();