RxJava2 toList() 从不发出

Question

所以我有以下 Disposable 不起作用。我正在使用 Room 将表中的所有行作为列表获取，将它们中的每一个映射到某物并创建一个列表，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();

Answer 1

来自第 3 方的这些类型的数据源通常是无限源，但toList()需要有限源。我猜您想处理该集合storedSuggestions并将其保存在一起。您可以通过内部转换来实现这一点：

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();