Java 如何将 Mono<List<String>> 转换为 Flux<String>
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How to convert Mono<List<String>> into Flux<String>
提问by Mavo
I'm converting small project written in RxJava 1.x to Reactor 3.x. All is good, except that I could not find out how to replace flatMap(Observable::from)with an appropriate counterpart. I have Mono<List<String>>and I need to convert it to Flux<String>.
我正在将用 RxJava 1.x 编写的小项目转换为 Reactor 3.x。一切都很好，只是我不知道如何flatMap(Observable::from)用合适的对应物替换。我有Mono<List<String>>，我需要将其转换为Flux<String>.
采纳答案by Simon Baslé
In Reactor 3, the fromoperator has been specialized into a few variants, depending on the original source (array, iterable, etc...).
在 Reactor 3 中，from运算符被专门化为几个变体，具体取决于原始来源（数组、可迭代等...）。
Use yourMono.flatMapMany(Flux::fromIterable)in your case.
yourMono.flatMapMany(Flux::fromIterable)在你的情况下使用。
回答by Anthony Pi?ero
Thanks Simon, I implemented something like this:
谢谢西蒙，我实现了这样的事情：
List<Object> dbObjects = ListObjectsBD();
List<Dao> daos = mapperObjToDao(dbObjects);
Flux<Dao> daoFlux = Mono.just(daos).flatMapMany(Flux::fromIterable);
回答by Michail Alexakis
I think that probably Flux::mergeSequentialstatic factory fits better here:
我认为Flux::mergeSequential静态工厂可能更适合这里：
Iterable<Mono<String>> monos = ...
Flux<String> f = Flux.mergeSequential(monos);
This kind of merge (sequential) will maintain the ordering inside given source iterable, and will also subscribe/request eagerly from all participating sources (so more parallelization expected while computing mono results).
这种合并（顺序）将保持给定源可迭代内的排序，并且还会从所有参与的源急切地订阅/请求（因此在计算单声道结果时需要更多的并行化）。
