英文:
Execute Monos sequentially/parallel
问题
我有这样的情况：
Mono<List<String>> resultAMono = listA();
Mono<Void> resultBMono = loadB(resultAMono.block());
Mono<Void> resultCMono = loadC();
我想按顺序执行resultAMono，而resultBMono和resultCMono并行执行，因为resultBMono依赖于resultAMono的结果。
英文:
I have this situation:
Mono<List<String>> resultAMono = listA();
Mono<Void> resultBMono = loadB(resultAMono.block());
Mono<Void> resultCMono = loadC();
I want to execute resultAMono sequentially, and the resultBMono and resultCMono parallelly, because resultBMono depends of the result of resultAMono
答案1
得分: 1
不需要在这里block。在响应式编程中，您需要构建一个结合不同操作符的流。
Mono<Void> res = resultAMono
.flatMap(a -> Flux.merge(loadB(a), loadC()).then());
在这里使用Flux.merge是因为loadB和loadC都返回Mono<Void>，而我们不需要结果。
如果需要结果，可以使用Mono.zip来并行解析发布者。
Mono<Tuple2<B, C>> res = resultAMono
.flatMap(a -> Mono.zip(loadB(a), loadC()));
Flux.merge和Mono.zip都会急切地订阅发布者。
英文:
You don't need to block here. In reactive you need to build a flow combining different operators
Mono<Void> res = resultAMono
.flatMap(a -> Flux.merge(loadB(a), loadC()).then());
We are using Flux.merge here because both loadB & loadC return Mono<Void> and we don't need results.
In case results are needed Mono<B> loadB(A a) & Mono<C> loadC() we can use Mono.zip to resolve publishers in parallel.
Mono<Tuple2<B, C>> res = resultAMono
.flatMap(a -> Mono.zip(loadB(a), loadC()));
Both Flux.merge & Mono.zip subscribes to publishers eagerly.
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