英文:
What is the proper way of zipping Flux with Mono?
问题
以下是翻译好的内容：
有一个 Mono<A> 和 Flux<B>，我们需要创建一个类似这样的元组流：
Mono<A> monoA = createMono(); // {a}
Flux<B> fluxB = createFlux(); // {b1, b2, ... b100, ...}
Flux<Tuple<A,B>> zippedTuples = magicZip(monoA, fluxB); // { (a:b1), (a:b2), ... (a:b100), ...}
编写 magicZip 函数的正确（或标准）方法是什么？
英文:
There's a Mono<A> and Flux<B>, and we need to create a flux of tuples like this:
Mono<A> monoA = createMono(); // {a}
Flux<B> fluxB = createFlux(); // {b1, b2, ... b100, ...}
Flux<Tuple<A,B>> zippedTuples = magicZip(monoA, fluxB); // { (a:b1), (a:b2), ... (a:b100), ...}
What is the proper (or standard) way to write the magicZip function?
答案1
得分: 1
你可以创建这个方法：
private <T> Flux<Tuple2<T, T>> magicZip(Mono<T> mono, Flux<T> flux) {
Flux<T> repeatableMono = mono.repeat();
return flux.zipWith(repeatableMono);
}
字符串类型的示例：
Flux<Tuple2<String, String>> test = magicZip(getMono(), getFlux()).doOnNext(objects -> System.out.println(objects.getT1() + objects.getT2()));
test.blockLast();
英文:
You can create this method:
private <T>Flux<Tuple2<T, T>> magicZip(Mono<T> mono, Flux<T> flux) {
Flux<T> repeatableMono = mono.repeat();
return flux.zipWith(repeatableMono);
}
Example for the String type:
Flux<Tuple2<String, String>> test = magicZip(getMono(), getFlux()).doOnNext(objects -> System.out.println(objects.getT1() + objects.getT2()));
test.blockLast();
答案2
得分: 0
我认为使用zip函数是不可能的，因为它生成的元素数量与两者中最小的数量相同。
我认为你可以实现这样的方式：
Flux<A> fluxA = monoA.flux();
Flux<Tuple2<A,B>> zippedTuples = fluxB.flatMap(b -> fluxA.map(a -> Tuples.of(a,b)));
英文:
I think that with zip function is not possible because it produces as many elements as the smallest of two.
The way I think you can achive this is:
Flux<A> fluxA = monoA.flux();
Flux<Tuple2<A,B>> zippedTuples =fluxB.flatMap(b -> fluxA.map(a -> Tuples.of(a,b)));
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