从 Mono 的列表中创建 Flux 的正确方法

Question

假设我有一个使用 CustomObjects 列表的 API 操作。对于这些对象中的每一个，它都会调用一个创建 Mono 的服务方法。如何以惯用的非阻塞方式从这些 Mono 对象创建 Flux？

我现在想到的是这个。我更改了方法名称以更好地反映它们的预期目的。

fun myApiMethod(@RequestBody customObjs: List<CustomObject>): Flux<CustomObject> {

    return Flux.create { sink ->
        customObjs.forEach {

            service.persistAndReturnMonoOfCustomObject(it).map {
                sink.next(it)
            }
        }
        sink.complete()
    }
}

此外，我需要订阅 flux 才能真正做到 return 吗？

Answer 1

我相信你可以使用 concat() 代替：

/**
 * Concatenate all sources provided as a vararg, forwarding elements emitted by the
 * sources downstream.
 * <p>
 * Concatenation is achieved by sequentially subscribing to the first source then
 * waiting for it to complete before subscribing to the next, and so on until the
 * last source completes. Any error interrupts the sequence immediately and is
 * forwarded downstream.
 * <p>
 * <img class="marble" src="https://raw.githubusercontent.com/reactor/reactor-core/v3.1.3.RELEASE/src/docs/marble/concat.png" alt="">
 * <p>
 * @param sources The {@link Publisher} of {@link Publisher} to concat
 * @param <T> The type of values in both source and output sequences
 *
 * @return a new {@link Flux} concatenating all source sequences
 */
@SafeVarargs
public static <T> Flux<T> concat(Publisher<? extends T>... sources) {

或merge():

/**
 * Merge data from {@link Publisher} sequences contained in an array / vararg
 * into an interleaved merged sequence. Unlike {@link #concat(Publisher) concat},
 * sources are subscribed to eagerly.
 * <p>
 * <img class="marble" src="https://raw.githubusercontent.com/reactor/reactor-core/v3.1.3.RELEASE/src/docs/marble/merge.png" alt="">
 * <p>
 * Note that merge is tailored to work with asynchronous sources or finite sources. When dealing with
 * an infinite source that doesn't already publish on a dedicated Scheduler, you must isolate that source
 * in its own Scheduler, as merge would otherwise attempt to drain it before subscribing to
 * another source.
 *
 * @param sources the array of {@link Publisher} sources to merge
 * @param <I> The source type of the data sequence
 *
 * @return a merged {@link Flux}
 */
@SafeVarargs
public static <I> Flux<I> merge(Publisher<? extends I>... sources) {

从 Mono 的列表中创建 Flux 的正确方法

Proper way to create a Flux from a list of Mono's

mono

spring

flux

kotlin

project-reactor