Question

我正在使用RxJava 2和改造库，并希望执行700多个查询，其中每个查询都是一个改进的API调用。

如何使用zip操作符在RXJava 2中实现此目的？

Answer 1

700或1000万没关系，你通常有一个8核的CPU，所以只有8个并行处理它将被执行。

您可以将flatMap与MaxConcurrent一起使用

/**
 * FlatMap operator allow you to specify the max number of concurrent operation that this operator can process.
 * This is another way to make back pressure if you know your system cannot process more than specific number
 * of elements in the pipeline.
 */
@Test
public void asyncFlatMapWithMaxConcurrent() {
    Observable.from(Arrays.asList(1, 2, 3, 4, 5, 6, 7, 8, 9, 10))//Your number of request
            .flatMap(value -> Observable.just(value)
                            .map(number -> {
                                 System.out.println("Make your request here");

                                return number;
                            }).subscribeOn(Schedulers.newThread())
                    , 8)//Max concurrent operations
            .subscribe();
    new TestSubscriber()
            .awaitTerminalEvent(15, TimeUnit.SECONDS);
}

您可以在此处查看有关RxJava的更多示例https://github.com/politrons/reactive

RxJava 2 Observable zip和Retrofit

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