RxJava：是否可以将可完成的执行构建到doOnSuccess或类似物中

Question

当需要在Rx操作链中执行一些非同步操作时：

    private static Single<String> process(final String s) {
        return Single.just(s)
                .map(String::toUpperCase)
                .doOnSuccess(str -> saveValue1(str) // asynchronous op 1)
                .doOnSuccess(str -> saveValue2(str) // asynchronous op 2);
    }

没有flatMap或blockingGet就可以编写Rx链的替代方案，它可以将可完成方法saveValue1，saveValue2等的执行建立到链中。

    private static Single<String> process(final String s) {
        return Single.just(s)
                .map(String::toUpperCase)
                .doOnSuccess(str -> saveValue1(str).blockingGet() // asynchronous op 1)
                .doOnSuccess(str -> saveValue2(str).blockingGet() // asynchronous op 2);
    }

或

private static Single<String> process(final String s) {
    return Single.just(s)
            .map(String::toUpperCase)
            .flatMap(str -> saveValue1(str).toSingleDefault(str) // asynchronous op 1)
            .flatMap(str -> saveValue2(str).toSingleDefault(str) // asynchronous op 2);
}

复制示例：

package com.eisgroup.genesis.policy.core.lifecycle.commands.quote;

import io.reactivex.Completable;
import io.reactivex.Single;
import org.springframework.util.Assert;

public class Test {
    // some variables to check that asynchronous save works
    static String s1;
    static String s2; 

    public static void main(String[] args) {
        process("test").blockingGet();
        Assert.hasText(s1, "TEST");
        Assert.hasText(s2, "TEST");
    }   


    // some fake examples to show asynchronous save
    private static Completable saveValue1(String s) {
        return Completable.defer(() -> {
            s1 = s;
            return Completable.complete();
        });
    }

    private static Completable saveValue2(String s) {
        return Completable.defer(() -> {
            s2 = s;
            return Completable.complete();
        });
    }
}

Answer 1

尝试一下：

public static void main(String[] args) {
        process("test").blockingAwait();
        System.out.println(s1.equals("TEST"));
        System.out.println(s2.equals("TEST"));
    }

    private static Completable process(final String s) {
        return Single.just(s)
                .map(String::toUpperCase)
                .flatMapCompletable(string -> Completable.mergeArray(
                        saveValue1(string).subscribeOn(Schedulers.io()),
                        saveValue2(string).subscribeOn(Schedulers.io())));
    }

[saveValue1和saveValue2将并行运行好吗？不知道这是否是您想要的，但是出于效率和速度的考虑，我还是鼓励这样做。