如果希望每个重复的通量位于不同的线程上，可以移动 publishOn 以前，像这样：

Flux.just(1,2,3,4,5,6,7,8)
        .publishOn(Schedulers.parallel()) // <- before
        .flatMap(integer -> {
           System.out.println("val:" + integer + ", thread:" + Thread.currentThread().getId()); 
           return Mono.just(integer);
        }, 5)
        .repeat()
        .subscribe();

现在的输出是这样的：

val:1, thread:20
val:2, thread:20
val:3, thread:20
val:4, thread:20
val:5, thread:20
val:6, thread:20
val:7, thread:20
val:8, thread:20
val:1, thread:13
val:2, thread:13
val:3, thread:13
val:4, thread:13
val:5, thread:13
val:6, thread:13
val:7, thread:13
val:8, thread:13

如果希望每个整数位于不同的线程中，可以执行以下操作：

Flux.just(1,2,3,4,5,6,7,8)
        .publishOn(Schedulers.parallel()) // <- Each flux can be published in a different thread
        .flatMap(integer -> {
            return Mono.fromCallable(() -> {
                 System.out.println("val:" + integer + ", thread:" + Thread.currentThread().getId()); 
                 return integer;
            }).publishOn(Schedulers.parallel()); // <- Each Mono processing every integer can be processed in a different thread
         })
        .repeat()
        .subscribe();

输出变为：

val:3, thread:16
val:2, thread:15
val:7, thread:20
val:8, thread:13
val:5, thread:18
val:6, thread:19
val:3, thread:17
val:5, thread:19
val:6, thread:20
val:1, thread:15
val:8, thread:14
val:4, thread:18
val:7, thread:13

你需要使用 parallel 操作如下：

Flux.just(1,2,3,4,5,6,7,8)
    .parallel(2) // mention number of threads
    .runOn(Schedulers.parallel())
    .map(integer -> {
             System.out.println("val:" + integer + ", thread:" + Thread.currentThread().getId()); 
             return integer;
        })   
    .subscribe();