如何在Mutiny反应式编程中调用长时间运行的阻塞空返回方法？

Question

我在Mutiny的Uni上有一个异步和同步方法调用链，有些方法是一个返回类型为void的长时间运行的进程。

在不阻塞下游的情况下调用它们的正确方式是什么？

下面是简单的类比代码。

class Root {

    public static void main(String[] args) {

        final Response response = getResponsePayload(); // Gets the the Payload from upstream service
        Uni.createFrom().item(response)
            .onItem().invoke(() -> System.out.println("Process Started"))
            .onItem().call(res -> {
            longRunningMethodAsync(res);    // long running blocking method, I want to run on a worker thread
            return Uni.createFrom().voidItem(); // This line I created, because of the ppipeline will be broken if the Uni is not returned from here
        })
            .onItem().transform(item -> item.hello + " mutiny")
            .onItem().transform(String::toUpperCase)
            .subscribe().with(
            item -> System.out.println(">> " + item));  // This is printed to the console
    }



    // Boilerplate method - I created to invoke/call the actual method actual method - `longRunningMethod`, this method basically an adapter
    // This is the best apprach I could come up, but I'm looking for better thatn this as I'm not conviced I'm doing it right
    private static UniSubscribe<Void> longRunningMethodAsync(final Response response) {

        
        return Uni.createFrom().voidItem().invoke(() -> longRunningMethod(response))
            .runSubscriptionOn(Infrastructure.getDefaultExecutor()).subscribe();
    }


    // Important - this is the method I want to run asynchronously independently of main *event-loop* thread.
    private static void longRunningMethod(final Response response) {
  
        System.out.println("Long running process started"); // Doesn't get printed, which means this is never called at all, not even in the blocked manner by the main even-loop thread
    }




   // Not as importatnt, I provded this in case if you like to run on your local box
    private static Response getResponsePayload() {
        return new Response();
    }

    private static class Response {
        public final String hello = "hello";
    }
}

Answer 1

尝试：

private static UniSubscribe<Void> longRunningMethodAsync(final Response response) {
            
   return Uni.createFrom()
  .voidItem()
  .invoke(() -> longRunningMethod(response))
  .runSubscriptionOn(Infrastructure.getDefaultExecutor())
  .subscribeAsCompletionStage().join();
}

Answer 2

通常，使用runSubscriptionOn并传递特定的executor：

longRunningMethodAsync
   .runSubscriptionOn(executor);

请注意，它会将并发限制为executor中可用线程的数量。

参考资料：

• https://smallrye.io/smallrye-mutiny/guides/emit-subscription
• https://smallrye.io/smallrye-mutiny/guides/emission-threads