如何使用Spring Reactive修改助焊剂的结果

Question

I'm trying to convert the result of a Flux to confirm with Json API spec. I have my standard method which converts the POJO into the correct format and returns the json as a string. This works perfectly with a Mono .

However I can't workout how to handle that with the Flux .

@GetMapping(value = "/", produces = "application/vnd.api+json")
@ResponseBody
public Flux<String> findAll() {
    return feedItemRepository.findAll().flatMap(this::createJsonApiSpec);
}

private Mono<String> createJsonApiSpec(FeedItem item) {
    ObjectMapper objectMapper = new ObjectMapper();
    ResourceConverter resourceConverter = new ResourceConverter(objectMapper, FeedItem.class);

    JSONAPIDocument<Object> document = new JSONAPIDocument<>(item);

    byte[] bytes = new byte[0];
    try {
        bytes = resourceConverter.writeDocument(document);
    } catch (DocumentSerializationException e) {
        e.printStackTrace();
    }

    return Mono.justOrEmpty(new String(bytes));
}

The find all gets all the FeedItem s from a reactive mongo database. Then I get confused. Using flatMap I can process each item at a time and convert it to a String (that confirms to the spec using the method below) but this doesn't work because each item is processed independently meaning invalid Json is returned (No , between each item).

Is there another way to modify outgoing data from a Spring controller?

Edit: I know I can make this work using

public Mono<String> findAll() {
    List<FeedItem> feedItems = feedItemRepository.findAll().collectList().block();

    return writeList(feedItems);
}

But I'm sure that's a terrible idea to be using block() like that.

Answer 1

What does returning Mono<FeedItem> give you? Just a pointer: If FeedItem is a json pojo, you should be able to just return that. No need to do the json marshalling/unmarshalling programmatically, the framework will do that for you.

Like here: https://dzone.com/articles/spring-webflux-first-steps and many other examples along the lines.