Question

我正在尝试使用SpringBoot 2.0和新的反应式webFlux库。我想知道如何将通过无阻塞WebClient进行的两次调用的结果返回给我的Springboot API的调用者。我的代码是：

@RequestMapping("/search")
public CombinedResults perfomSearch(@RequestParam final String searchTerm) {
    Mono<SearchResponse> fasMono = searchService.getSearchResults(searchTerm, "fh");
    Mono<SearchResponse> esMono = searchService.getSearchResults(searchTerm, "es");
    CombinedResults combinedResults = new CombinedResults(fasMono, esMono);
    return combinedResults;

}

CombinedResult对象只是一个POJO：

public class CombinedResults {

private Mono<SearchResponse> fasSearchResponse;

private Mono<SearchResponse> esSearchResponse;

public CombinedResults(final Mono<SearchResponse> fasSearchResponse, final Mono<SearchResponse> esSearchResponse) {
    this.fasSearchResponse = fasSearchResponse;
    this.esSearchResponse = esSearchResponse;
}

public Mono<SearchResponse> getFasSearchResponse() {
    return fasSearchResponse;
}

public void setFasSearchResponse(final Mono<SearchResponse> fasSearchResponse) {
    this.fasSearchResponse = fasSearchResponse;
}

public Mono<SearchResponse> getEsSearchResponse() {
    return esSearchResponse;
}

public void setEsSearchResponse(final Mono<SearchResponse> esSearchResponse) {
    this.esSearchResponse = esSearchResponse;
}

但是如果我称之为我回复的回复是

{
  "fasSearchResponse": {
    "scanAvailable": true
  },
  "esSearchResponse": {
    "scanAvailable": true
  }
}

而不是SearchResponse对象的内容。我觉得我可能会错过这个应该如何运作的基本要点！我的想法是，因为WebClient没有阻塞，我可以解雇两个Web服务调用，然后将它们组合起来而不需要可完成的期货等？

Answer 1

Spring WebFlux不支持嵌套的响应类型。你应该这样做：

Mono<SearchResponse> fasMono = searchService.getSearchResults(searchTerm, "fh");
Mono<SearchResponse> esMono = searchService.getSearchResults(searchTerm, "es");
Mono<CombinedResults> results = fasMono.zipWith(esMono, 
    (fas, es) -> {return new CombinedResults(fas, es);});

Answer 2

我认为您应该返回一个对象的Mono，该对象代表通过此操作响应的模型。假设CombinedResults是您的模型。这个类应该是这样的：

public class CombinedResults {

    private SearchResponse fasSearchResponse;

    private SearchResponse esSearchResponse;

    public CombinedResults(final SearchResponse fasSearchResponse, final SearchResponse esSearchResponse) {
        this.fasSearchResponse = fasSearchResponse;
        this.esSearchResponse = esSearchResponse;
    }

    //... getters AND/OR setters
}

并且，在您的控制器上，您可以执行以下操作：

@RequestMapping("/search")
public Mono<CombinedResults> perfomSearch(@RequestParam final String searchTerm) {
    Mono<SearchResponse> fasMono = searchService.getSearchResults(searchTerm, "fh");
    Mono<SearchResponse> esMono = searchService.getSearchResults(searchTerm, "es");
    Mono<CombinedResults> combinedResults = 
        fasMono
          .flatMap(fh -> esMono.map(es -> new CombinedResults(fh, es)));
    return combinedResults;
}

通过这种方式，您将返回一个Mono对象，其中包含您想要的响应。当两个Mono都发出项目时，带有fasMono.flatMap的操作链esMono.map会构建CombinedResults。当尝试将两个Monos连接成一个时，这种组合很常见。我想你也可以使用zip运算符加入Monos。所有这些都与WebClient无关。如果您的getSearchResults仅执行异步非阻塞操作，则所有内容都是异步非阻塞。

在Spring Webflux中返回多个Monos

2 个答案: