如何在 Spring WebClient 构建器实例中设置 onStatus

Question

我有很多方法使用 Spring 的 WebClient 中的 onStatus API：

@Override
public Mono<Accommodation> createAccommodation(CreateAccommodation create) {
    return webClient
            .post()
            .contentType(APPLICATION_JSON)
            .bodyValue(create)
            .retrieve()
            .onStatus(HttpStatus::isError,
                    clientResponse -> clientResponse
                            .bodyToMono(ApiErrorResponse.class)
                            .flatMap(errorResponse -> Mono.error(new ResponseStatusException(
                                    HttpStatus.valueOf(errorResponse.getStatus()),
                                    errorResponse.getMessage()
                            ))))
            .bodyToMono(Accommodation.class);
}

我想做的是避免在每个 WebClient 调用中都使用“onStatus”。

在构建 WebClient 实例时有没有办法设置它？你能举一些例子吗？

这是我的 WebClient 实例：

    public AccommodationServiceClientImpl(WebClient.Builder builder) {
         this.webClient = builder
            .baseUrl("lb://accommodation-service/api/v1/accommodations")
            .build();
    }

Answer 1

找到了一个解决方案： ExchangeFilterFunction.ofResponseProcessor 似乎是我正在寻找的。

@Configuration
public class WebClientConfig {

    @Bean
    @LoadBalanced
    public WebClient.Builder webClientBuilder(){
        return WebClient
                .builder()
                .filter(ExchangeFilterFunction.ofResponseProcessor(this::renderApiErrorResponse));
    }

    private Mono<ClientResponse> renderApiErrorResponse(ClientResponse clientResponse) {
        if(clientResponse.statusCode().isError()){
            return clientResponse.bodyToMono(ApiErrorResponse.class)
                    .flatMap(apiErrorResponse -> Mono.error(new ResponseStatusException(
                            clientResponse.statusCode(),
                            apiErrorResponse.getMessage()
                    )));
        }
        return Mono.just(clientResponse);
    }

}