如何注销对Spring WebFlux WebClient请求的失败响应的主体，同时将响应返回给调用方？

Question

我对反应编程非常陌生，我有一个REST服务，它接受一个请求，然后使用WebFlux WebClient调用另一个API。当API以4xx或5xx响应进行响应时，我希望将响应体记录在我的服务中，然后将响应传递给调用方。我已经找到了许多处理响应日志记录的方法，但它们通常会将mono.error返回给调用方，这不是我想要做的。我几乎可以做到这一点，但是当我向我的服务发出请求时，当我返回API返回的4xx代码时，我的客户机只是挂起等待响应的正文，而服务似乎从未完成对流的处理。我使用的是Spring Boot版本2.2.4.发行版。

我得到的是：

@PostMapping(path = "create-order")
public Mono<ResponseEntity<OrderResponse>> createOrder(@Valid @RequestBody CreateOrderRequest createOrderRequest) {
    return orderService.createOrder(createOrderRequest);
}

public Mono<ResponseEntity<OrderResponse>> createOrder(CreateOrderRequest createOrderRequest) {
    return this.webClient
            .mutate()
            .filter(OrderService.errorHandlingFilter(ORDERS_URI, createOrderRequest))
            .build()
            .post()
            .uri(ORDERS_URI)
            .contentType(MediaType.APPLICATION_JSON)
            .bodyValue(createOrderRequest)
            .exchange()
            .flatMap(response -> response.toEntity(OrderResponse.class));
}

public static ExchangeFilterFunction errorHandlingFilter(String uri, CreateOrderRequest request) {
    return ExchangeFilterFunction.ofResponseProcessor(clientResponse -> {
        if (clientResponse.statusCode() != null && (clientResponse.statusCode().is5xxServerError() || clientResponse.statusCode().is4xxClientError())) {
            return clientResponse.bodyToMono(String.class)
                    .flatMap(errorBody -> OrderService.logResponseError(clientResponse, uri, request, errorBody));
        } else {
            return Mono.just(clientResponse);
        }
    });
}

static Mono<ClientResponse> logResponseError(ClientResponse response, String attemptedUri, CreateOrderRequest orderRequest, String responseBody) {
    ObjectMapper objectMapper = new ObjectMapper();
    objectMapper.setSerializationInclusion(JsonInclude.Include.NON_NULL);
    try {
        log.error("Response code {} received when attempting to hit {}, request:{}, response:{}",
                response.rawStatusCode(), attemptedUri, objectMapper.writeValueAsString(orderRequest),
                responseBody);
    } catch (JsonProcessingException e) {
        log.error("Error attempting to serialize request object when reporting on error for request to {}, with code:{} and response:{}",
                attemptedUri, response.rawStatusCode(), responseBody);
    }
    return Mono.just(response);
}

return Mono.error(new Exception("Some error"));

我在客户端收到一个500，请求就完成了。如果有人知道为什么当我尝试发回响应时它不会完成，我很想知道我做错了什么。

Answer 1

鱼与熊掌不可兼得！

这里的问题是，您正在尝试消耗响应的主体两次，这是不允许的。通常这样做会得到一个错误。

一旦进入

return clientResponse.bodyToMono(String.class) 

response.toEntity(OrderResponse.class)

@Override
public <T> Mono<ResponseEntity<T>> toEntity(Class<T> bodyType) {
    return WebClientUtils.toEntity(this, bodyToMono(bodyType));
}

public Mono<ResponseEntity<OrderResponse>> createOrder(CreateOrderRequest createOrderRequest) {
    return this.webClient
            //no need for mutate unless you already have things specified in 
            //base webclient?
            .post()
            .uri(ORDERS_URI)
            .contentType(MediaType.APPLICATION_JSON)
            .bodyValue(createOrderRequest)
            .exchange()
            //Here you map the response to an entity first
            .flatMap(response -> response.toEntity(OrderResponse.class))
            //Then run the errorHandler to do whatever
            //Use doOnNext since there isn't any reason to return anything
            .doOnNext(response -> 
                errorHandler(ORDERS_URI,createOrderRequest,response));

}

//Void doesn't need to return
public static void  errorHandler(String uri, CreateOrderRequest request,ResponseEntity<?> response) {
    if( response.getStatusCode().is5xxServerError() 
        || response.getStatusCode().is4xxClientError())
            //run log method if 500 or 400
            OrderService.logResponseError(response, uri, request);
}

//No need for redundant final param as already in response
static void logResponseError(ResponseEntity<?> response, String attemptedUri, CreateOrderRequest orderRequest) {
    //Do the log stuff
    ObjectMapper objectMapper = new ObjectMapper();
    objectMapper.setSerializationInclusion(JsonInclude.Include.NON_NULL);
    try {
        log.error("Response code {} received when attempting to hit {}, request:{}, response:{}",
                response.getStatusCodeValue(), attemptedUri, objectMapper.writeValueAsString(orderRequest),
                response.getBody());
    } catch (JsonProcessingException e) {
        log.error("Error attempting to serialize request object when reporting on error for request to {}, with code:{} and response:{}",
                attemptedUri, response.getStatusCodeValue(), response.getBody());
    }
}

请注意，没有使用ExchangeFilter的真正理由，因为您实际上并没有执行任何筛选，只是根据响应执行一个操作