WebExceptionHandler : How to write a body with Spring Webflux

问题

I want to handle the Exception of my api by adding a WebExceptionHandler. I can change the status code, but I am stuck when i want to change the body of the response : ex adding the exception message or a custom object.

Does anyone have exemple ?

How I add my WebExceptionHandler :

HttpHandler httpHandler = WebHttpHandlerBuilder.webHandler(toHttpHandler(routerFunction))
  .prependExceptionHandler((serverWebExchange, exception) -> {

      exchange.getResponse().setStatusCode(myStatusGivenTheException);
      exchange.getResponse().writeAndFlushWith(??)
      return Mono.empty();

  }).build();

回答1:

WebExceptionHandler is rather low level, so you have to directly deal with the request/response exchange.

Note that:

the Mono<Void> return type should signal the end of the response handling; this is why it should be connected to the Publisher writing the response
at this level, you're dealing directly with data buffers (no serialization support available)

Your WebExceptionHandler could look like this:

(serverWebExchange, exception) -> {

  exchange.getResponse().setStatusCode(myStatusGivenTheException);
  byte[] bytes = "Some text".getBytes(StandardCharsets.UTF_8);
  DataBuffer buffer = exchange.getResponse().bufferFactory().wrap(bytes);
  return exchange.getResponse().writeWith(Flux.just(buffer));
}

回答2:

ServerResponse has a method writeTo which can be used to write your body to ServerExchange (Spring framework does it this way). Only problem is that you have to provide Contextas a second parameter, so I have just copied HandlerStrategiesResponseContext from framework implementation.

Make sure that you are using at least Spring Boot 2.0.0 M2, before this version WebExceptionHandler was not registered while using RouterFunctions.

import org.springframework.http.HttpStatus
import org.springframework.http.HttpStatus.*
import org.springframework.http.codec.HttpMessageWriter
import org.springframework.stereotype.Component
import org.springframework.web.reactive.function.server.HandlerStrategies
import org.springframework.web.reactive.function.server.ServerResponse
import org.springframework.web.reactive.result.view.ViewResolver
import org.springframework.web.server.ServerWebExchange
import org.springframework.web.server.WebExceptionHandler


@Component
class GlobalErrorHandler() : WebExceptionHandler {

    override fun handle(exchange: ServerWebExchange, ex: Throwable): Mono<Void> =
        handle(ex)
                .flatMap {
                    it.writeTo(exchange, HandlerStrategiesResponseContext(HandlerStrategies.withDefaults()))
                }
                .flatMap {
                    Mono.empty<Void>()
                }

    fun handle(throwable: Throwable): Mono<ServerResponse> {

        return when (throwable) {
            is EntityNotFoundException -> {
                createResponse(NOT_FOUND, "NOT_FOUND", "Entity not found, details: ${throwable.message}")
            }
            else -> {
                createResponse(INTERNAL_SERVER_ERROR, "GENERIC_ERROR", "Unhandled exception")
            }
        }
    }

    fun createResponse(httpStatus: HttpStatus, code: String, message: String): Mono<ServerResponse> =
        ServerResponse.status(httpStatus).syncBody(ApiError(code, message))
}

private class HandlerStrategiesResponseContext(val strategies: HandlerStrategies) : ServerResponse.Context {

    override fun messageWriters(): List<HttpMessageWriter<*>> {
        return this.strategies.messageWriters()
    }

    override fun viewResolvers(): List<ViewResolver> {
        return this.strategies.viewResolvers()
    }
}

回答3:

Given the answer, to serialize object I use this way :

 Mono<DataBuffer> db = commonsException.getErrorsResponse().map(errorsResponse -> {

     ObjectMapper objectMapper = new ObjectMapper();
     try {
         return objectMapper.writeValueAsBytes(errorsResponse);
     } catch (JsonProcessingException e) {
          return e.getMessage().getBytes();
     }
}).map(s -> exchange.getResponse().bufferFactory().wrap(s));

exchange.getResponse().getHeaders().add("Content-Type", "application/json");
exchange.getResponse().setStatusCode(commonsException.getHttpStatus());
return exchange.getResponse().writeWith(db);

回答4:

for anyone looking at a way to write JSON response body, here's a Kotlin code sample:

fun writeBodyJson(body: Any, exchange: ServerWebExchange) =
    exchange.response.writeWith(
        Jackson2JsonEncoder().encode(
            Mono.just(body),
            exchange.response.bufferFactory(),
            ResolvableType.forInstance(body),
            MediaType.APPLICATION_JSON_UTF8,
            Hints.from(Hints.LOG_PREFIX_HINT, exchange.logPrefix))
    )

not 100% sure that's the way to go though, would like to get some opinions.

来源：https://stackoverflow.com/questions/45211431/webexceptionhandler-how-to-write-a-body-with-spring-webflux

标签

Spring

exception-handling

spring-webflux