为什么WebClient释放原始异常消息?

问题描述 投票:0回答:1

我想将异常从我的“数据库” REST API重新引发到我的“后端” REST API,但是我松开了原始异常的消息。

这是我通过邮递员从我的“数据库” REST API获得的内容:

{
    "timestamp": "2020-03-18T15:19:14.273+0000",
    "status": 400,
    "error": "Bad Request",
    "message": "I'm DatabaseException (0)",
    "path": "/database/api/vehicle/test/0"
}

这部分还可以。

这是我通过邮递员从我的“后端” REST API获得的内容:

{
    "timestamp": "2020-03-18T15:22:12.801+0000",
    "status": 400,
    "error": "Bad Request",
    "message": "400 BAD_REQUEST \"\"; nested exception is org.springframework.web.reactive.function.client.WebClientResponseException$BadRequest: 400 Bad Request from GET http://localhost:8090/database/api/vehicle/test/0",
    "path": "/backend/api/vehicle/test/0"
}

您可以看到原来的“消息”字段丢失了。

我使用:

  • Spring boot 2.2.5.RELEASE
  • spring-boot-starter-web
  • spring-boot-starter-webflux

后端和数据库以Tomcat(web and webflux in the same application)开头。

这是后端:

    @GetMapping(path = "/test/{id}")
    public Mono<Integer> test(@PathVariable String id) {
        return vehicleService.test(id);
    }

使用vehicleService.test:

    public Mono<Integer> test(String id) {
        return WebClient
            .create("http://localhost:8090/database/api")
            .get()
            .uri("/vehicle/test/{id}", id)
            .accept(MediaType.APPLICATION_JSON)
            .retrieve()
            .bodyToMono(Integer.class);
    }

这是数据库:

    @GetMapping(path = "/test/{id}")
    public Mono<Integer> test(@PathVariable String id) throws Exception {

        if (id.equals("0")) {
            throw new DatabaseException("I'm DatabaseException (0)");
        }

        return Mono.just(Integer.valueOf(2));
    }

我也尝试过return Mono.error(new DatabaseException("I'm DatabaseException (0)"));

和DatabaseException:

public class DatabaseException extends ResponseStatusException {

    private static final long serialVersionUID = 1L;

    public DatabaseException(String message) {
        super(HttpStatus.BAD_REQUEST, message);

    }
}

似乎我的后端转换了响应,并且在互联网上找不到任何答案。

java spring-boot spring-mvc exception spring-webflux
1个回答
0
投票

代替retrieveWebClient,可以使用exchange来处理错误并传播自定义异常以及从服务响应中检索到的消息。

private void execute()
{
    WebClient webClient = WebClient.create();

    webClient.get()
             .uri("http://localhost:8089")
             .exchange()
             .flatMap(this::handleResponse)
             .doOnNext(System.out::println)
             .block();  // not required, just for testin purposes
}

private Mono<Response> handleResponse(ClientResponse clientResponse)
{
    if (clientResponse.statusCode().isError())
    {
        return clientResponse.bodyToMono(Response.class)
                             .flatMap(response -> Mono.error(new RuntimeException(response.message)));
    }

    return clientResponse.bodyToMono(Response.class);
}

private static class Response
{
    private String message;

    public Response()
    {
    }

    public String getMessage()
    {
        return message;
    }

    public void setMessage(String message)
    {
        this.message = message;
    }

    @Override
    public String toString()
    {
        return "Response{" +
                "message='" + message + '\'' +
                '}';
    }
}
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