将REST API的404响应更改为200空响应

Question

我有一个用Java编写的Spring Boot应用程序，它是一个REST API。该服务（Svc A）调用REST API服务（Svc B）,该服务也是用Java编写的Spring Boot应用程序。当没有找到数据时，Svc B返回404状态码。我需要将此响应更改为200状态代码，并返回一个空响应对象。我不确定是否或如何做到这一点。

我可以捕捉错误，并确定404是否是这个没有数据发现的错误。但是，我不知道如何将响应更改为200空响应。我正在使用FeignClient调用服务。这是捕获404的错误代码：

@Component
public class FeignErrorDecoder implements ErrorDecoder {

    Logger logger = LoggerFactory.getLogger(this.getClass());

    @Override
    public Exception decode(String methodKey, Response response) {
        Reader reader = null;
        String messageText = null;
        switch (response.status()){
            case 400:
                logger.error("Status code " + response.status() + ", methodKey = " + methodKey);
            case 404:
            {
                logger.error("Error took place when using Feign client to send HTTP Request. Status code " + response.status() + ", methodKey = " + methodKey);
                try {
                    reader = response.body().asReader();
                    //Easy way to read the stream and get a String object
                    String result = CharStreams.toString(reader);
                    logger.error("RESPONSE BODY: " + result);
                    ObjectMapper mapper = new ObjectMapper();
                    //just in case you missed an attribute in the Pojo
                    mapper.disable(DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES);
                    //init the Pojo
                    ExceptionMessage exceptionMessage = mapper.readValue(result,
                            ExceptionMessage.class);

                    messageText = exceptionMessage.getMessage();
                    logger.info("message: " + messageText);

                } catch(IOException ex) {
                    logger.error(ex.getMessage());
                }
                finally {
                    try {

                        if (reader != null)
                            reader.close();

                    } catch (IOException e) {
                        e.printStackTrace();
                    }
                }

                return new ResponseStatusException(HttpStatus.valueOf(200), messageText);
            }
            default:
                return new Exception(response.reason());
        }
    }
}

{
   "statusCd" : "200",
   "message" : "The Location not found for given Location Number and Facility Type Code",
   "detailDesc" : "The Location not found for given Location Number and Facility Type Code. Error Timestamp : 2020-01-31 18:19:13"
}

{
  "facilityNumber": "923",
  "facilityTimeZone": null,
  "facilityAbbr": null,
  "scheduledOperations": []
}

Answer 1

在404的情况下，只需尝试

return new ResponseStatusException(HttpStatus.valueOf(200));