RestTemplate 401 获取错误信息的处理方案

RestTemplate 401错误

调用第三方api 若是服务返回状态码不为200，默认会执行DefaultResponseErrorHandler

异常处理

@Override
public void handleError(ClientHttpResponse response) throws IOException {
 HttpStatus statusCode = getHttpStatusCode(response);
 switch (statusCode.series()) {
  case CLIENT_ERROR:
   throw new HttpClientErrorException(statusCode, response.getStatusText(),
     response.getHeaders(), getResponseBody(response), getCharset(response));
  case SERVER_ERROR:
   throw new HttpServerErrorException(statusCode, response.getStatusText(),
     response.getHeaders(), getResponseBody(response), getCharset(response));
  default:
   throw new RestClientException("Unknown status code [" + statusCode + "]");
 }
}

判断是否异常

protected boolean hasError(HttpStatus statusCode) {
 return (statusCode.series() == HttpStatus.Series.CLIENT_ERROR ||
   statusCode.series() == HttpStatus.Series.SERVER_ERROR);
}

通常会直接已异常形势抛出，若不特殊处理无法获取返回提示信息。

需要捕捉HttpClientErrorException 异常，则可获取返回信息

try{
     ......
   }catch (HttpClientErrorException e) {
               String resBody = e.getResponseBodyAsString();
               log.info("客户端异常返回:{}", resBody);
               return new ResponseEntity<>(JSON.parseObject(resBody, res), e.getStatusCode());
           }

一开始我这样写，死活返回的都是null

原来跟我设置的requestFactory有关

采用SimpleClientHttpRequestFactory 无法获取提示

需要换成 HttpComponentsClientHttpRequestFactory

RestTemplate通过对象传参，response的body为空讨论

代码复现

实体类

@Entity
@Table(name = "a",schema = "a")
@JsonIgnoreProperties(value = {"a"})
@Setter
@Generated
public class C {
   @Id
   @GeneratedValue
   private Integer id;
   @Column(name = "diseaseName",length = 255,nullable = false,unique = true)
   private String diseaseName;
   @Column(name = "description",length = 255,nullable = false,unique = true)
   private String description;
   @Column(name = "department",length = 255,nullable = false,unique = true)
   private String department;
}
controller
@ResponseBody
   @RequestMapping(value = "",method = RequestMethod.POST)
   public Response APIcreate(@RequestBody C c) {
       String json = JSONUtil.toJSONString(c);
       HttpHeaders headers = new HttpHeaders();
       headers.setContentType(MediaType.APPLICATION_JSON_UTF8);
       HttpEntity<String> entity = new HttpEntity<>(json, headers);
       String url = "http://localhost:3001/c";
       ResponseEntity<Commondisease> responseEntity = restTemplate.postForEntity(url, entity, C.class);
       return new ResponseData(ExceptionMsg.SUCCESS, responseEntity);
   }

返回结果截图：

返回结果为空的讨论：返回的C类是jpa封装后的类，即使通过json工具，也无法转换成功

解决办法一：实体类转成普通类

import lombok.AllArgsConstructor;
import lombok.Data;
import lombok.NoArgsConstructor;
@Data
@AllArgsConstructor
@NoArgsConstructor
public class C {
   private Integer id;
   private String diseaseName;
   private String description;
   private String department;
}
@ResponseBody
   @RequestMapping(value = "",method = RequestMethod.POST)
   public Response APIcreate(@RequestBody C c) {
       //C c = new Commondisease(1,"zhangsan","11","2222");
       String json = JSONUtil.toJSONString(c);
       HttpHeaders headers = new HttpHeaders();
       headers.setContentType(MediaType.APPLICATION_JSON_UTF8);
       HttpEntity<String> entity = new HttpEntity<>(json, headers);
       String url = "http://localhost:3001/c/";
       ResponseEntity<Commondisease> responseEntity =         restTemplate.postForEntity(url,entity,C.class);
       return new ResponseData(ExceptionMsg.SUCCESS,responseEntity);
}

返回成功

解决办法二：添加注解

@Data

来源：https://blog.csdn.net/xuyw10000/article/details/88790391