RestTemplate 401 获取错误信息的处理方案

目录

• RestTemplate 401错误
• 异常处理
• 判断是否异常
• RestTemplate通过对象传参，response的body为空讨论
• 代码复现
• 解决办法一：实体类转成普通类
• 解决办法二：添加注解

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

以上为个人经验，希望能给大家一个参考，也希望大家多多支持我们。