RestTemplate 401 獲取錯誤信息的處理方案

更新時間：2021年08月19日 11:07:20 作者：xyw10000

這篇文章主要介紹了RestTemplate 401 獲取錯誤信息的處理方案，具有很好的參考價值，希望對大家有所幫助。如有錯誤或未考慮完全的地方，望不吝賜教

RestTemplate 401錯誤

調(diào)用第三方api 若是服務返回狀態(tài)碼不為200，默認會執(zhí)行DefaultResponseErrorHandler

異常處理

@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 + "]");
  }
 }

判斷是否異常

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

通常會直接已異常形勢拋出，若不特殊處理無法獲取返回提示信息。

需要捕捉HttpClientErrorException 異常，則可獲取返回信息

try{
      ......
    }catch (HttpClientErrorException e) {
                String resBody = e.getResponseBodyAsString();
                log.info("客戶端異常返回:{}", resBody);
                return new ResponseEntity<>(JSON.parseObject(resBody, res), e.getStatusCode());
            }

一開始我這樣寫，死活返回的都是null

原來跟我設置的requestFactory有關

采用SimpleClientHttpRequestFactory 無法獲取提示

需要換成 HttpComponentsClientHttpRequestFactory

RestTemplate通過對象傳參，response的body為空討論

代碼復現(xiàn)

實體類

@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);
    }

返回結果截圖：

返回結果為空的討論：返回的C類是jpa封裝后的類，即使通過json工具，也無法轉換成功

解決辦法一：實體類轉成普通類

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