RestTemplate(401,获取错误信息的处理方案)

RestTemplate,401,获取错误信息的处理方案，久久派带你了解更多相关信息。

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@Generatedpublic 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@NoArgsConstructorpublic 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趣讯吧。