RestTemplate|RestTemplate 401 获取错误信息的处理方案 RestTemplate401获取错误信息的处

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 = https://www.it610.com/article/{"a"})@Setter@Generatedpublic class C {@Id@GeneratedValueprivate 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(valuehttps://www.it610.com/article/= "",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 entity = new HttpEntity<>(json, headers); String url = "http://localhost:3001/c"; ResponseEntity responseEntity = restTemplate.postForEntity(url, entity, C.class); return new ResponseData(ExceptionMsg.SUCCESS, responseEntity); }

返回结果截图：

RestTemplate|RestTemplate 401 获取错误信息的处理方案

文章图片

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(valuehttps://www.it610.com/article/= "",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 entity = new HttpEntity<>(json, headers); String url = "http://localhost:3001/c/"; ResponseEntity 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
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