RestTemplate 401错误
xyw10000 人气:0想了解RestTemplate 401 获取错误信息的处理方案的相关内容吗,xyw10000在本文为您仔细讲解RestTemplate 401错误的相关知识和一些Code实例,欢迎阅读和指正,我们先划重点:RestTemplate,401错误,错误信息,下面大家一起来学习吧。
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。
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