我有一个端点
@GetMapping(value = "/accounts/{accountId}/placementInfo", headers = "version=1")
@ResponseStatus(HttpStatus.OK)
public List<PlacementDetail> findPlacementDetailByPlacementInfoAtTime(@PathVariable("accountId") Long accountId,
@RequestParam(value = "searchDate", required = false)
@DateTimeFormat(iso = DateTimeFormat.ISO.DATE) LocalDate searchDate) {}
我正在使用rest模板发送请求
placementResponseEntity = restTemplate.exchange(placementUriBuilder(accountId, searchDate), HttpMethod.GET,
apiRequestEntity,new ParameterizedTypeReference<List<PlacementDetail>>() {});
使用辅助方法
private String placementUriBuilder(long accountId, LocalDate searchDate) throws IOException {
String resourceUri = ACCOUNT_RESOURCE_URI_START + accountId + PLACEMENT_DETAIL_RESOURCE_URI_END;
String url;
if(searchDate != null) {
url = UriComponentsBuilder.fromUri(serverUri.getUri()).path(resourceUri).queryParam("searchDate", searchDate.format(DateTimeFormatter.ISO_DATE)).build().toUriString();
} else {
url = UriComponentsBuilder.fromUri(serverUri.getUri()).path(resourceUri).build().toUriString();
}
return url;
}
当我查看SO时,人们谈论发送对象并失败,因为创建的JSON格式错误,但这是一个get api,我不明白问题的根源。
答案 0 :(得分:1)
这通常是由RestTemplate
上缺少错误处理程序引起的。您的服务器响应错误,客户端尝试将其反序列化为List<PlacementDetail>
。为了解决这个问题,您应该正确处理HTTP错误代码。
请参见下面的代码段。
@Configuration
public class ClientConfig {
@Bean
public RestTemplate restTemplate(RestTemplateBuilder restTemplateBuilder) {
return restTemplateBuilder
.errorHandler(new ClientErrorHandler())
.build();
}
public class ClientErrorHandler implements ResponseErrorHandler {
@Override
public boolean hasError(ClientHttpResponse httpResponse) throws IOException {
// check if HTTP status signals an error response
return !HttpStatus.OK.equals(httpResponse.getStatusCode());
}
@Override
public void handleError(ClientHttpResponse httpResponse) throws IOException {
// handle exception case as you see fit
throw new RuntimeException("Error while making request");
}
}
}