如何在 spring 启动时从 Rest Template 捕获错误响应？

Question

我有 2 springboot REST APIs REST-A 和 REST-B。 REST-B 正在与 mongodb 交互以进行 CRUD 操作。 REST-A 出于不同原因调用 REST-B 端点。

REST-B 中的控制器（客户 API）

public class CustomerController {
    
    @Autowired
    private CustomerRepository customerRepository;
    
    @GetMapping(value = "/customers/{id}")
    public ResponseEntity<Customer> getCustomerByExternalReferenceId(@PathVariable(value = "id") String id)
            throws ResourceNotFoundException {
        System.out.println("Customer id received :: " + id);
        Customer customer = customerRepository.findByExternalCustomerReferenceId(id)
                .orElseThrow(() -> new ResourceNotFoundException("Customer not found for this id :: " + id));
        return ResponseEntity.ok().body(customer);
    }
}

如果我从邮递员处调用此端点，无论是在数据库中找到客户还是在数据库中找不到客户，都可以正常工作。

现在，如果我尝试从 REST-A 调用相同的端点，并且如果在数据库中找到客户，我可以获得响应。

        String url = "http://localhost:8086/customer-api/customers/{id}";
        String extCustRefId = 
        setupRequest.getPayload().getCustomer().getCustomerReferenceId();
        
        // URI (URL) parameters
        Map<String, String> urlParams = new HashMap<>();
        urlParams.put("id", extCustRefId); // here I tried with id that exists in DB and getting 200 ok response
        
        HttpHeaders headers = new HttpHeaders();
        headers.set("X-GP-Request-Id", "abc-xyz-123");
        headers.set("Content-Type", "application/json");
        headers.set("Accept", "application/json");
        headers.set("Content-Length", "65");
        
        String searchurl = UriComponentsBuilder.fromUriString(url).buildAndExpand(urlParams).toString();
        
        System.out.println(searchurl);
        
        HttpEntity request = new HttpEntity(headers);
        RestTemplate restTemplate = new RestTemplate();
        try {
            ResponseEntity<String> response = restTemplate.exchange(
                    searchurl,
                    HttpMethod.GET,
                    request,
                    String.class
            );
        } catch (Exception e) {
            e.printStackTrace();
        }

但是如果没有从 REST-B 中找到客户（客户 API），那么我会得到

http://localhost:8086/customer-api/customers/customer-528f2331-d0c8-46f6-88c2-7445ee6f4821
Customer id received :: customer-528f2331-d0c8-46f6-88c2-7445ee6f4821
org.springframework.web.client.HttpClientErrorException: 404 null
        at org.springframework.web.client.DefaultResponseErrorHandler.handleError(DefaultResponseErrorHandler.java:78)
        at org.springframework.web.client.RestTemplate.handleResponse(RestTemplate.java:700)
        at org.springframework.web.client.RestTemplate.doExecute(RestTemplate.java:653)
        at org.springframework.web.client.RestTemplate.execute(RestTemplate.java:613)
        at org.springframework.web.client.RestTemplate.exchange(RestTemplate.java:531)

如何从一个 springboot 应用程序调用休息端点到另一个应用程序并正确处理响应？

Answer 1

您可以从HttpClientErrorException获取响应正文如下：

try {
    ResponseEntity<String> response = restTemplate.exchange(
            searchurl,
            HttpMethod.GET,
            request,
            String.class
    );
} catch (HttpClientErrorException e) {
    String errorResponseBody = e.getResponseBodyAsString();
    e.printStackTrace();
}

然后您可以使用 Jackson ObjectMapper 将字符串映射到 Java 对象。

如何在 spring 启动时从 Rest Template 捕获错误响应？

How to capture error responses from Rest Template in spring boot?

java

resttemplate

spring-boot