相关疑难解决方法(0)

如何在纯资源服务器上获取我的自定义ResponseEntityExceptionHandler或OAuth2ExceptionRenderer处理Spring安全性引发的异常？

我们实施了一个

@ControllerAdvice
@RestController
public class GlobalExceptionHandler extends ResponseEntityExceptionHandler {

因此,只要资源服务器上出现错误,我们就希望它能够回答

{
  "message": "...",
  "type": "...",
  "status": 400
}

资源服务器使用application.properties设置:

security.oauth2.resource.userInfoUri: http://localhost:9999/auth/user

对我们的auth服务器进行身份验证和授权请求.

但是,任何弹簧安全性错误都将始终绕过我们的异常处理程

    @ExceptionHandler(InvalidTokenException.class)
    public ResponseEntity<Map<String, Object>> handleInvalidTokenException(InvalidTokenException e) {
        return createErrorResponseAndLog(e, 401);
    }

并生产

{
  "timestamp": "2016-12-14T10:40:34.122Z",
  "status": 403,
  "error": "Forbidden",
  "message": "Access Denied",
  "path": "/api/templates/585004226f793042a094d3a9/schema"
}

要么

{
  "error": "invalid_token",
  "error_description": "5d7e4ab5-4a88-4571-b4a4-042bce0a076b"
}

那么如何配置资源服务器的安全性异常处理呢？我找到的只是如何通过实现自定义来自定义Auth服务器的示例OAuth2ExceptionRenderer.但我无法找到将其连接到资源服务器的安全链的位置.

我们唯一的配置/设置是这样的:

@SpringBootApplication
@Configuration
@ComponentScan(basePackages = {"our.packages"})
@EnableAutoConfiguration
@EnableResourceServer

Url,Credentials在RestClient UI以及Curl中工作,因为当通过Spring RestTemplate访问相同内容时,我遇到"500"错误.

我使用以下代码:

MultiValueMap<String, Object> map = new LinkedMultiValueMap<String, Object>();
map.add("name", user);
map.add("password", password);
restTemplate.postForObject(url, request, Employee.class, map);

请告诉我您的建议或意见以解决问题.

我已经创建了一个用于创建新用户的测试：

private static String USERS_ENDPOINT = "http://localhost:8080/users/";
private static String GROUPS_ENDPOINT = "http://localhost:8080/groups/";

@Test
@DirtiesContext(classMode = ClassMode.BEFORE_EACH_TEST_METHOD)
public void whenCreateAppUser() {

    AppUser appUser = new AppUser();
    appUser.setUsername("test@example.com");
    appUser.setPassword("password");

    // Throws java.net.HttpRetryException
    template.postForEntity(USERS_ENDPOINT, appUser, AppUser.class);

    ResponseEntity<AppUser> appUserResponse = template.getForEntity(USERS_ENDPOINT + "1/", AppUser.class);

    assertEquals("Username is incorrect. AppUser not created?",
            appUser.getUsername(), appUserResponse.getBody().getUsername());
}

但是，由于某种原因，我得到了：

Caused by: java.net.HttpRetryException: cannot retry due to server authentication, in streaming mode
    at sun.net.www.protocol.http.HttpURLConnection.getInputStream0(HttpURLConnection.java:1692)
    at sun.net.www.protocol.http.HttpURLConnection.getInputStream(HttpURLConnection.java:1492)
    at java.net.HttpURLConnection.getResponseCode(HttpURLConnection.java:480)
    at org.springframework.http.client.SimpleClientHttpResponse.getRawStatusCode(SimpleClientHttpResponse.java:55)
    at org.springframework.web.client.DefaultResponseErrorHandler.hasError(DefaultResponseErrorHandler.java:49)
    at org.springframework.web.client.RestTemplate.handleResponse(RestTemplate.java:735)
    at org.springframework.web.client.RestTemplate.doExecute(RestTemplate.java:700)
    ... 34 more …