我想知道是否可以自定义以下授权错误:
sh '''
cd path/to/your/submodule
git pull $ZUUL_URL/$ZUUL_PROJECT +$ZUUL_REF:refs/heads/zuul
'''
我在用户请求没有权限时得到它。我想将它定制为与Spring Boot错误非常相似:
{
"error": "unauthorized",
"error_description": "Full authentication is required to access this resource"
}
可能吗?
非常感谢。
答案 0 :(得分:6)
我明白了:)
https://stackoverflow.com/a/37132751/2520689
我需要创建一个实现“AuthenticationEntryPoint”的新类,如下所示:
public class AuthExceptionEntryPoint implements AuthenticationEntryPoint
{
@Override
public void commence(HttpServletRequest request, HttpServletResponse response, AuthenticationException arg2) throws IOException, ServletException
{
final Map<String, Object> mapBodyException = new HashMap<>() ;
mapBodyException.put("error" , "Error from AuthenticationEntryPoint") ;
mapBodyException.put("message" , "Message from AuthenticationEntryPoint") ;
mapBodyException.put("exception", "My stack trace exception") ;
mapBodyException.put("path" , request.getServletPath()) ;
mapBodyException.put("timestamp", (new Date()).getTime()) ;
response.setContentType("application/json") ;
response.setStatus(HttpServletResponse.SC_UNAUTHORIZED) ;
final ObjectMapper mapper = new ObjectMapper() ;
mapper.writeValue(response.getOutputStream(), mapBodyException) ;
}
}
并将其添加到我的ResourceServerConfigurerAdapter实现中:
@Configuration
@EnableResourceServer
public class ResourceServerConfiguration extends ResourceServerConfigurerAdapter
{
@Override
public void configure(HttpSecurity http) throws Exception
{
http.exceptionHandling().authenticationEntryPoint(new AuthExceptionEntryPoint()) ;
}
}
你可以找到我的GitHub项目,它实现了你需要的一切:
答案 1 :(得分:0)
使用Oauth2,接受的答案对我不起作用。经过研究,exception translator solution可以正常工作。
基本上,您需要创建一个WebResponseExceptionTranslator
并将其注册为您的异常翻译器。
首先,创建一个WebResponseExceptionTranslator
bean:
@Slf4j
@Configuration
public class Oauth2ExceptionTranslatorConfiguration {
@Bean
public WebResponseExceptionTranslator oauth2ResponseExceptionTranslator() {
return new DefaultWebResponseExceptionTranslator() {
@Override
public ResponseEntity<OAuth2Exception> translate(Exception e) throws Exception {
ResponseEntity<OAuth2Exception> responseEntity = super.translate(e);
OAuth2Exception body = responseEntity.getBody();
HttpStatus statusCode = responseEntity.getStatusCode();
body.addAdditionalInformation("timestamp", dateTimeFormat.format(clock.instant()))
body.addAdditionalInformation("status", body.getHttpErrorCode().toString())
body.addAdditionalInformation("message", body.getMessage())
body.addAdditionalInformation("code", body.getOAuth2ErrorCode().toUpperCase())
HttpHeaders headers = new HttpHeaders();
headers.setAll(responseEntity.getHeaders().toSingleValueMap());
// do something with header or response
return new ResponseEntity<>(body, headers, statusCode);
}
};
}
}
现在,您需要更改Oauth2配置以注册bean WebResponseExceptionTranslator
:
@Slf4j
@Configuration
public class OAuth2Config extends AuthorizationServerConfigurerAdapter {
@Autowired
private AuthenticationManager authenticationManager;
@Autowired
private ClientDetailsServiceBuilder builder;
@Autowired
private WebResponseExceptionTranslator oauth2ResponseExceptionTranslator;
@Autowired
private UserDetailsService userDetailsService;
@Override
public void configure(ClientDetailsServiceConfigurer clients) {
clients.setBuilder(builder);
}
@Override
public void configure(AuthorizationServerEndpointsConfigurer endpoints) {
TokenEnhancerChain tokenEnhancerChain = new TokenEnhancerChain();
tokenEnhancerChain.setTokenEnhancers(
Arrays.asList(tokenEnhancer(), accessTokenConverter()));
endpoints.tokenStore(tokenStore())
.tokenEnhancer(tokenEnhancerChain)
.authenticationManager(authenticationManager)
.userDetailsService(userDetailsService)
.exceptionTranslator(oauth2ResponseExceptionTranslator);
}
}
最终结果将是:
{
"error": "unauthorized",
"error_description": "Full authentication is required to access this resource",
"code": "UNAUTHORIZED",
"message": "Full authentication is required to access this resource",
"status": "401",
"timestamp": "2018-06-28T23:55:28.86Z"
}
您可以看到我没有从OAuth2Exception的原始正文中删除error
和error_description
。我建议维护它们,因为这两个字段遵循OAuth2规范。有关更多详细信息,请参见the RFC和OAuth2 API definitions。
您还可以自定义结果:覆盖error
或error_description
(仅调用addAdditionalInformation
),使用instance of
标识特定的异常以返回不同的json结果,等等。但是也有一些限制:如果您想将某个字段定义为integer
,我认为这是不可能的,因为addAdditionalInformation
方法只接受String
作为类型。>
答案 2 :(得分:0)
故事短片:https://github.com/melardev/JavaSpringBootOAuth2JwtCrudPagination.git
在阅读@pakkk响应后,我不同意,所以我决定尝试自己的方式,但同样失败了,所以我决定看看Spring Security源代码本身,这是怎么回事:
有一个非常早就被调用的过滤器,即OAuth2AuthenticationProcessingFilter。
如果引发异常,此过滤器将尝试从标头中提取JWT。
它的authenticationEntryPoint.commence()(@pakk在这里)
我尝试添加一个过滤器来检查Jwt无效或存在时是否调用了它,但是没有,因此,添加自定义过滤器以更改响应将不起作用。
然后,我查看了OAuth2AuthenticationProcessingFilter的配置位置,发现它是在ResourceServerSecurityConfigurer :: configure(HttpSecurity http)上设置的。
如此说来,让我们看看如何融入流程。
事实证明这很容易,因为您将在资源服务器应用程序中扩展ResourceServerConfigurerAdapter类:
@Configuration
@EnableResourceServer
public class OAuth2ResourceServerConfig extends ResourceServerConfigurerAdapter {
// ....
}
您继续操作并覆盖:
@Override
public void configure(ResourceServerSecurityConfigurer resources) throws Exception {
super.configure(resources);
}
如您所见,是的!您可以访问ResourceServerSecurityConfigurer,那么现在呢? 好吧,让我们用我们的替换默认的入口点:
@Autowired
private AuthenticationEntryPoint oauthEntryPoint;
@Override
public void configure(ResourceServerSecurityConfigurer resources) throws Exception {
super.configure(resources);
resources.authenticationEntryPoint(oauthEntryPoint);
}
有关示例的完整源代码,请参见: https://github.com/melardev/JavaSpringBootOAuth2JwtCrudPagination.git
没有此步骤,至少对我来说这是行不通的,@ pakkk提供的响应对我不起作用,我在调试器上进行了检查,默认情况下使用的入口点不是我们的入口点,即使使用: / p>
http.and().exceptionHandling().authenticationEntryPoint(oauthEntryPoint)
这是我测试的第一件事,要使其正常运行,您必须直接从ResourceServerSecurityConfigurer类中更改入口点。
这是我的切入点:请注意,我正在发送属于我自己的类的ErrorResponse对象,因此我可以完全控制响应:
@Component
public class OAuthEntryPoint implements AuthenticationEntryPoint {
@Autowired
ObjectMapper mapper;
@Override
public void commence(HttpServletRequest httpServletRequest, HttpServletResponse httpServletResponse, AuthenticationException e) throws IOException, ServletException {
ServletServerHttpResponse res = new ServletServerHttpResponse(httpServletResponse);
res.setStatusCode(HttpStatus.FORBIDDEN);
res.getServletResponse().setHeader(HttpHeaders.CONTENT_TYPE, MediaType.APPLICATION_JSON_VALUE);
res.getBody().write(mapper.writeValueAsString(new ErrorResponse("You must authenticated")).getBytes());
}
}
答案 3 :(得分:-1)
我认为您可以使用@ControllerAdvice
来捕获未经授权的异常,然后根据您的期望格式化响应并将其返回。像这样:
@ResponseBody
@ExceptionHandler(CustomException.class)
@ResponseStatus(value=HttpStatus.UNAUTHORIZED, reason="Exception message")
public JsonResponse unAuthorised(HttpServletRequest request, Exception ex) {
return new JsonResponse("ERROR", 401, "Unauthorised Request");
}
希望得到这个帮助。