我正在使用okta来进行身份验证,我们公司的okta禁用了 "默认 "授权服务器。我们公司的okta禁用了 "默认 "授权服务器。所以现在我不能使用 "okta-spring-security-starter "来简单地验证url头传递的token。
import org.springframework.context.annotation.Configuration;
import org.springframework.security.config.annotation.web.builders.HttpSecurity;
import org.springframework.security.config.annotation.web.configuration.WebSecurityConfigurerAdapter;
@Configuration
public class OktaOAuth2WebSecurityConfig extends WebSecurityConfigurerAdapter {
@Override
protected void configure(HttpSecurity http) throws Exception {
http.authorizeRequests()
.antMatchers("/health").permitAll()
.anyRequest().authenticated()
.and()
.oauth2ResourceServer().jwt();
http.cors();
Okta.configureResourceServer401ResponseBody(http);
}
}
所以我需要使用okta内测端点(https:/developer.okta.comdocsreferenceapioidc#introspect。)来验证。所以我想知道,我是否可以将这个过程整合到配置的 WebSecurityConfigurerAdapter
......也许是这样的?
import com.okta.spring.boot.oauth.Okta;
import org.springframework.context.annotation.Configuration;
import org.springframework.security.config.annotation.web.builders.HttpSecurity;
import org.springframework.security.config.annotation.web.configuration.WebSecurityConfigurerAdapter;
@Configuration
public class OktaOAuth2WebSecurityConfig extends WebSecurityConfigurerAdapter {
@Override
protected void configure(HttpSecurity http) throws Exception {
http.authorizeRequests()
.antMatchers("/health").permitAll()
.anyRequest().authenticated()
.and()
/*add something there*/
http.cors();
}
}
我看到了一些类似于override AuthenticationProvider(具有Spring安全和Java配置的自定义身份验证提供商。),并使用httpbasic auth。如果我使用.oauth2ResourceServer().jwt(),能不能做类似的事情。
我的想法是重写认证提供者,然后在提供者中打okta introspect端点,这样可以吗?
Spring Security 5.2支持自省端点。请看一下 不透明令牌样本 在GitHub repo中。
不过在这里简单回答一下,你可以这样做。
http
.authorizeRequests(authz -> authz
.anyRequest().authenticated()
)
.oauth2ResourceServer(oauth2 -> oauth2
.opaqueToken(opaque -> opaque
.introspectionUri("the-endpoint")
.introspectionClientCredentials("client-id", "client-password")
)
);
如果你使用的是Spring Boot,那就简单一些。你可以在你的 application.yml
:
spring:
security:
oauth2:
resourceserver:
opaquetoken:
introspection-uri: ...
client-id: ...
client-secret: ...
然后你的DSL就可以指定 opaqueToken
:
http
.authorizeRequests(authz -> authz
.anyRequest().authenticated()
)
.oauth2ResourceServer(oauth2 -> oauth2
.opaqueToken(opaque -> {})
);
我没有使用Okta,因此我不知道它到底是如何工作的。但我有两个假设。
在这2个假设的前提下,如果我必须手动实现自定义安全逻辑认证,我会做以下步骤。
CustomAuthenticationProvider
注册自定义认证提供者。
// In OktaOAuth2WebSecurityConfig.java
@Override
protected void configure(AuthenticationManagerBuilder auth) throws Exception {
auth.authenticationProvider(customAuthenticationProvider());
}
@Bean
CustomAuthenticationProvider customAuthenticationProvider(){
return new CustomAuthenticationProvider();
}
CustomAuthenticationProvider:
public class CustomAuthenticationProvider implements AuthenticationProvider {
private static final Logger logger = LoggerFactory.getLogger(CustomAuthenticationProvider.class);
@Override
public Authentication authenticate(Authentication authentication) throws AuthenticationException {
logger.debug("Authenticating authenticationToken");
OktaTokenAuthenticationToken auth = (OktaTokenAuthenticationToken) authentication;
String accessToken = auth.getToken();
// You should make a POST request to ${oktaBaseUrl}/v1/introspect
// to determine if the access token is good or bad
// I just put a dummy if here
if ("ThanhLoyal".equals(accessToken)){
List<GrantedAuthority> authorities = Collections.singletonList(new SimpleGrantedAuthority("USER"));
logger.debug("Good access token");
return new UsernamePasswordAuthenticationToken(auth.getPrincipal(), "[ProtectedPassword]", authorities);
}
logger.debug("Bad access token");
return null;
}
@Override
public boolean supports(Class<?> clazz) {
return clazz == OktaTokenAuthenticationToken.class;
}
}
注册过滤器,从请求中提取accessToken。
// Still in OktaOAuth2WebSecurityConfig.java
@Override
protected void configure(HttpSecurity http) throws Exception {
http
.addFilterAfter(accessTokenExtractorFilter(), UsernamePasswordAuthenticationFilter.class)
.authorizeRequests().anyRequest().authenticated();
// And other configurations
}
@Bean
AccessTokenExtractorFilter accessTokenExtractorFilter(){
return new AccessTokenExtractorFilter();
}
以及过滤器本身:
public class AccessTokenExtractorFilter extends OncePerRequestFilter {
private static final Logger logger = LoggerFactory.getLogger(AccessTokenExtractorFilter.class);
@Override
protected void doFilterInternal(HttpServletRequest request, HttpServletResponse response, FilterChain filterChain) throws ServletException, IOException {
logger.debug("Filtering request");
Authentication authentication = getAuthentication(request);
if (authentication == null){
logger.debug("Continuing filtering process without an authentication");
filterChain.doFilter(request, response);
} else {
logger.debug("Now set authentication on the request");
SecurityContextHolder.getContext().setAuthentication(authentication);
filterChain.doFilter(request, response);
}
}
private Authentication getAuthentication(HttpServletRequest request) {
String accessToken = request.getHeader("Authorization");
if (accessToken != null){
logger.debug("An access token found in request header");
List<GrantedAuthority> authorities = Collections.singletonList(new SimpleGrantedAuthority("USER"));
return new OktaTokenAuthenticationToken(accessToken, authorities);
}
logger.debug("No access token found in request header");
return null;
}
}
我在这里上传了一个简单的项目,方便大家参考。https:/github.comMrLoyalspring-security-custom-authentication。
它是如何工作的。
另外,你的问题应该包含 spring-security
标签。
更新1: 关于认证入口点
安 AuthenticationEntryPoint
声明当一个未经认证的请求到达时该怎么做(在我们的例子中,当请求不包含一个有效的 "授权 "头时该怎么做)。
在我的REST API中,我只是简单地响应401 HTTP状态码给客户端。
// CustomAuthenticationEntryPoint
@Override
public void commence(HttpServletRequest request, HttpServletResponse response, AuthenticationException authException) throws IOException, ServletException {
response.reset();
response.setStatus(401);
// A utility method to add CORS headers to the response
SecUtil.writeCorsHeaders(request, response);
}
Spring的 LoginUrlAuthenticationEntryPoint
重定向用户到登录页面(如果配置了登录页面)。
因此,如果您想将未经认证的请求重定向到Okta的登录页面,您可以使用AuthenticationEntryPoint。