Spring 安全 401 在不安全端点上未经授权

Question

我正尝试在 Spring 引导应用程序上配置 Spring 安全性，如下所示：

@Configuration
@EnableWebSecurity
@EnableGlobalMethodSecurity(prePostEnabled = true)
public class SecurityConfig extends WebSecurityConfigurerAdapter {

@Autowired
private RestAuthenticationEntryPoint unauthorizedHandler;

@Bean
public JwtAuthenticationFilter authenticationTokenFilterBean() throws Exception {
    JwtAuthenticationFilter authenticationTokenFilter = new JwtAuthenticationFilter();
    authenticationTokenFilter.setAuthenticationManager(authenticationManagerBean());
    return authenticationTokenFilter;
}

@Override
protected void configure(HttpSecurity httpSecurity) throws Exception {

    //@formatter:off
     httpSecurity
      .csrf()
        .disable()
      .exceptionHandling()
        .authenticationEntryPoint(this.unauthorizedHandler)
        .and()
      .sessionManagement()
        .sessionCreationPolicy(SessionCreationPolicy.STATELESS)
        .and()
      .authorizeRequests()
        .antMatchers(HttpMethod.OPTIONS, "/**").permitAll()
        .antMatchers("/login", "/singup", "/subscribers").permitAll()
        .anyRequest().authenticated();

        // Custom JWT based security filter 
    httpSecurity            
        .addFilterBefore(authenticationTokenFilterBean(), UsernamePasswordAuthenticationFilter.class);   

    //@formatter:on

}
}

我的 unauthorizedHandler 是：

public class RestAuthenticationEntryPoint implements AuthenticationEntryPoint {

private static final Logger LOGGER = LoggerFactory.getLogger(RestAuthenticationEntryPoint.class);

@Override
public void commence(HttpServletRequest request, HttpServletResponse response,
        AuthenticationException authException) throws IOException, ServletException {
    response.sendError(HttpServletResponse.SC_UNAUTHORIZED, "Unauthorized");
}

}

最后，/subscribers 的 REST 控制器是：

@RestController
public class SubscriberRestController {

@Autowired
ISubscribersService subscribersService;

@RequestMapping(value = RequestMappingConstants.SUBSCRIBERS, method = RequestMethod.GET)
@ResponseBody
public Number subscriberCount() {

    return subscribersService.subscribersCount();
}

@RequestMapping(value = RequestMappingConstants.SUBSCRIBERS, method = RequestMethod.POST)
public String subscriberPost(@RequestBody SubscriberDocument subscriberDocument) {

    return subscribersService.subscribersInsert(subscriberDocument);
}

@RequestMapping(value = "/test", method = RequestMethod.GET)
public String test() {

    return "This is a test";
}

}

我使用 postman 测试端点，当我执行 POST 到 "localhost:8080/subscribers" 时，我得到：

我想打开没有任何安全控制或凭据检查的端点（/订阅者），用于注册和登录的端点以及用于经过身份验证的用户的安全端点。

谢谢！ :)

Answer 1

经过一番研究，这里是解决方案：

@SpringBootApplication(exclude = {SecurityAutoConfiguration.class })
@ComponentScan(basePackages = { PackageConstants.PACKAGE_CONTROLLERS_REST, PackageConstants.PACKAGE_SERVICES,
        PackageConstants.PACKAGE_SERVICES_IMPL, PackageConstants.PACKAGE_MONGO_REPOSITORIES,
        PackageConstants.PACKAGE_MONGO_REPOSITORIES_IMPL, PackageConstants.PACKAGE_UTILS })
public class Application {

    // Clase principal que se ejecuta en el bootrun

    public static void main(String[] args) {

        SpringApplication.run(Application.class, args);
    }
}

主线是 @SpringBootApplication(exclude = {SecurityAutoConfiguration.class }) 它告诉我们不要使用 Spring 引导安全自动配置配置。这不是完整的答案，因为现在你必须告诉 Spring 用户你的 Spring 安全配置 class。我还建议您使用 init Root Config 类创建 Initializer class，使用 ApplicationConfiguration 并拒绝使用 SpringBoot 应用程序。像这样：

应用程序配置：

@Configuration
@EnableWebMvc
@ComponentScan("com.trueport.*")
@PropertySource("classpath:app.properties")
public class ApplicationConfig extends WebMvcConfigurerAdapter {
    ....
}

应用程序安全配置：

@Configuration
@EnableWebSecurity
@EnableGlobalMethodSecurity(prePostEnabled = true, securedEnabled = true)
public class ApplicationSecurityConfig extends WebSecurityConfigurerAdapter {
    ....
}

初始化程序：

public class Initializer implements WebApplicationInitializer {

    private static final String DISPATCHER_SERVLET_NAME = "dispatcher";

    @Override
    public void onStartup(ServletContext servletContext) throws ServletException {
        AnnotationConfigWebApplicationContext ctx = new AnnotationConfigWebApplicationContext();
        ....
        DispatcherServlet dispatcherServlet = new DispatcherServlet(ctx);
        dispatcherServlet.setThrowExceptionIfNoHandlerFound(true);
        ctx.register(ApplicationConfig.class);
        ServletRegistration.Dynamic servlet =     servletContext.addServlet(DISPATCHER_SERVLET_NAME,
            dispatcherServlet);
        servlet.addMapping("/");
        servlet.setLoadOnStartup(1);
        servlet.setAsyncSupported(true);
    }
}

Answer 2

Spring 引导未应用配置，因为找不到它。 Application.java 配置包未包含在 @ComponentScan 注释中。

Answer 3

您需要将以下内容添加到您的配置方法中 /error 是应用程序由于任何异常而发生错误时的默认回退，并且默认情况下是安全的。

protected void configure(HttpSecurity httpSecurity) throws Exception {
//disable CRSF
httpSecurity
        //no authentication needed for these context paths
        .authorizeRequests()
        .antMatchers("/error").permitAll()
        .antMatchers("/error/**").permitAll()
        .antMatchers("/your Urls that dosen't need security/**").permitAll()

还有下面的代码片段

     @Override
       public void configure(WebSecurity webSecurity) throws Exception
         {
          webSecurity
          .ignoring()
           // All of Spring Security will ignore the requests
           .antMatchers("/error/**")
          }

现在，当 permitAll Urls 发生异常时，您将不会得到 401 和 500 异常详细信息

Answer 4

如果您的应用程序只是简单地保存 API，并且您包含了 spring 安全性的依赖项 - 出于任何其他原因（我的是启用 headers X-Frame 和 Content-Security-Policy)，那么默认情况下 Spring 包括用于 csrf 保护的 servlet 过滤器。如果不禁用此功能，所有请求都会失败并出现 HTTP 401 错误。

要禁用它，您创建一个配置 class 扩展 WebSecurityConfigurerAdapter 并用 EnableWebSecurity

注释

@EnableWebSecurity
@Configuration
public class WebSecurityConfig extends
   WebSecurityConfigurerAdapter {

  @Override
  protected void configure(HttpSecurity http) throws Exception {
    http
      .csrf().disable();
  }
}

这篇 article 值得一读 - 非常详细。

Spring 安全 401 在不安全端点上未经授权

Spring security 401 Unauthorized on unsecured endpoint

java

spring

spring-security

spring-boot

postman