我想用一個custom authentication filter
說:配置Spring Security沒有在Spring XML 4
- 驗證之後捕獲加密頭標記
- ,提取用戶的詳細信息,並將它們添加到當前請求的安全上下文以無狀態的方式
我希望能夠使用此安全上下文持有人獲取有關當前請求用戶正確處理其請求的詳細信息。
@RequestMapping(value = "/simple", method = RequestMethod.POST)
@ResponseBody
@Transactional
@Preauthorize(...)
public String simple(){
//collect the user's current details from the getPrinciple() and complete the transaction...
Object principal = SecurityContextHolder.getContext().getAuthentication().getPrincipal();
return "Simple";
}
我在XML做過像這樣:
<beans xmlns="http://www.springframework.org/schema/beans"
xmlns:security="http://www.springframework.org/schema/security"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://www.springframework.org/schema/beans
http://www.springframework.org/schema/beans/spring-beans-3.2.xsd
http://www.springframework.org/schema/security
http://www.springframework.org/schema/security/spring-security-3.2.xsd">
<security:global-method-security
secured-annotations="enabled" />
<security:http pattern="/**"
auto-config="true" disable-url-rewriting="true" use-expressions="true">
<security:custom-filter ref="authenticationTokenProcessingFilter"
position="FORM_LOGIN_FILTER" />
<security:intercept-url pattern="/authenticate"
access="permitAll" />
<security:intercept-url pattern="/secure/**"
access="isAuthenticated()" />
</security:http>
<bean id="CustomAuthenticationEntryPoint" class="org.foo.CustomAuthenticationEntryPoint" />
<bean class="org.foo.AuthenticationTokenProcessingFilter" id="authenticationTokenProcessingFilter">
<constructor-arg ref="authenticationManager" />
</bean>
</beans>
不過,我想這與非XML WebSecurityConfigurerAdapter喜歡在他們的春天啓動的例子更新Spring Boot
應用工作文件:
@Bean
public ApplicationSecurity applicationSecurity() {
return new ApplicationSecurity();
}
@Order(Ordered.LOWEST_PRECEDENCE - 8)
protected static class ApplicationSecurity extends WebSecurityConfigurerAdapter {
@Override
protected void configure(HttpSecurity http) throws Exception {
// this is obviously for a simple "login page" not a custom filter!
http.authorizeRequests().anyRequest().fullyAuthenticated().and().formLogin()
.loginPage("/login").failureUrl("/login?error").permitAll();
}
}
任何意見或類似的例子嗎?