XML configuration of Spring Security in Spring Boot

Question

I'd like to use XML based configuration to Spring Security. The first idea was to use SHA-256 or any other hashing function for user passwords. I could not find a nice way to solve this with plain java., so I started to configure things in xml. That was the point, when it started to get interesting.

My configuration:

• spring-boot 1.1.8.RELEASE
• spring-boot-starter-* at 1.1.8
• tomcat-embed-jasper:8.0.8

spring-security.xml:

<beans:beans xmlns="http://www.springframework.org/schema/security"
             xmlns:beans="http://www.springframework.org/schema/beans"
             xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
             xmlns:jdbc="http://www.springframework.org/schema/jdbc"
             xsi:schemaLocation="http://www.springframework.org/schema/beans
           http://www.springframework.org/schema/beans/spring-beans-3.0.xsd
           http://www.springframework.org/schema/security
           http://www.springframework.org/schema/security/spring-security.xsd>

    <http pattern="/css/**" security="none"/>
    <http pattern="/login.html*" security="none"/>

    <http>
        <intercept-url pattern="/**" access="ROLE_USER" />
        <form-login login-page='/login.html'/>
    </http>

    <authentication-manager>

        <authentication-provider>
            <user-service>
                <user name="admin" password="admin"
                      authorities="ROLE_USER, ROLE_ADMIN"/>
                <user name="bob" password="bob"
                      authorities="ROLE_USER"/>
            </user-service>
        </authentication-provider>
    </authentication-manager>

</beans:beans>

I load the xml file in the class, where the public static void main can be found:

@Configuration
@ComponentScan
@EnableAutoConfiguration
@Order(HIGHEST_PRECEDENCE)
@ImportResource({
        "/spring-security.xml"
})
public class PhrobeBootApplication extends SpringBootServletInitializer {
...
}

But I get the following exception on any pageload:

[ERROR] org.apache.catalina.core.ContainerBase.[Tomcat].[localhost].[/].[dispatcherServlet] - Servlet.service() for servlet [dispatcherServlet] in context with path [] threw exception
org.springframework.security.authentication.AuthenticationCredentialsNotFoundException: An Authentication object was not found in the SecurityContext 
...

So it seems like the configuration from resources/WEB-INF/web.xml doesn't load, if I have a good understanding from the documentation, I should use it when using just plain spring, without the boot. (the filters should be configured). Am I right?

Why is this error happens? Is there a better way to use xml based configuration for spring-security in spring-boot? Does web.xml even load by tomcat?

Answer 1

I was getting the same issue then I changed the path of XML file and kept it as in src/main/resources/spring . It's working fine.

@SpringBootApplication

@ImportResource("classpath:/spring/spring-security.xml")

Answer 2

I haven't tried this (because there really isn't anything you can't do with Java configuration), but you would have to eliminate the Spring Boot-provided WebSecurityConfigurers (and @EnableWebSecurity ). I think to do that is maybe more complicated than it needs to be (but then again no-one needs to use XML). You would have to exclude SecurityAutoConfiguration in your @EnableAutoConfiguration and then deal with problems as they arise (you might need a bean of type SecurityProperties still for example, and you probably can't use the Actuator without some more messing about).

I'm not sure what you mean about "configuration from resources/WEB-INF/web.xml " since a) it's a Spring Boot app and b) even if it wasn't there wouldn't be a resources/WEB-INF/web.xml .

Answer 3

According to the statement by Dave Syer in the recent version of spring boot the best way to configure the spring security is with Java configuration.

I needed an SHA-256 encoder, but I haven't find any simple and good solution for implementing one. You just need to configure the jdbcAuthentication with the passwordEncoder. This is really really simple:

@EnableWebSecurity
public class SpringSecurityConfigurer extends WebMvcConfigurerAdapter {

    @Override
    public void addViewControllers(ViewControllerRegistry registry) {
        registry.addViewController("/login").setViewName("login");
    }

    @Bean
    public ApplicationSecurity applicationSecurity() {
        return new ApplicationSecurity();
    }

    @Order(SecurityProperties.ACCESS_OVERRIDE_ORDER)
    protected static class ApplicationSecurity extends WebSecurityConfigurerAdapter {

        @Autowired
        private SecurityProperties security;

        @Autowired
        private DataSource dataSource;

        @Override
        protected void configure(HttpSecurity http) throws Exception {
            http.authorizeRequests().antMatchers("/css/**").permitAll().anyRequest().fullyAuthenticated()
                    .and().formLogin().loginPage("/login").failureUrl("/login?error").permitAll()
                    .and().logout().logoutRequestMatcher(new AntPathRequestMatcher("/logout")).logoutSuccessUrl("/login");
        }

        PasswordEncoder sha256PasswordEncoder = new PasswordEncoder() {
            @Override
            public String encode(CharSequence rawPassword) {
                return Hashing.sha256().hashString(rawPassword, Charsets.UTF_8).toString();
            }

            @Override
            public boolean matches(CharSequence rawPassword, String encodedPassword) {
                return encodedPassword.equals(Hashing.sha256().hashString(rawPassword, Charsets.UTF_8).toString());
            }
        };

        @Override
        public void configure(AuthenticationManagerBuilder auth) throws Exception {
            auth.jdbcAuthentication()
                    .dataSource(this.dataSource)
                    .passwordEncoder(sha256PasswordEncoder);
        }

    }

}