Login/logout in REST with Spring 3

Question

We are developing RESTful webservices with Spring 3 and we need to have the functionality of login/logout, something like /webservices/login/<username>/<password>/ and /webservices/logout. The session should be stored in the context until the session is timed out or logged out to allow consumption of other webservices. Any request to access webservices without session information should be rejected. Looking for state-of-the-art solution for this scenario.

I am actually resurrecting the question asked here Spring Security 3 programmatically login, which is still not properly answered. Please specify the changes needed in web.xml as well.

Solution

I would suggest defining your Spring Security filters completely manually. It's not that difficult, and you get full control over your login/logout behaviour.

First of all, you will need standard web.xml blurb to delegate filter chain handling to Spring (remove async-supported if you are not on Servlet API ver 3):

<filter>
    <filter-name>springSecurityFilterChain</filter-name>
    <async-supported>true</async-supported>
    <filter-class>
        org.springframework.web.filter.DelegatingFilterProxy
    </filter-class>
</filter>



<filter-mapping>
    <filter-name>springSecurityFilterChain</filter-name>
    <url-pattern>/*</url-pattern>
</filter-mapping>

Now, in security context you will define filters separately for each path. Filters can authenticate user, log out user, check security credentials etc.

<bean id="springSecurityFilterChain" class="org.springframework.security.web.FilterChainProxy">
    <sec:filter-chain-map path-type="ant">
        <sec:filter-chain pattern="/login" filters="sif,wsFilter"/>
        <sec:filter-chain pattern="/logout" filters="sif,logoutFilter" />
        <sec:filter-chain pattern="/rest/**" filters="sif,fsi"/>
    </sec:filter-chain-map>
</bean>

The XML above tells Spring to pass requests to specific context-relative URLs through filter chains. First thing in any of the filter chains is establishing security context - 'sif' bean takes care of that.

<bean id="sif" class="org.springframework.security.web.context.SecurityContextPersistenceFilter"/>

Next filter in chain can now either add data to the security context (read: log in/log out user), or make a decision as to whether allow access based on said security context.

For your login URL you will want a filter that reads authentication data from the request, validates it, and in turn stores it in security context (which is stored in session):

<bean id="wsFilter" class="my.own.security.AuthenticationFilter">
  <property name="authenticationManager" ref="authenticationManager"/>
  <property name="authenticationSuccessHandler" ref="myAuthSuccessHandler"/>
  <property name="passwordParameter" value="pass"></property>
  <property name="usernameParameter" value="user"></property>
  <property name="postOnly" value="false"></property>

You can use Spring generic UsernamePasswordAuthenticationFilter but the reason I use my own implementation is to continue filter chain processing (default implementation assumes user will get redirected on successful auth and terminates filter chain), and being able to process authentication every time username and password is passed to it:

public class MyAuthenticationFilter extends UsernamePasswordAuthenticationFilter {

@Override
protected boolean requiresAuthentication(HttpServletRequest request, HttpServletResponse response) {
    return ( StringUtils.hasText(obtainUsername(request)) && StringUtils.hasText(obtainPassword(request)) );
}

@Override
protected void successfulAuthentication(HttpServletRequest request, HttpServletResponse response, FilterChain chain,
        Authentication authResult) throws IOException, ServletException{
    super.successfulAuthentication(request, response, chain, authResult);
    chain.doFilter(request, response);
}

You can add any number of your own filter implementations for /login path, such as authentication using HTTP basic auth header, digest header, or even extract username/pwd from the request body. Spring provides a bunch of filters for that.

I have my own auth success handler who overrides the default redirect strategy:

public class AuthenticationSuccessHandler extends SimpleUrlAuthenticationSuccessHandler {

   @PostConstruct
   public void afterPropertiesSet() {
       setRedirectStrategy(new NoRedirectStrategy());
   }

    protected class NoRedirectStrategy implements RedirectStrategy {

        @Override
        public void sendRedirect(HttpServletRequest request,
                HttpServletResponse response, String url) throws IOException {
            // no redirect

        }

    }

}

You don't have to have custom auth success handler (and probably custom auth filter as well) if you're ok with user being redirected after successful login (redirect URL can be customized, check docs)

Define authentication manager who will be responsible for retrieving user's details:

<sec:authentication-manager alias="authenticationManager">
    <sec:authentication-provider ref="myAuthAuthProvider"/>
</sec:authentication-manager>

 <bean id="myAuthAuthProvider" class="org.springframework.security.web.authentication.preauth.PreAuthenticatedAuthenticationProvider">
    <property name="preAuthenticatedUserDetailsService">
        <bean id="userDetailsServiceWrapper" class="org.springframework.security.core.userdetails.UserDetailsByNameServiceWrapper">
            <property name="userDetailsService" ref="myUserDetailsImpl"/>
        </bean>
    </property>
</bean>

You will have to provide your own user details bean implementation here.

Logout filter: responsible for clearing security context

<bean id="logoutFilter" class="org.springframework.security.web.authentication.logout.LogoutFilter">
    <constructor-arg>
        <list>
            <bean class="org.springframework.security.web.authentication.logout.SecurityContextLogoutHandler"/>
        </list>
    </constructor-arg>
</bean>

Generic authentication stuff:

<bean id="httpRequestAccessDecisionManager" class="org.springframework.security.access.vote.AffirmativeBased">
    <property name="allowIfAllAbstainDecisions" value="false"/>
    <property name="decisionVoters">
        <list>
            <ref bean="roleVoter"/>
        </list>
    </property>
</bean>

<bean id="roleVoter" class="org.springframework.security.access.vote.RoleVoter"/>

<bean id="securityContextHolderAwareRequestFilter" class="org.springframework.security.web.servletapi.SecurityContextHolderAwareRequestFilter"/>

Access control filter (should be self-explanatory):

<bean id="fsi" class="org.springframework.security.web.access.intercept.FilterSecurityInterceptor">
    <property name="authenticationManager" ref="myAuthenticationManager"/>
    <property name="accessDecisionManager" ref="httpRequestAccessDecisionManager"/>
    <property name="securityMetadataSource">
        <sec:filter-invocation-definition-source>
            <sec:intercept-url pattern="/rest/**" access="ROLE_REST"/>
        </sec:filter-invocation-definition-source>
    </property>
</bean>

You should also be able to secure your REST services with @Secured annotations on methods.

Context above was plucked from existing REST service webapp - sorry for any possible typos.

It is also possible to do at least most of what is implemented here by using stock sec Spring tags, but I prefer custom approach as that gives me most control.

Hope this at least gets you started.

Comments

[FKhan]

Thank you so much for this convincing and detailed answer.

[Franklin]

How would you invoke the Logout URL to logout a user?

[rootkit]

@Franklin, note the paths definitions on springSecurityFilterChain bean: path /logout is set to invoke logoutFilter, which will clear the security context and log out the user

[msangel]

best answer about spring sec flow controll

[OhadR]

great and detailed answer. as for the logout - i did it a bit different, i think it worth a look: codeproject.com/Tips/521847/Logout-Spring-s-LogoutFilter

[We are Borg]

Hi, I have implemented your solution, it does not give any errors, but I don't know how to use it, can you please check this link stackoverflow.com/questions/29276806/…