How can I point Spring MVC to our database for security? - spring-mvc

Question

I'm new to Spring. I just recently got an example working with annotations and basic security. Right now, we have to define our users in the security-app-context.xml file. What we want to do is connect to our existing database and use the users there.
Our user passwords are stored as such (MD5):
username salt password
moe blah bb3e4e328a64e9745a98728468aacbb0
The user is moe and the password is howard so we use: md5(moehowardblah) or md5(Username + Password + Salt)
Here is our web.xml:
<?xml version="1.0" encoding="UTF-8"?>
<web-app version="2.5" xmlns="http://java.sun.com/xml/ns/javaee"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd">
<!--
- Location of the XML file that defines the root application context
- Applied by ContextLoaderListener.
-->
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>
/WEB-INF/applicationContext-business.xml
/WEB-INF/security-app-context.xml
</param-value>
</context-param>
<filter>
<filter-name>springSecurityFilterChain</filter-name>
<filter-class>org.springframework.web.filter.DelegatingFilterProxy</filter-class>
</filter>
<filter-mapping>
<filter-name>springSecurityFilterChain</filter-name>
<url-pattern>/*</url-pattern>
</filter-mapping>
<listener>
<listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>
<!-- Processes application requests -->
<servlet>
<servlet-name>dispatcher</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<init-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/dispatcher-context.xml</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>dispatcher</servlet-name>
<url-pattern>/</url-pattern>
</servlet-mapping>
</web-app>
Our security-app-context.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"
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-3.1.xsd">
<http use-expressions="true">
<intercept-url pattern="/assets/**" access="isAuthenticated()" />
<intercept-url pattern="/**" access="permitAll" />
<form-login />
<logout />
</http>
<authentication-manager>
<authentication-provider>
<user-service>
<user name="rod" password="koala" authorities="supervisor, teller, user" />
<user name="dianne" password="emu" authorities="teller, user" />
<user name="scott" password="wombat" authorities="user" />
<user name="peter" password="opal" authorities="user" />
</user-service>
</authentication-provider>
</authentication-manager>
</beans:beans>
Also, the example I have is using the built-in form for the user/pass which is fine for now but I will need to know how to use our own custom login. But I can save that for another question.
Thanks.

Answer 1

All you need - is to create your own userDetailsService implementation. As you already have existing database, I assume, you have some kind of service or DAO with takes data from DB.
Extend this service from UserDetailsService interface and override the only method:
UserDetails loadUserByUsername(String username) throws UsernameNotFoundException;
This assuming that your User object also implement the UserDetails interface.
Aftre this you only need to point spring use your service:
<authentication-manager>
<authentication-provider user-service-ref="yourUserService" />
</authentication-manager>
Where is "yourUserService" — іs a reference to your userDetailsService implementation bean.