Thalaivar Thalaivar - 1 month ago 18
Java Question

Spring login form example

I tried searching in Google, but I could not find any good examples where a username and password are checked with a database for authentication purposes.

In further simple words, how can I create a simple login form using Spring and Hibernate where the credentials are checked with the database.

Update

Cam anyone come up with a simple example where I can see how the flow goes and how the input data is passed to hibernate?

Answer

At first you should define this file WEB-INF/spring/serurity-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-2.0.xsd
                                 http://www.springframework.org/schema/security http://www.springframework.org/schema/security/spring-security-2.0.1.xsd">

    <http auto-config="true" />

    <beans:bean id="myUserService" class="org.my.UserService" />
    <authentication-provider user-service-ref="myUserService" />

</beans:beans>

Now you should create org.my.UserService class and implement interface org.springframework.security.core.userdetails.UserDetailsService. This interface has one method:

UserDetails loadUserByUsername(String username) throws UsernameNotFoundException, org.springframework.dao.DataAccessException

And in this method you can use Hibernate in order to load user by userName. If user does not exists - just throw UsernameNotFoundException, otherwise return new intialized UserDetails instance (there you can provide a lot of stuff like user roles, account expiration date, etc...).

Now comes web.xml:

<web-app 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"
         version="2.5">

    <display-name>My Webapp</display-name>

    <context-param>
        <param-name>contextConfigLocation</param-name>
        <param-value>
            /WEB-INF/spring/*-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>

    <servlet>
        <servlet-name>dispatcher</servlet-name>
        <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
        <load-on-startup>1</load-on-startup>
    </servlet>

    <servlet-mapping>
        <servlet-name>dispatcher</servlet-name>
        <url-pattern>/*</url-pattern>
    </servlet-mapping>

</web-app>

If you have any questions or something goes wrong, feel free to ask :)

PS: So with UserDetailsService you don't have to check password of whether user account is active, etc. You just provide spring-security information about user with provided userName and framework validates user itself. If you encode your passwords with MD5 for example, than you can use password-encoder like this:

<beans:bean id="myUserService" class="org.my.UserService" />
<authentication-provider user-service-ref="myUserService">
    <password-encoder hash="md5"/>
</authentication-provider>

Update

Now we will dive more deeper in UserService - my (simplified) real world example.

UserService class:

import org.my_company.my_app.domain.User

public class UserService implements UserDetailsService {
    private UserDao userDao;

    public void setUserDao(UserDao userDao) {
        this.userDao = userDao;
    }

    public UserDetails loadUserByUsername(String username) throws UsernameNotFoundException, DataAccessException {
        // load user
        User user = userDao.getUser(username);

        if (user != null) {

            // convert roles
            List<GrantedAuthority> roles = new ArrayList<GrantedAuthority>();
            for (Privilege p : user.getPrivileges()) {
                roles.add(new GrantedAuthorityImpl(p.getName()));
            }

            // initialize user
            SecurityUser securityUser = new SecurityUser(
                user.getUsername(),
                user.getLdapAuth() ? getLdapPassword(user.getUsername()) : user.getPassword(),
                user.getStatus() != User.Status.NOT_COMMITED, user.getStatus() != User.Status.BLOCKED, true, true,
                roles.toArray(new GrantedAuthority[0])
            );

            securityUser.setUser(user);

            return securityUser;
        } else {
            throw new UsernameNotFoundException("No user with username '" + username + "' found!");
        }
    }
}

Now SecurityUser:

import org.my_company.my_app.domain.User

public class SecurityUser extends org.springframework.security.core.userdetails.User {

    private User user;

    public User getUser() {
        return user;
    }

    public void setUser(User user) {
        this.user = user;
    }

    public SecurityUser(String username, String password, boolean enabled, boolean accountNonExpired, boolean credentialsNonExpired, boolean accountNonLocked, GrantedAuthority[] authorities) throws IllegalArgumentException {
        super(username, password, enabled, accountNonExpired, credentialsNonExpired, accountNonLocked, authorities);
    }
}

And finally UserDao:

import org.my_company.my_app.domain.User

public class UserDao extends HibernateDaoSupport {

    public User getUser(String username) {
        List users = getHibernateTemplate().find("from User where username = ?", username);
        return users == null || users.size() <= 0 ? null : (User) users.get(0);
    }
}

As you can see I used HibernateTemplate here.