User Details Service integration to the gateway.

[Randika Hapugoda]

In 'double' project we use

auth.inMemoryAuthentication()
                .withUser("user").password("password").roles("USER")
            .and()
                .withUser("admin").password("admin").roles("USER", "ADMIN", "READER", "WRITER")
            .and()
                .withUser("audit").password("audit").roles("USER", "ADMIN", "READER");

and we can login users with these details. But when we add

auth.userDetailsService(customUserDetailsService).passwordEncoder(passwordEncoder());

it gives an error with

java.lang.ClassNotFoundException: <pakage>.CustomUserDetailsService$UserRepositoryUserDetails

How to solve this issue. Or even get user details from database?

[lzy]

this is my way:

@Service
public class CustomUserDetailsService implements UserDetailsService
{
    @Autowired
    private UserRepository userRepository;

    @Override
    public UserDetails loadUserByUsername(String userName)
            throws UsernameNotFoundException {


        UsersEntity usersEntity = userRepository.findOne(userName);

        if(usersEntity == null){
            throw new UsernameNotFoundException("UserName "+userName+" not found");
        }

        SecurityUser user = new SecurityUser();
        user.setUsername(usersEntity.getUsername());
        user.setPassword(usersEntity.getPassword());
        user.setEnabled(usersEntity.isEnabled());
        return user;
    }

@Entity
@Table(name = "users", schema = "", catalog = "spring_security")
public class UsersEntity {
    private String username;
    private String password;
    private boolean enabled;

    @Id
    @Column(name = "username", nullable = false, insertable = true, updatable = true, length = 50)
    public String getUsername() {
        return username;
    }

    public void setUsername(String username) {
        this.username = username;
    }

    @Basic
    @Column(name = "password", nullable = false, insertable = true, updatable = true, length = 50)
    public String getPassword() {
        return password;
    }

    public void setPassword(String password) {
        this.password = password;
    }

    @Basic
    @Column(name = "enabled", nullable = false, insertable = true, updatable = true)
    public boolean isEnabled() {
        return enabled;
    }

    public void setEnabled(boolean enabled) {
        this.enabled = enabled;
    }

    @Override
    public boolean equals(Object o) {
        if (this == o) return true;
        if (o == null || getClass() != o.getClass()) return false;

        UsersEntity that = (UsersEntity) o;

        if (enabled != that.enabled) return false;
        if (username != null ? !username.equals(that.username) : that.username != null) return false;
        if (password != null ? !password.equals(that.password) : that.password != null) return false;

        return true;
    }

    @Override
    public int hashCode() {
        int result = username != null ? username.hashCode() : 0;
        result = 31 * result + (password != null ? password.hashCode() : 0);
        result = 31 * result + (enabled ? 1 : 0);
        return result;
    }
public interface UserRepository extends CrudRepository<UsersEntity, String> {



}

I use spring data jpa ,you must design the db for user table that have

mysql:

CREATE TABLE `users` (
  `username` varchar(50) NOT NULL,
  `password` varchar(60) NOT NULL,
  `enabled` bit(1) NOT NULL,
  PRIMARY KEY (`username`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;

2.this is get password way

        String password = "123456";
        BCryptPasswordEncoder passwordEncoder = new BCryptPasswordEncoder();
        System.out.println(passwordEncoder.encode(password));

user password(123456)
admin $2a$10$qeW7Nv.L53Et7.UOTJPipeIUQIvaOH8yw76FWbe2xy.qhzeRlZ0O6 1

[Dave Syer]

Seems like a general question on Spring Security. Stack Overflow probably best source for answer.

I'll think about adding a new blog on custom user details services.