Spring Security avoid create a table for roles - spring

Question

I'm learning Spring Security. I have my login system ready and I want to add roles. I've seen many tutorials and docs about it and I couldn't find what I'm looking for.
I don't want to create an extra table for Roles, because my table user has a column named "type" and I want to use it for authorization. The value of that column can be "person", "teacher" or "organization". So, I want to based the role system on that column, not in a OneToMany o ManyToMany relationship with a table named "role".
How can I configure that?
Thanks
UPDATED
I forgot, I'm using Spring Data. This is the code I'm using
#Configuration
#EnableWebSecurity
public class SpringSecurityConfig extends WebSecurityConfigurerAdapter {
private AuthenticationProvider authenticationProvider;
#Autowired
#Qualifier("daoAuthenticationProvider")
public void setAuthenticationProvider(AuthenticationProvider authenticationProvider) {
this.authenticationProvider = authenticationProvider;
}
#Bean
public PasswordEncoder passwordEncoder(BCryptPasswordEncoder passwordEncoder){
return passwordEncoder;
}
#Bean
public DaoAuthenticationProvider daoAuthenticationProvider(BCryptPasswordEncoder passwordEncoder,
UserDetailsService userDetailsService){
DaoAuthenticationProvider daoAuthenticationProvider = new DaoAuthenticationProvider();
daoAuthenticationProvider.setPasswordEncoder(passwordEncoder);
daoAuthenticationProvider.setUserDetailsService(userDetailsService);
return daoAuthenticationProvider;
}
#Override
protected void configure(HttpSecurity http) throws Exception {
http.csrf().ignoringAntMatchers("/h2-console").disable()
.authorizeRequests().antMatchers("/").authenticated()
.antMatchers("/console/**").permitAll()
.antMatchers("/static/**").permitAll()
.antMatchers("/profile").hasAuthority("PERSON")
.and().formLogin().loginPage("/login").permitAll()
.and().exceptionHandling().accessDeniedPage("/login")
.and().logout().permitAll()
http.headers().frameOptions().disable();
}
#Autowired
public void configureAuthManager(AuthenticationManagerBuilder authenticationManagerBuilder) throws Exception{
authenticationManagerBuilder
.jdbcAuthentication().authoritiesByUsernameQuery("select type from users where username = ?").and()
.authenticationProvider(authenticationProvider);
}
}

Answer 1

You can define a UserDetailsService with a PasswordEncoder in Java Config like following:
#Configuration
#EnableWebSecurity
public class SecurityConfig extends WebSecurityConfigurerAdapter {
#Autowired private PersonRepository personRepository;
#Override
#Autowired
protected void configure(AuthenticationManagerBuilder auth) throws Exception {
auth
.userDetailsService(username -> {
Person person = personRepository.findByUsername(username);
if (person == null) throw new UsernameNotFoundException("Invalid user");
return new User(person.getUsername(),
person.getPassword(),
Collections.singleton(new SimpleGrantedAuthority(person.getType())));
})
.passwordEncoder(passwordEncoder())
}
// Rest of the configuration
}
In the above example, i supposed you have a PersonRespository that has access to your user information. With this UserDetailsService you won't need your AuthenticationProvider. Also, User resides in org.springframework.security.core.userdetails package.