通过数据库身份验证，没有映射ID为“ null”的PasswordEncoder

Question

I successfully build in-memory authentication. But when I going to build it with database comes this error.

There is no PasswordEncoder mapped for the id "null"

This is followed tutorial - Spring Boot Tutorial for Beginners, 10 - Advanced Authentication using Spring Security | Mighty Java

There are classes

SpringSecurityConfiguration.java

@Configuration
@EnableWebSecurity
public class SpringSecurityConfiguration extends 
WebSecurityConfigurerAdapter{

@Autowired
private AuthenticationEntryPoint entryPoint;

@Autowired
private MyUserDetailsService userDetailsService;

@Override
protected void configure(AuthenticationManagerBuilder auth) throws Exception {
    auth.userDetailsService(userDetailsService);
}

@Override
protected void configure(HttpSecurity http) throws Exception {
    http.authorizeRequests().anyRequest().authenticated().and().httpBasic()
        .authenticationEntryPoint(entryPoint);
}

}

AuthenticationEntryPoint.java

@Configuration
public class AuthenticationEntryPoint extends BasicAuthenticationEntryPoint{


@Override
public void commence(HttpServletRequest request, HttpServletResponse response,
        AuthenticationException authException) throws IOException, ServletException {

    response.addHeader("WWW-Authenticate", "Basic realm -" +getRealmName());
    response.setStatus(HttpServletResponse.SC_UNAUTHORIZED);
    PrintWriter writer = response.getWriter();
    writer.println("Http Status 401 "+authException.getMessage());
}

@Override
public void afterPropertiesSet() throws Exception {
    setRealmName("MightyJava");
    super.afterPropertiesSet();
}

}

MyUserDetailsService .java

@Service
public class MyUserDetailsService implements UserDetailsService{

@Autowired
private UserRepository userRepository;

@Override
public UserDetails loadUserByUsername(String username) throws UsernameNotFoundException {
    User user = userRepository.findByUsername(username);
    if(user == null){
        throw new UsernameNotFoundException("User Name "+username +"Not Found");
    }
    return new org.springframework.security.core.userdetails.User(user.getUserName(),user.getPassword(),getGrantedAuthorities(user));
}

private Collection<GrantedAuthority> getGrantedAuthorities(User user){

    Collection<GrantedAuthority> grantedAuthority = new ArrayList<>();
    if(user.getRole().getName().equals("admin")){
        grantedAuthority.add(new SimpleGrantedAuthority("ROLE_ADMIN"));
    }
    grantedAuthority.add(new SimpleGrantedAuthority("ROLE_USER"));
    return grantedAuthority;
}
}

UserRepository interface

public interface UserRepository extends JpaRepository<User, Long>{

@Query("FROM User WHERE userName =:username")
User findByUsername(@Param("username") String username);

}

Role.java

@Entity
public class Role extends AbstractPersistable<Long>{

private String name;

@OneToMany(targetEntity = User.class , mappedBy = "role" , fetch = FetchType.LAZY ,cascade = CascadeType.ALL)
private Set<User> users;

//getter and setter
}

User.java

@Entity
public class User extends AbstractPersistable<Long>{

//AbstractPersistable class ignore primary key and column annotation(@Column)

private String userId;
private String userName;
private String password;

@ManyToOne
@JoinColumn(name = "role_id")
private Role role;

@OneToMany(targetEntity = Address.class, mappedBy = "user",fetch= FetchType.LAZY ,cascade =CascadeType.ALL)
private Set<Address> address; //Instead of Set(Unordered collection and not allow duplicates) we can use list(ordered and allow duplicate values) as well

//getter and setter}

If you have any idea plese inform. Thank you.

Answer 1

I changed MyUserDetailsService class adding passwordEncoder method.

Added lines

BCryptPasswordEncoder encoder = passwordEncoder();

Changed Line

//changed, user.getPassword() as encoder.encode(user.getPassword())
return new org.springframework.security.core.userdetails.User(--)

MyUserDetailsService.java

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

    BCryptPasswordEncoder encoder = passwordEncoder();
    User user = userRepository.findByUsername(username);
    if(user == null){
        throw new UsernameNotFoundException("User Name "+username +"Not Found");
    }
    return new org.springframework.security.core.userdetails.User(user.getUserName(),encoder.encode(user.getPassword()),getGrantedAuthorities(user));
}

@Bean
public BCryptPasswordEncoder passwordEncoder() {
    return new BCryptPasswordEncoder();
}

Answer 2

As from Spring Security 5.x, Spring Security enforces you to use a password encoder if you're working with other than in-memory (production) databases.

Spring Security enforces this by activating the default DelegatingPasswordEncoder , which looks for PasswordEncoder beans. By adding a BCryptPasswordEncoder , the DelegatingPasswordEncoder will return that instance to encrypt passwords.

@Bean
public BCryptPasswordEncoder passwordEncoder() {
    return new BCryptPasswordEncoder();
}

I don't recommend you to do this, but if you really want to, you can override password encoding by adding {noop} to the password value. This will treat the password by activating the NoOpPasswordEncoder instead of the default DelegatingPasswordEncoder and will treat your password as plain text.

Please note that this is not recommended if you deploy your app to a production environment!