没有为 id“null"映射的 PasswordEncoder带数据库认证 [英] There is no PasswordEncoder mapped for the id "null" with database authentication

问题描述

我成功构建了内存中的身份验证.但是当我打算用数据库构建它时会出现这个错误.

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

没有为 id "null" 映射的 PasswordEncoder

There is no PasswordEncoder mapped for the id "null"

这是遵循教程 - Spring Boot 初学者教程，10 - 使用 Spring Security 的高级身份验证|强大的Java

有课

@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();
}

}

@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 接口

public interface UserRepository extends JpaRepository<User, Long>{

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

}

角色.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
}

用户.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}

如果您有任何想法，请告知.谢谢.

If you have any idea plese inform. Thank you.

推荐答案

我更改了 MyUserDetailsS​​ervice 类，添加了 passwordEncoder 方法.

I changed MyUserDetailsService class adding passwordEncoder method.

添加行

BCryptPasswordEncoder encoder = passwordEncoder();

换行

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

MyUserDetailsS​​ervice.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();
}

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