Spring Starter安全性未验证

Spring Starter安全性未验证,spring,spring-mvc,spring-boot,spring-security,spring-data-jpa,Spring,Spring Mvc,Spring Boot,Spring Security,Spring Data Jpa,当我添加具有“角色\用户”权限的用户时,我无法进行身份验证。尝试使用用户名:“username”和密码“password”进行身份验证时,会一致返回401 我可以在JSON的输出中看到BCryptPasswordEncoder正在按其应该的方式编码密码,但无论我使用的是原始密码还是编码版本,我仍然无法进行身份验证 我已经为此工作了几天,但都没有用。我有什么遗漏吗 代码如下-- 数据库加载器: User user = new User("first", "last", "username", "p

当我添加具有“角色\用户”权限的用户时,我无法进行身份验证。尝试使用用户名:“username”和密码“password”进行身份验证时,会一致返回401

我可以在JSON的输出中看到BCryptPasswordEncoder正在按其应该的方式编码密码,但无论我使用的是原始密码还是编码版本,我仍然无法进行身份验证

我已经为此工作了几天,但都没有用。我有什么遗漏吗

代码如下--

数据库加载器:

User user = new User("first", "last", "username", "password", "email", "phone", new String[] {"ROLE_USER"});
userRepository.save(user);
详细信息服务:

@Component
public class DetailsService implements UserDetailsService {
    @Autowired
    UserRepository users;

    @Override
    public UserDetails loadUserByUsername(String userUsername) throws UsernameNotFoundException {
        User user = users.findByUsername(userUsername);
        if (user == null) {
            throw new UsernameNotFoundException(userUsername + " was not found");
        }
        return new org.springframework.security.core.userdetails.User(
                user.getUsername(),
                user.getUserPassword(),
                AuthorityUtils.createAuthorityList(user.getUserRoles())
        );
    }
}
Web安全配置:

@Configuration
@EnableWebSecurity
@EnableGlobalMethodSecurity(prePostEnabled = true)
public class WebSecurityConfiguration extends WebSecurityConfigurerAdapter {
    @Autowired
    DetailsService userDetailsService;

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

    @Override
    protected void configure(HttpSecurity http) throws Exception {
        http
                .authorizeRequests()
                    .anyRequest().authenticated()
                .and()
                .httpBasic()
                .and()
                .csrf().disable();
    }
}
用户:


你的问题对这个问题不是很清楚。但我猜你被SpringStarter安全性的用户身份验证卡住了。
你应该检查这个

你是对的,这个问题不够具体。它与密码编码器更相关-将重新措辞并提出一个新问题。谢谢它确实帮助我了解了更多关于SpringStarter安全性的知识。我的问题具体到密码编码器——编码的密码永远不会匹配。我完全放弃了它,现在使用的是存储在字符串中的纯文本密码。将在需要时进一步研究密码加密。谢谢
@Entity
public class User {
    public static final PasswordEncoder PASSWORD_ENCODER = new BCryptPasswordEncoder();
    private long userId;
    private String userFirstName;
    private String userLastName;
    private String username;
    @JsonIgnore
    private String userPassword;
    private String userPhone;
    private String userEmail;
    @JsonIgnore
    private String[] userRoles;

    public User() {}

    public User(String userFirstName, String userLastName, String username, String userPassword, String userPhone, String userEmail, String[] userRoles) {
        this.userFirstName = userFirstName;
        this.userLastName = userLastName;
        this.username = username;
        setUserPassword(userPassword);
        this.userPhone = userPhone;
        this.userEmail = userEmail;
        this.userRoles = userRoles;
    }

    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    public long getUserId() {
        return userId;
    }

    public void setUserId(long userId) {
        this.userId = userId;
    }

    @Column
    public String getUserFirstName() {
        return userFirstName;
    }

    public void setUserFirstName(String userFirstName) {
        this.userFirstName = userFirstName;
    }

    @Column
    public String getUserLastName() {
        return userLastName;
    }

    public void setUserLastName(String userLastName) {
        this.userLastName = userLastName;
    }

    @Column
    public String getUsername() {
        return username;
    }

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

    @Column
    public String getUserPassword() {
        return userPassword;
    }

    public void setUserPassword(String userPassword) {
        this.userPassword = PASSWORD_ENCODER.encode(userPassword);
    }

    @Column
    public String getUserPhone() {
        return userPhone;
    }

    public void setUserPhone(String userPhone) {
        this.userPhone = userPhone;
    }

    @Column
    public String getUserEmail() {
        return userEmail;
    }

    public void setUserEmail(String userEmail) {
        this.userEmail = userEmail;
    }

    @Column
    public String[] getUserRoles() {
        return userRoles;
    }

    public void setUserRoles(String[] userRoles) {
        this.userRoles = userRoles;
    }
}