Spring security Userdetails 无法转换为我自己的用户实现答案

【问题标题】：Spring security Userdetails cannot be cast to my own user implementationSpring security Userdetails 无法转换为我自己的用户实现
【发布时间】：2021-12-20 20:47:27
【问题描述】：

我有一个User 类，它实现了弹簧安全UserDetails。但是，在对登录请求进行身份验证期间，它会给出一个 ClassCastException，即不能将 UserDetails 强制转换为我的 User 类。

完整的错误：

java.lang.ClassCastException: 类 org.springframework.security.core.userdetails.User 不能被强制转换为类 nl.teamrepositories.vliegmaatschappij.security.domain.User （org.springframework.security.core.userdetails.User 未命名加载器“app”的模块； nl.teamrepositories.vliegmaatschappij.security.domain.User 在 loader的未命名模块 org.springframework.boot.devtools.restart.classloader.RestartClassLoader @6ef8cb6)

用户：

package nl.teamrepositories.vliegmaatschappij.security.domain;

import lombok.Getter;
import lombok.NoArgsConstructor;
import lombok.Setter;
import org.springframework.security.core.GrantedAuthority;
import org.springframework.security.core.authority.SimpleGrantedAuthority;
import org.springframework.security.core.userdetails.UserDetails;

import javax.persistence.*;
import java.util.Collection;
import java.util.List;

@Entity
@Table(name = "users")
@Getter @Setter @NoArgsConstructor
public class User implements UserDetails {

    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    private Long id;

    @Column(nullable = false, unique = true)
    private String firstName;
    private String lastName;
    private String username;
    private String password;

    private boolean enabled;
    private boolean tokenExpired;

    public User(String firstName, String lastName, String username, String password) {
        this.firstName = firstName;
        this.lastName = lastName;
        this.username = username;
        this.password = password;
    }

    @ManyToMany
    @JoinTable(
            name = "users_roles",
            joinColumns = @JoinColumn(
                    name = "user_id", referencedColumnName = "id"),
            inverseJoinColumns = @JoinColumn(
                    name = "role_id", referencedColumnName = "id"))
    private Collection<Role> roles;

    @Override
    public Collection<? extends GrantedAuthority> getAuthorities() {
        return List.of(new SimpleGrantedAuthority("ROLE_USER"));
    }

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

    @Override
    public boolean isAccountNonExpired() {
        return true;
    }

    @Override
    public boolean isAccountNonLocked() {
        return true;
    }

    @Override
    public boolean isCredentialsNonExpired() {
        return true;
    }
    
    @Override
    public boolean isEnabled() {
        return true;
    }
}

在 JwtAuthenticationFilter 中：

@Override
protected void successfulAuthentication(HttpServletRequest request, HttpServletResponse response,
                                        FilterChain filterChain, Authentication authentication) {
    User user = (User) authentication.getPrincipal(); // this is where the error occurs
    // more code
}

我不明白为什么我的 User 类不能转换为 Spring 的 UserDetails 类。

我应该改变什么？

提前致谢。

【问题讨论】：

标签： java spring authentication spring-security jwt

【解决方案1】：

在JwtAuthenticationFilter 中检查您的导入。我猜你正在使用import org.springframework.security.core.userdetails.User 而不是你自己的班级import nl.teamrepositories.vliegmaatschappij.security.domain.User：

import nl.teamrepositories.vliegmaatschappij.security.domain.User;
(...)

public class JwtAuthenticationFilter {
    (...)

    @Override
    protected void successfulAuthentication(HttpServletRequest request, HttpServletResponse response,
                                            FilterChain filterChain, Authentication authentication) {
        User user = (User) authentication.getPrincipal(); // this is where the error occurs
        // more code
    }
}

无论是在这里还是在代码中的其他地方，您都使用了错误的 User 类（Spring 类而不是您的类）。

【讨论】：