如何创建角色用户登录？

Question

我想创建一个角色用户来登录。所以我仍然很困惑如何创建用户角色并使用角色用户登录。我已经制作了一些源代码如下：

User.java：

package com.practice.login.entity;

import javax.persistence.*;
import java.util.Set;

@Entity
@Table(name = "user")
public class User {
    private Long id;
    private String username;
    private String password;
    private String passwordConfirm;
    private Set<Role> roles;

    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    public Long getId() {
        return id;
    }

    public void setId(Long idUser) {
        this.id = idUser;
    }

    public String getUsername() {
        return username;
    }

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

    public String getPassword() {
        return password;
    }

    public void setPassword(String password) {
        this.password = password;
    }

    @Transient
    public String getPasswordConfirm() {
        return passwordConfirm;
    }

    public void setPasswordConfirm(String passwordConfirm) {
        this.passwordConfirm = passwordConfirm;
    }

    @ManyToMany
    @JoinTable(
            name = "user_role", 
            joinColumns = @JoinColumn(
                    name = "user_id"), 
            inverseJoinColumns = @JoinColumn(
                    name = "role_id"))
    public Set<Role> getRoles() {
        return roles;
    }

    public void setRoles(Set<Role> roles) {
        this.roles = roles;
    }
}

Role.java

package com.practice.login.entity;

import javax.persistence.*;
import java.util.Set;

@Entity
@Table(name = "role")
public class Role {
    private Long id;
    private String name;
    private Set<User> users;

    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    public Long getId() {
        return id;
    }

    public void setId(Long id) {
        this.id = id;
    }

    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }

    @ManyToMany(mappedBy = "roles")
    public Set<User> getUsers() {
        return users;
    }

    public void setUsers(Set<User> users) {
        this.users = users;
    }
}

UserService.java：

package com.practice.login.service;

import com.practice.login.entity.User;

public interface UserService {

    void save(User user);

    User findByUsername(String username);

}

UserServiceImpl.java

package com.practice.login.service;

import com.practice.login.entity.User;
import com.practice.login.repository.RoleRepository;
import com.practice.login.repository.UserRepository;
import org.springframework.beans.factory.annotation.Autowired;
import org.springframework.security.crypto.bcrypt.BCryptPasswordEncoder;
import org.springframework.stereotype.Service;

import java.util.HashSet;

@Service
public class UserServiceImpl implements UserService {
    @Autowired
    private UserRepository userRepository;
    @Autowired
    private RoleRepository roleRepository;
    @Autowired
    private BCryptPasswordEncoder bCryptPasswordEncoder;

    @Override
    public void save(User user) {
        user.setPassword(bCryptPasswordEncoder.encode(user.getPassword()));
        user.setRoles(new HashSet<>(roleRepository.findAll()));
        userRepository.save(user);
    }

    @Override
    public User findByUsername(String username) {
        return userRepository.findByUsername(username);
    }
}

UserDetailServiceImpl.java

package com.practice.login.service;

import org.springframework.beans.factory.annotation.Autowired;
import org.springframework.security.core.GrantedAuthority;
import org.springframework.security.core.authority.SimpleGrantedAuthority;
import org.springframework.security.core.userdetails.UserDetails;
import org.springframework.security.core.userdetails.UserDetailsService;
import org.springframework.security.core.userdetails.UsernameNotFoundException;
import org.springframework.stereotype.Service;
import org.springframework.transaction.annotation.Transactional;

import com.practice.login.entity.Role;
import com.practice.login.entity.User;
import com.practice.login.repository.UserRepository;

import java.util.HashSet;
import java.util.Set;

@Service
public class UserDetailsServiceImpl implements UserDetailsService {
    @Autowired
    private UserRepository userRepository;

    @Override
    @Transactional(readOnly = true)
    public UserDetails loadUserByUsername(String username) throws UsernameNotFoundException {
        User user = userRepository.findByUsername(username);
        Set<GrantedAuthority> grantedAuthorities = new HashSet<>();
        for (Role role : user.getRoles()) {
            grantedAuthorities.add(new SimpleGrantedAuthority(role.getName()));
        }

        return new org.springframework.security.core.userdetails.User(user.getUsername(), user.getPassword(),
                grantedAuthorities);
    }
}

UserRepository.java

package com.practice.login.repository;

import org.springframework.data.jpa.repository.JpaRepository;
import org.springframework.stereotype.Repository;

import com.practice.login.entity.User;

@Repository
public interface UserRepository extends JpaRepository<User, Long> {
    User findByUsername(String username);
}

RoleRepository.java

package com.practice.login.repository;

import org.springframework.data.jpa.repository.JpaRepository;
import org.springframework.stereotype.Repository;

import com.practice.login.entity.Role;

@Repository
public interface RoleRepository extends JpaRepository<Role, Long> {
}

如何使用角色权限登录？

Answer 1

这主要取决于您的HttpSecurity配置：

示例SecurityConfig.java：

float voltage2 = batteryIn * 5.0 / 1023.0;