Question

在我的项目中，人们有基于角色的访问权限。一个人可以在多个部门工作。

我的角色表

Role_id Role
1       Manager
2       Employee

我的部门表

Departmant_id Departmant
1             Production
2             Research
3             Marketing

我的用户表

User_id User_name
1       Jennifer
2       Kate
3       David

我想要的是一个新表格，用于指定哪些人处于何种状态以及他们在该部门中扮演的角色。

User_id Departmant_id Role_id
x       x             x

我尝试的是

Class User{
     @ManyToOne(cascade = CascadeType.ALL)
     @JoinTable(name = "user_department_role",joinColumns = {@JoinColumn(name = "department_id",referencedColumnName = "department_id"),@JoinColumn(name = "user_id",referencedColumnName = "user_id")}, inverseJoinColumns = {@JoinColumn(name = "role_id")})
     private Set<Department> departmentList;
}

Answer 1

你需要一个关联表，通常是在JPA中构建的，原因有多种，主要是为了控制表中的内容，或者在这种情况下映射n-way M：N关系。

创建所有实体：

@Entity
public class User {
    @Id @GeneratedValue(strategy=GenerationType.AUTO) 
    private Integer id;
    private String userName;
    @OneToMany(mappedBy="user")
    private Set<UserDepartmentRoleAssociation> associations;
... etc
}

和

@Entity
public class Department {
    @Id @GeneratedValue(strategy=GenerationType.AUTO) 
    private Integer id;
    private String department;
    @OneToMany(mappedBy="department")
    private Set<UserDepartmentRoleAssociation> associations;
    ... etc
}

和

@Entity
public class Role {
    @Id @GeneratedValue(strategy=GenerationType.AUTO) 
    private Integer id;
    private String role;
    ... etc
}

并创建关联表和id类。

@Entity
public class UserDepartmentRoleAssociation {
    @EmbeddedId private UserDepartmentRoleAssociationId id;
    @ManyToOne @MapsId("userId")
    private User user;
    @ManyToOne @MapsId("departmentId")
    private Department department;
    @ManyToOne @MapsId("roleId")
    private Role role;
    public UserDepartmentRoleAssociation() {
        id = new UserDepartmentRoleAssociationId();
    }
    ... etc
}

和

@Embeddable
public class UserDepartmentRoleAssociationId implements Serializable {
    private Integer userId;
    private Integer departmentId;
    private Integer roleId;
    ... etc
}

然后坚持关系...

        User user = new User();
        user.setUserName("user1");

        Department department = new Department();
        department.setDepartment("department 1");

        Role role = new Role();
        role.setRole("Manager");

        UserDepartmentRoleAssociation association = new UserDepartmentRoleAssociation();
        association.setUser(user);
        association.setDepartment(department);
        association.setRole(role);

        em.persist(user);
        em.persist(department);
        em.persist(role);
        em.persist(association);

然后使用join fetch读取它

User user = em.createQuery("select u from User u left join fetch u.associations ass left join fetch ass.department left join fetch ass.role where u.id = :id", User.class).setParameter("id", 1).getSingleResult();

请注意，我在Set和List中使用Department而不是User，这导致在这些情况下出现的问题要少得多。此外，当我坚持关系时，我不必创建associations，因为UserDepartmentRoleAssociation是拥有实体，因此持久化。 associations集是由JPA在读取记录时创建的。

2个外键进入不同实体的新表Hibernate

1 个答案: