小编典典

2个外键从不同实体进入新表休眠

spring-boot

在我的项目中,人员具有基于角色的访问权限。一个人可以在多个部门中工作。

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,该表指定哪些人在哪个部门中以及他们在该部门中所扮演的角色。

User_id Departmant_id Role_id
x       x             x

我试过的是

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;
}

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2020-05-30

共1个答案

小编典典

您需要一个关联表,通常出于各种原因通常在JPA中构建该关联表,主要是为了控制表中的内容,或者在这种情况下映射n路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
}

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

@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
}

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

并保持一段感情然后…

        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);

并通过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,在这些情况下,我使用了in Set而不是Listin
DepartmentUser它引起的问题要少得多。另外,associations当我保留关系时,我也不必创建,因为它UserDepartmentRoleAssociation是拥有实体,因此也可以保留。这些associations集是由JPA在读取记录时创建的。

2020-05-30