如何在spring引导中为三个不同角色设计类？

Question

我正在开发一个涉及分发管理的spring引导应用程序。有3种类型的用户谁将与系统互动-管理，经销商和零售商。

company.

• Retailer

1. 只能有一个公司的管理员，
2. 在一个特定的城市中只能有一个分销商，可以与特定城市的不同公司的分销商联系在一起。

现在，为了这些东西。我已经创建了5个类-用户、角色、管理员、分销商和零售商。请任何人建议我，在设计所需的类时，我应该如何处理这个问题。我最初尝试过，但我仍然认为这个类的结构本身是复杂的。有什么东西我遗漏了吗?还是有什么设计模式可以使事情变得更不稳定？

@Entity
@Table(name = "tbl_User")
public class User {

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY) 
    @Column(name = "user_id")
    private Long userId;
    
    @NotNull(message = "first Name cannot be null")
    @Column(name = "first_Name")
    private String firstName;
    
    @NotNull(message = "last Name cannot be null")
    @Column(name = "last_Name")
    private String lastName;
    
    @ManyToOne()
    @JoinColumn(name = "role_Id")
    private Role role;
    
    @NotNull
    @Size(min = 10, max = 10, message = "Contact must be of 10 characters")
    @Column(name = "contact")
    private String contact;
    
    @NotNull
    @Email(message = "Email should be valid")
    @Column(name = "email", unique = true)
    private String email;
    
    @NotNull
    @Size(min = 10, max = 255, 
    message = "Local address must be between 10 and 255 characters")
    @Column(name = "local_Address")
    private String localAddress;
    
    @ManyToOne()
    @JoinColumn(name = "city_Id")
    private City city;
    
    @NotNull
    @Column(name = "user_Name")
    private String userName;
    
    @NotNull
    @Column(name = "password")
    private String password;
    
    @Column(name = "registered_On")
    private LocalDateTime registeredOn;
    
    @Column(name = "updated_On")
    private LocalDateTime updatedOn;
    
    @Column(name = "approved_By")
    private Long approvedBy;
    
    @NotNull
    @Size(min = 1, max = 1, 
    message = "Approval status must be of 1 character.")
    @Column(name = "approval_Status")
    private String approvalStatus;

    @NotNull
    @Column(name = "active_Status")
    private boolean activeStatus;
    
    @Transient 
    private Company company;
}

@Entity
@Table(name = "tbl_Administrator")
public class Admin {
    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY) 
    private int adminId;
    
    @OneToOne()
    @JoinColumn(name = "user_Id")
    private User user;
    
    @OneToOne()
    @JoinColumn(name = "company_Id")
    private Company company;    
}

@Entity
@Table(name = "tbl_Distributor")
public class Distributor {
    
    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY) 
    private Long distributorId;
    
    @OneToOne()
    @JoinColumn(name = "user_Id")
    private User user;
    
    @ManyToOne()
    @JoinColumn(name = "company_Id")
    private Company company;
}

@Entity
@Table(name = "tbl_Retailer")
public class Retailer {

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY) 
    private Long retailerId;

    @OneToOne()
    @JoinColumn(name = "user_Id")
    private User user;
}

Answer 1

我将使用@MappedSuperclass，而不是创建User和实体本身。基本上，Admin、Distributor和Retailer将扩展这个映射的超类。为每个表创建了一个单独的表，但没有为超类(User)本身创建任何表。

@MappedSuperclass
public class User {

   // Your properties
   
}

@Entity
@Table(name = "tbl_Administrator")
public class Admin extends User {

    @OneToOne()
    @JoinColumn(name = "company_Id")
    private Company company;  
     
}

@Entity
@Table(name = "tbl_Distributor")
public class Distributor extends User {
    
    @ManyToOne()
    @JoinColumn(name = "company_Id")
    private Company company;
    
}

@Entity
@Table(name = "tbl_Retailer")
public class Retailer extends User {

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY) 
    private Long retailerId;
    
}

您还有其他可能性(请参阅https://medium.com/analytics-vidhya/jpa-hibernate-entity-inheritance-1f6aa7ea2eea中的更多内容)，如下所示：

用于超类和子类的所有entities

• Separate表的
• 单表。超类中定义的属性不包含在每个类的subclass.
• Table表中，这类似于映射的超类，其区别在于超类也有相应的表。

不过，我还是会按照描述的那样使用映射的Superclass 1。