如何在hibernate中定义复合外键映射？

Posted 2023-02-24

【中文标题】如何在hibernate中定义复合外键映射？

【英文标题】：How to define composite foreign key mapping in hibernate?

【发布时间】：2021-12-12 13:35:51

【问题描述】：

我有两个表：users和userdetails如下：

package com.example.easynotes.model;

import javax.persistence.*;
import java.io.Serializable;

@Entity
@Table(name = "users")
@IdClass(UserID.class)
public class User implements Serializable 

    @Id
    int id;

    @Id
    String name;

    String department;

    //getters and setters

userdetails 类是这样的：

public class UserDetails implements Serializable

int id;

String name;

String address;

String otherFields;

//getters and setters

users 中的 id 和 name 是复合主，我希望 userdetails 中的相同字段成为外键。如何在休眠中实现这一点？

【问题讨论】：

vladmihalcea.com/…

【参考方案1】：

我们需要将两个键都放入@Embeddable 以分离复合键，然后使用@EmbeddedId 将其放入User 实体中并使用Hibernate Relational Mapping 映射两个主键...

Composite Primary Key有两种选择：

使用@EmbeddedId 使用@IdClass()

下面是例子：

---------------------- 使用 EmbeddedId -------- ----------------------------------------

复合主键：

@Embeddable
public class UserIdName implements Serializable 
    int id;
    String name;
    
    // getter and setter

用户：

@Entity
@Table(name = "users")
public class USER
    @EmbeddedId
    private UserIdName id;

    String department;
    
    @OneToMany(cascade = CascadeType.ALL, mappedBy = "user")
    private Set<Userdetail> userdetail;

    // getter and setter

用户详情：

@Entity
@Table(name = "Userdetail")
public class Userdetail 

    @Id
    private int detail_id;

    @ManyToOne
    @JoinColumns( @JoinColumn(name = "id", referencedColumnName = "id"),
            @JoinColumn(name = "name", referencedColumnName = "name") )
    private USER user;

    String address;

    String otherFields;

    // getter setter

---------------------- 使用 IdClass -------- ----------------------------------------

复合主键：

public class UserIdName implements Serializable 
    int id;
    String name;
    
    // getter and setter

用户：

@Entity
@Table(name = "users")
@IdClass(UserIdName.class)
public class USER
    @Id
    int id;

    @Id
    String name;

    String department;
    
    @OneToMany(cascade = CascadeType.ALL, mappedBy = "user")
    private Set<Userdetail> userdetail;

    // getter and setter

用户详情：

@Entity
@Table(name = "Userdetail")
public class Userdetail 

    @Id
    private int detail_id;

    @ManyToOne
    @JoinColumns( @JoinColumn(name = "id", referencedColumnName = "id"),
            @JoinColumn(name = "name", referencedColumnName = "name") )
    private USER user;

    String address;

    String otherFields;

    // getter setter

->如果您想手动插入两个外键，请尝试以下代码

将此代码放入UserDetails

@ManyToOne
@JoinColumn(name = "id", referencedColumnName = "id", insertable = false, updatable = false)
@JoinColumn(name = "name", referencedColumnName = "name", insertable = false, updatable = false)
private USER user;

@Column(name="id")
private int id;

@Column(name="name")
private String name

// don't forget to put getter setter

用户表：

用户明细表：

【讨论】：

通过@IdClass 没有解决方案吗？你的这个解决方案在插入数据时也会保持约束吗？

如何在 userdetails 对象中设置 id 和 name ？

您要手动添加两个外键？

您有两个选项 @IdClass 和 @EmbeddedId 用于复合键，请参阅上面编辑的答案