JPA外键连接同名

Question

假设我有：

1. 表编辑（ID，BUSINESS_NAME，增值税）
2. 表作者（ID，姓名，电子邮件）
3. 表书籍（id，title，description，fk_author，fk_editor）。

比方说fk_editor是表格编辑器的ID字段和fk_author外键是表作者的ID字段的外键。

假设的关系书：作者是N：1周和书：编辑是N：1。

问题是：如何通过书籍表加入三张表表？ 这意味着，我必须在Book类中添加哪些代码才能让Hibernate理解并与作者和编辑者建立关系？请考虑我在作者和编辑者中拥有相同的id字段名称。 在这里它的示例代码，我需要纠正：

表作者：

package com.bytecode.jpaexample.SpringBootMySqlJpaRestExample; 

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

@Entity 
@Table(name = "authors") 
public class Author implements Serializable{ 

    @Id 
    @GeneratedValue(strategy = GenerationType.AUTO) 
    @Column(name = "id") 
    @OneToMany(fetch = FetchType.LAZY) 
    private int id; 

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

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

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

    /* constructors and getters and setters omitted intentionally */ 

} 

表编辑：

package com.bytecode.jpaexample.SpringBootMySqlJpaRestExample; 

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

@Entity 
@Table(name = "editors") 
public class Editor implements Serializable{ 

    @Id 
    @GeneratedValue(strategy = GenerationType.AUTO) 
    @Column(name = "id") 
    @OneToMany(fetch = FetchType.LAZY) 
    private int id; 

    @Column(name = "business_name") 
    private String businessName; 

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

    /* constructors and getters and setters omitted intentionally */ 

} 

表书：

package com.bytecode.jpaexample.SpringBootMySqlJpaRestExample; 

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

@Entity 
@Table(name="books") 
public class Book implements Serializable{ 

    @Id 
    @GeneratedValue(strategy = GenerationType.AUTO) 
    @Column(name = "id") 
    private int id; 

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

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

    @ManyToOne(fetch = FetchType.LAZY)  
    @JoinColumn(name = "id") //authors.id     
    @Column(name = "fk_author") 
    private int fk_editor; 

    @ManyToOne(fetch = FetchType.LAZY)  
    @JoinColumn(name = "id") //editors.id     
    @Column(name = "fk_editor") 
    private int fk_editor; 

    /* constructors and getters and setters omitted intentionally */ 
} 

两

Answer 1

几件事情：

的Author和Editor类不应该标注为@OneToMany他们id领域：

@Id 
    @GeneratedValue(strategy = GenerationType.AUTO) 
    @Column(name = "id") 
    private int id; 

相反，我猜你想在每一个类另一个领域：

@Entity 
@Table(name = "authors") 
public class Author implements Serializable { 
    ... 
    @OneToMany(mappedBy = "author", fetch = FetchType.LAZY) 
    private List<Book> books; 
    ... 
} 

@Entity 
@Table(name = "editors") 
public class Editor implements Serializable { 
    ... 
    @OneToMany(mappedBy = "editor", fetch = FetchType.LAZY) 
    private List <Book> books; 
    ... 
} 

然后，您需要以几种方式更改类Book：

• 更适当地重命名字段（例如， fk_author变成author）;这些必须与相应类别中的mappedBy设置相匹配，如上所述）
• 将字段类型更改为相应的类（例如Author），而不是外键类型（例如，int）
• 变化@JoinColumn批注指定在books表（将适当列不目标表），并添加referencedColumnName设置（尽管，这些是在这种情况下可选的;因为它们将默认为单个主键字段目标实体）
• 删除@Column注释

看看这是有道理的：

@Entity 
@Table(name="books") 
public class Book implements Serializable { 
    ... 
    @ManyToOne(fetch = FetchType.LAZY)  
    @JoinColumn(name = "fk_author", referencedColumnName="id") 
    private Author author; 

    @ManyToOne(fetch = FetchType.LAZY)  
    @JoinColumn(name = "fk_editor", referencedColumnName="id") 
    private Editor editor; 
    ... 
}