休眠中的OneToMany关系创建4个表

Question

i got a problem. I have two classes with the same @OneToMany relation. Hibernate creates 4 tables : product, product_categorie, categorie, categorie_product.

In my case i need only 3 tables: product, categorie and product_categorie.

This is my class diagram:

The code i wrote in Java:

@Entity
public class Product {
    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private int product_id;
    private String naam, omschrijving;
    private double prijs;
    @OneToMany(mappedBy = "product_m")
    private List<Aanbieding> aanbiedingen;
    @OneToMany
    private List<Categorie> categories;
}

@Entity
public class Categorie {
    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private int categorie_id;
    private String naam, omschrijving;
    @OneToMany
    private List<Product> producten;

}

In my case i need to archieve the following :

One Product belongs to 1 or more categories

One Categorie contains 0 or more products

Did i do somthing wrong in code?

It's my first time using hibernate, hope you understand.

Answer 1

Adam,

What you need is a ManyToMany relationship, not OneToMany. Together with a JoinTable to map the relationship between product and categories.

@Entity
public class Product {
    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private int product_id;
    private String naam;
    private String omschrijving;
    private double prijs;
    @OneToMany(mappedBy = "product_m")
    private List<Aanbieding> aanbiedingen;

    @ManyToMany(cascade = { CascadeType.ALL })
    @JoinTable(
        name = "product_categories", 
        joinColumns = { @JoinColumn(name = "product_id") }, 
        inverseJoinColumns = { @JoinColumn(name = "categorie_id") }
    )
    private List<Categorie> categories;
}

@Entity
public class Categorie {
    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private int categorie_id;
    private String naam;
    private String omschrijving;

    @ManyToMany(mappedBy = "categories")
    private List<Product> producten;

}