Question

我尝试使用多对多的方法产品与商店有多对多的关系的 Product.java

package models;
@Entity
public class Product extends Model {

    @Id
    @SequenceGenerator(name="product_gen", sequenceName="product_id_seq", allocationSize=1)
    @GeneratedValue(strategy=GenerationType.SEQUENCE, generator="product_gen")
    @Column(name="id")
    public Long id;

    @Required
    public String name;

    @Required
    public Float price;

    @ManyToMany(cascade = CascadeType.ALL)
    public List<Shop> shops = new ArrayList<Shop>();

    public Product(String name, float price) {
        this.name = name;
        this.price = price;
    }
    public static List<Product> all(){
        return find.all();
    }
    public static Model.Finder<Long, Product> find = new Model.Finder(Long.class, Product.class);

    public static Product create(String name,float price) {
        Product product = new Product(name, price);
        product.save();
        product.saveManyToManyAssociations("shops");
        return product;
    }




}

play创建了一个table_shop表，其中shop_id和product_id为外键，但在添加产品时无法在product_shop表中存储任何值任何帮助将不胜感激。

Answer 1

我有Apps和Users。每个应用程序可以有多个用户，每个用户可以使用多个应用程序。

在您的情况下，您可以将App替换为Product，将User替换为Shop

这是Play下的工作代码！ 2.2.2

DB的演变：

create table app_user (
  id                varchar(40) not null,
  constraint pk_fb_user primary key (id)
);

create table app (
  id                    varchar(40) not null,
  name                  text,
  constraint pk_app_id primary key (id)
);

create table membership (
  app_id                varchar(40) not null,
  app_user_id           varchar(40) not null,
  constraint fk_membership_app_id foreign key (app_id) references app,
  constraint fk_membership_app_user_id foreign key (app_user_id) references app_user
);

模型应用

@Entity
public class App extends Model {

    @Id
    public UUID id;

    @Column
    public String name;

    @ManyToMany(cascade = CascadeType.ALL)
    @JoinTable(name = "membership")
    public List<User> users;

    ....
}

模型用户

@Entity
@Table(name = "app_user")
// user is reserved keyword in DB
public class User extends Model {

    @Id
    public UUID id;

}

我不需要访问用户的应用，因此我在用户中没有应用字段。

Answer 2

尝试查看嵌入式ID。

为您的主键创建课程

@Embeddable
public class ProductShopKey {
 public Long productId;
 public Long shopId;

 public boolean equals(Object object) {
  if(object isInstanceOf ProductShopKey){
   ProductShopKey key = (ProductShopKey) object
   if(this.productId == key.productId
      && this.shopId == key.shopId
     ){
        return true;
   }
  }
  return false;
 }

 public int hashCode() {
  return productId.hashCode() + shopId.hashCode();
 }
}

然后为类

创建一个实体

@entity
public class ProductShop {
 @EmbeddedId
 public ProductShopKey psKey = new ProductShopKey();

 @ManyToOne
 @JoinColumn(name = "product_id")
 public Product product;

 @ManyToOne
 @JoinColumn(name = "shop_id")
 public Shop shop;

 //Extra value to go in your join table
 public String someValue;
 ...
}

然后Shop和Product与ProductShop

之间存在一对多的关系

在表格中插入条目是手动完成的

ProductShop ps = new ProductShop();
ps.psKey.productId = someProductId;
ps.psKey.shopId = someShopId;
ps.someValue = someValue;
ps.save();

play框架java中的多对多关系

2 个答案: