Hibernate 2主键(1是外键)

Question

我的问题是关于下面的代码：

Ingredient.java

@Entity
@Table(name = "recipes_ingredients")
public class Ingredient implements Serializable {
 
    @Id
    @Column(name = "ingredient_id")
    private Long id;
 
    @NotEmpty
    private String name;
 
    private Integer quantity;
 
    // Getters and Setters
}

Recipe.java

@Entity
@Table(name = "recipes")
public class Recipe implements Serializable {
 
    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    @Column(name = "recipe_id")
    private Long id;
 
    @NotEmpty
    private String name;
 
    @ManyToOne(fetch = FetchType.LAZY)
    private Client client;
 
    @OneToMany(fetch = FetchType.LAZY, cascade = CascadeType.ALL, orphanRemoval = true)
    @JoinColumn(name = "fk_recipe_id")
    private List<Ingredient> ingredients;
 
    // Getters and Setters
}

我想要做的是，PK (recipes_ingredients表)以某种方式由：ingredient_id和recipe_id组成(使用我已经拥有的，并使其也使用PK，或者只是以其他方式获取该ID，并将其改为PKE 215和ingredient_id)。

ingredient_id将是每个菜谱从1开始的数字，这样，如果我有一个recipe_id =1和3成分的配方，我的想法是，在配料表中有这样的内容：

ingredient_id(PK)    name      quantity   recipe_id(PK)
1                    Oil       150        1
2                    Salt      5          1
3                    Flour     500        1

知道我该怎么做吗？

Answer 1

    public class IngredientId {

        private Recipe recipe;

        private Long id;

        public IngredientId(Recipe recipe, Long id) {
            this.recipe = recipe;
            this.id = id;
        }

        public IngredientId() {
        }

        //Getters and setters are omitted for brevity

        public Recipe getRecipe() {
            return recipe;
        }

        public void setRecipe(Recipe recipe) {
            this.recipe = recipe;
        }

        public Long getId() {
            return id;
        }

        public void setId(Long id) {
            this.id = id;
        }

        @Override
        public boolean equals(Object o) {
            if ( this == o ) {
                return true;
            }
            if ( o == null || getClass() != o.getClass() ) {
                return false;
            }
            IngredientId ingredientId = (IngredientId) o;
            return Objects.equals( recipe, ingredientId.recipe ) && Objects.equals( id, ingredientId.id );
        }

        @Override
        public int hashCode() {
            return Objects.hash( recipe, id );
        }
    }

    @Entity
    @IdClass( IngredientId.class )
    @Table(name = "recipes_ingredients")
    public class Ingredient {

        @Id
        @ManyToOne(fetch = FetchType.LAZY)
        @JoinColumn(name = "recipe_id")
        private Recipe recipe;

        @Id
        @GeneratedValue
        @Column(name = "ingredient_id")
        private Long id;

        private String name;

        private Integer quantity;

        public IngredientId getId() {
            return new IngredientId( recipe, id );
        }

        public void setId(IngredientId id) {
            this.recipe = id.getRecipe();
            this.id = id.getId();
        }
    }

    @Entity
    @Table(name = "recipes")
    public class Recipe {

        @Id
        @GeneratedValue(strategy = GenerationType.IDENTITY)
        @Column(name = "recipe_id")
        private Long id;

        private String name;

        @OneToMany(fetch = FetchType.LAZY, cascade = CascadeType.ALL, orphanRemoval = true)
        @JoinColumn(name = "recipe_id")
        private List<Ingredient> ingredients = new ArrayList<>();

    }

需要注意的是，复合键不能具有空值。因此，只有当你知道每一种成分都与食谱有关时，这才有意义。见这个答案在StackOverflow上和这个冬眠的Jira

有关更多细节，您可以查看Hibernate ORM文档。