java 具有复合主键的一部分的休眠外键
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Hibernate foreign key with a part of composite primary key
提问by Jagger
I have to work with Hibernate and I am not very sure how to solve this problem, I have 2 tables with a 1..n relationship like this:
我必须与 Hibernate 一起工作，但我不太确定如何解决这个问题，我有 2 个具有 1..n 关系的表，如下所示：
------- TABLE_A ------- col_b (pk) col_c (pk) [other fields] ------- TABLE_B ------- col_a (pk) col_b (pk) (fk TABLE_A.col_b) col_c (fk TABLE_A.col_c) [other fields]
How can I manage this with Hibernate?
如何使用 Hibernate 管理此问题？
I do not have any idea how to declare a foreign key that would contain a part of primary key.
我不知道如何声明一个包含主键一部分的外键。
My database schema is generated from the Hibernate model.
我的数据库模式是从 Hibernate 模型生成的。
采纳答案by Jagger
I have found two solutions to this problem.
我找到了两个解决这个问题的方法。
The first one is rather a workaround and is not so neat as the second one.
第一个是一种解决方法，不如第二个那么整洁。
Define the primary key of the Bentity as composite key containing col_a, col_b, and col_cand what was supposed to be the primary key in the first place, define as unique constraint. The disadvantage is that the column col_cis not really conceptually a part of primary key.
将B实体的主键定义为包含col_a,col_b和 的复合键，以及col_c最初应该是主键的内容，定义为唯一约束。缺点是该列col_c在概念上并不是主键的一部分。
@Entity
class A {
@Id
private int b;
@Id
private int c;
}
@Entity
@Table(uniqueConstraints = {@UniqueConstraint(columnNames = { "a", "b" }) })
class B {
@Id
private int a;
@Id
@ManyToOne(optional = false)
@JoinColumns(value = {
@JoinColumn(name = "b", referencedColumnName = "b"),
@JoinColumn(name = "c", referencedColumnName = "c") })
private A entityA;
}
The second uses @EmbeddedIdand @MapsIdannotations and does exactly what I wanted to be done at the very beginning.
第二个使用@EmbeddedId和@MapsId注释并完全按照我一开始就想做的事情做。
@Entity
class A {
@Id
private int b;
@Id
private int c;
}
@Embeddable
class BKey {
private int a;
private int b;
}
@Entity
class B {
@EmbeddedId
private BKey primaryKey;
@MapsId("b")
@ManyToOne(optional = false)
@JoinColumns(value = {
@JoinColumn(name = "b", referencedColumnName = "b"),
@JoinColumn(name = "c", referencedColumnName = "c") })
private A entityA;
}
回答by Xiaofeng Li
Jagger's second solution that is my first reaction, with @EmbededId and @MapsId. The following is another way based on his second solution but without using @MapsId. `
Jagger 的第二个解决方案是我的第一反应，@EmbededId 和 @MapsId。以下是基于他的第二个解决方案但不使用@MapsId 的另一种方法。`
@Entity
class A {
@Id
private int b;
@Id
private int c;
}
@Embeddable
class BKey {
private int a;
private int b;
}
@Entity
class B {
@EmbeddedId
private BKey primaryKey;
@ManyToOne(optional = false)
@JoinColumns(value = {
@JoinColumn(name = "b", referencedColumnName = "b", insertable= false, updatable= false),
@JoinColumn(name = "c", referencedColumnName = "c") }, )
private A entityA;
}
