如何使休眠在添加子关系(@ManyToMany集合)时不递增父实体的版本列

Question

>Tables：

create table myuser.A (
id number primary key,
notes varchar2(50),
version number);
create table myuser.B (
id number primary key,
username varchar2(50),
version number);
create table myuser.A_B (
a_id number,
b_id number,
constraint a_fk foreign key (a_id) references A(id),
constraint b_fk foreign key (b_id) references B(id)
);
insert into myuser.A values ('1', 'notes', 1);
insert into myuser.B values ('44', 'username', 1);
insert into myuser.A_B values ('1', '44');

注释休眠映射：

@Entity
class A {
@Id
@Column
public String id;
@Column
public String notes;
@ManyToMany(fetch = FetchType.EAGER, cascade = CascadeType.ALL)
@JoinTable(name = "A_B",
joinColumns = { @JoinColumn(name = "A_ID") },
inverseJoinColumns = { @JoinColumn(name = "B_ID") })
public Set<B> bDomains;
@Version
public int version;
}
@Entity
class B {
@Id
@Column
public String id;
@Column
public String username;
@ManyToMany(fetch = FetchType.EAGER, mappedBy = "bDomains")
public Set<A> aDomains;
@Version
public int version;
}

获取现有的"A"对象，并将新的子对象添加到 ManyToMany 集合。所以"A"对象不会改变，但Hibernate会为它生成一个新的版本号并更新"A"表：

A a = session.get(A.class, "1");
int versionBefore = a.version;
B b = new B();
b.id = "48";
b.username = "asd";
a.bDomains.add(b);
session.save(a);
session.flush();
int versionAfter = a.version;
System.out.println("version before: " + versionBefore);
System.out.println("version after: " + versionAfter);

输出：

Hibernate: select a0_.id as id1_0_0_, a0_.notes as notes2_0_0_, a0_.version as version3_0_0_, bdomains1_.A_ID as A_ID1_1_1_, b2_.id as B_ID2_1_1_, b2_.id as id1_2_2_, b2_.username as username2_2_2_, b2_.version as version3_2_2_ from A a0_, A_B bdomains1_, B b2_ where a0_.id=bdomains1_.A_ID(+) and bdomains1_.B_ID=b2_.id(+) and a0_.id=?
Hibernate: select adomains0_.B_ID as B_ID2_1_0_, adomains0_.A_ID as A_ID1_1_0_, a1_.id as id1_0_1_, a1_.notes as notes2_0_1_, a1_.version as version3_0_1_ from A_B adomains0_, A a1_ where adomains0_.A_ID=a1_.id and adomains0_.B_ID=?
Hibernate: select b_.id, b_.username as username2_2_, b_.version as version3_2_ from B b_ where b_.id=?
Hibernate: insert into B (username, version, id) values (?, ?, ?)
Hibernate: update A set notes=?, version=? where id=? and version=?
Hibernate: insert into A_B (A_ID, B_ID) values (?, ?)
version before: 1
version after: 2

我不需要对由休眠生成的表进行 UPDATE 查询，因为在此逻辑中不会更改表。 据我所知，Hibernate在更改实体的任何字段时都会增加版本列，即使该字段只是对另一个表的引用。如何配置休眠以防止在仅更改其关系时递增实体的版本？

Answer 1

我找到了解决方案。 需要在@ManyToMany|@OneToMany属性上添加@org.hibernate.annotations.OptimisticLock(excluded = true)注释。

因此，我示例中A实体现在如下所示：

@Entity
class A {
@Id
@Column
public String id;
@Column
public String notes;
@org.hibernate.annotations.OptimisticLock(excluded = true)
@ManyToMany(fetch = FetchType.EAGER, cascade = CascadeType.ALL)
@JoinTable(name = "A_B",
joinColumns = { @JoinColumn(name = "A_ID") },
inverseJoinColumns = { @JoinColumn(name = "B_ID") })
public Set<B> bDomains;
@Version
public int version;
}

生成的 SQL：

Hibernate: select a0_.id as id1_0_0_, a0_.notes as notes2_0_0_, a0_.version as version3_0_0_, bdomains1_.A_ID as A_ID1_1_1_, b2_.id as B_ID2_1_1_, b2_.id as id1_2_2_, b2_.username as username2_2_2_, b2_.version as version3_2_2_ from A a0_, A_B bdomains1_, B b2_ where a0_.id=bdomains1_.A_ID(+) and bdomains1_.B_ID=b2_.id(+) and a0_.id=?
Hibernate: select adomains0_.B_ID as B_ID2_1_0_, adomains0_.A_ID as A_ID1_1_0_, a1_.id as id1_0_1_, a1_.notes as notes2_0_1_, a1_.version as version3_0_1_ from A_B adomains0_, A a1_ where adomains0_.A_ID=a1_.id and adomains0_.B_ID=?
Hibernate: select b_.id, b_.username as username2_2_, b_.version as version3_2_ from B b_ where b_.id=?
Hibernate: insert into B (username, version, id) values (?, ?, ?)
Hibernate: insert into A_B (A_ID, B_ID) values (?, ?)
version before: 1
version after: 1

唯一的问题是OptimisticLock注释来自Hibernate，而不是来自JPA。