hibernate - JPA @Version行为

Question

即时通讯与Hibernate 3.6.x一起使用JPA2

我已经对@Version进行了简单的测试。

假设我们有2个实体，

实体团队拥有玩家实体列表，双向关系，延迟获取类型，级联类型所有

两个实体都有@Version

这是场景:

每当对团队/球员实体之一进行修改时，刷新/提交时，团队/球员的版本都会增加(修改记录的版本会增加)。

使用persist将新的玩家实体添加到团队的集合中，在persist之后将分配团队版本的实体(添加一个新实体，该新实体将获取其版本)。

每当对一个玩家实体进行添加/修改/删除时，刷新/提交时团队的版本将增加。 (添加/修改/删除 child 的记录， parent 的版本也增加了)

我可以理解数字1和2，但是我不能理解数字3，为什么团队的版本增加了？

这让我想到了其他问题:

如果我得到了 parent <-> child <->亲子关系船该怎么办。对子孙的增加或修改会增加 child 和 parent 的版本吗？

在方案2中，如何在提交之前在团队中获取版本，例如使用flush？在对 child 做一些事情之后，是否建议获取 parent 的版本？

这是来自我的实验的代码示例，证明当ReceivingGoodDetail是拥有方时，刷新后版本在ReceivingGood中得到了增加。抱歉，这使用其他实体，但是ReceivingGood像Team，ReceivingGoodDetail像Player。 1个ReceivevingGood/Team，很多ReceivingGoodDetail/Player。

/*
Hibernate: select receivingg0_.id as id9_14_, receivingg0_.creationDate as creation2_9_14_, .. too long
Hibernate: select product0_.id as id0_4_, product0_.creationDate as creation2_0_4_, .. too long
before persisting the new detail, version of header is : 14
persisting the detail 1c9f81e1-8a49-4189-83f5-4484508e71a7
printing the size of the header : 
Hibernate: select details0_.receivinggood_id as receivi13_9_8_, details0_.id as id8_, details0_.id as id10_7_, .. too long
7
after persisting the new detail, version of header is : 14
Hibernate: insert into ReceivingGoodDetail (creationDate, modificationDate, usercreate_id, usermodify_id, version, buyQuantity, buyUnit, internalQuantity, internalUnit, product_id, receivinggood_id, supplierLotNumber, id) values (?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?)
Hibernate: update ReceivingGood set creationDate=?, modificationDate=?, usercreate_id=?, usermodify_id=?, version=?, purchaseorder_id=?, supplier_id=?, transactionDate=?, transactionNumber=?, transactionType=?, transactionYearMonth=?, warehouse_id=? where id=? and version=?
after flushing, version of header is now : 15
    */
public void addDetailWithoutTouchingCollection() {
    String headerId = "3b373f6a-9cd1-4c9c-9d46-240de37f6b0f";
    ReceivingGood receivingGood = em.find(ReceivingGood.class, headerId);

    // create a new detail
    ReceivingGoodDetail receivingGoodDetailCumi = new ReceivingGoodDetail();
    receivingGoodDetailCumi.setBuyUnit("Drum");
    receivingGoodDetailCumi.setBuyQuantity(1L);
    receivingGoodDetailCumi.setInternalUnit("Liter");
    receivingGoodDetailCumi.setInternalQuantity(10L);
    receivingGoodDetailCumi.setProduct(getProduct("b3e83b2c-d27b-4572-bf8d-ac32f6de5eaa"));
    receivingGoodDetailCumi.setSupplierLotNumber("Supplier Lot 1");
    decorateEntity(receivingGoodDetailCumi, getUser("3978fee3-9690-4377-84bd-9fb05928a6fc"));
    receivingGoodDetailCumi.setReceivingGood(receivingGood);

    System.out.println("before persisting the new detail, version of header is : " + receivingGood.getVersion());

    // persist it
    System.out.println("persisting the detail " + receivingGoodDetailCumi.getId());
    em.persist(receivingGoodDetailCumi);

    System.out.println("printing the size of the header : ");
    System.out.println(receivingGood.getDetails().size());

    System.out.println("after persisting the new detail, version of header is : " + receivingGood.getVersion());

    em.flush();

    System.out.println("after flushing, version of header is now : " + receivingGood.getVersion());
}

Answer 1

它看起来像是Hibernate中的错误。

JPA规范说:

All non-relationship fields and properties and all relationships owned by the entity are included in version checks

但是，Hibernate还会在更改非所有关系属性后增加版本(例如，EclipseLink不会这样做)。可以通过在属性上设置@OptimisticLock(exclude = true)来禁用此行为。

请注意，它仅适用于关系属性本身的更改，不适用于引用对象状态的更改，因此不会由于父级集合的更改而更改父级的版本。