函数指针使用场景和选择_在N + 1场景中使用@NamedEntityGraph更有选择地加载JPA实体...

函数指针使用场景和选择

N + 1问题是使用ORM解决方案时的常见问题。当您将某些@OneToMany关系的fetchType设置为lazy时，就会发生这种情况，以便仅在访问Set / List时才加载子实体。假设我们有一个具有两个关系的Customer实体：每个客户的一组订单和一组地址。

@OneToMany(mappedBy = "customer", cascade = CascadeType.ALL, fetch = FetchType.LAZY)
private Set<OrderEntity> orders;@OneToMany(mappedBy = "customer", cascade = CascadeType.ALL, fetch = FetchType.LAZY)
private Set<AddressEntity> addresses;

要加载所有客户，我们可以发出以下JPQL语句，然后加载每个客户的所有订单：

List<CustomerEntity> resultList = entityManager.createQuery("SELECT c FROM CustomerEntity AS c", CustomerEntity.class).getResultList();
for(CustomerEntity customerEntity : resultList) {Set<OrderEntity> orders = customerEntity.getOrders();for(OrderEntity orderEntity : orders) {...}
}

Hibernate 4.3.5（与JBoss AS Wildfly 8.1.0CR2一起提供）将从数据库中仅为两个（！）客户生成以下一系列SQL语句：

Hibernate: selectcustomeren0_.id as id1_1_,customeren0_.name as name2_1_,customeren0_.numberOfPurchases as numberOf3_1_ fromCustomerEntity customeren0_
Hibernate: selectorders0_.CUSTOMER_ID as CUSTOMER4_1_0_,orders0_.id as id1_2_0_,orders0_.id as id1_2_1_,orders0_.campaignId as campaign2_2_1_,orders0_.CUSTOMER_ID as CUSTOMER4_2_1_,orders0_.timestamp as timestam3_2_1_ fromOrderEntity orders0_ whereorders0_.CUSTOMER_ID=?
Hibernate: selectorders0_.CUSTOMER_ID as CUSTOMER4_1_0_,orders0_.id as id1_2_0_,orders0_.id as id1_2_1_,orders0_.campaignId as campaign2_2_1_,orders0_.CUSTOMER_ID as CUSTOMER4_2_1_,orders0_.timestamp as timestam3_2_1_ fromOrderEntity orders0_ whereorders0_.CUSTOMER_ID=?

如我们所见，第一个查询从表CustomerEntity中选择所有客户。接下来的两个选择先提取，然后在第一个查询中加载我们已加载的每个客户的订单。当我们有100个客户而不是2个客户时，我们将获得101个查询。一个初始查询可加载所有客户，然后针对100个客户中的每个客户，另外查询一个订单。这就是为什么将此问题称为N + 1的原因。

解决此问题的常见习惯是强制ORM生成内部联接查询。在JPQL中，这可以通过使用JOIN FETCH子句来完成，如以下代码片段所示：

entityManager.createQuery("SELECT c FROM CustomerEntity AS c JOIN FETCH c.orders AS o", CustomerEntity.class).getResultList();

正如预期的那样，ORM现在使用OrderEntity表生成一个内部联接，因此只需要一个SQL语句即可加载所有数据：

selectcustomeren0_.id as id1_0_0_,orders1_.id as id1_1_1_,customeren0_.name as name2_0_0_,orders1_.campaignId as campaign2_1_1_,orders1_.CUSTOMER_ID as CUSTOMER4_1_1_,orders1_.timestamp as timestam3_1_1_,orders1_.CUSTOMER_ID as CUSTOMER4_0_0__,orders1_.id as id1_1_0__
fromCustomerEntity customeren0_
inner joinOrderEntity orders1_on customeren0_.id=orders1_.CUSTOMER_ID

在您知道必须为每个客户加载所有订单的情况下，JOIN FETCH子句将SQL语句的数量从N + 1减少到1。这当然具有缺点，即您现在要转移一个订单的所有订单。客户一次又一次的客户数据（由于查询中的其他客户列）。

JPA规范从版本2.1引入，即所谓的NamedEntityGraphs。通过此批注，您可以描述JPQL查询应加载的图形，而不是JOIN FETCH子句可以执行的图形，从而为N + 1问题提供了另一种解决方案。下面的示例演示了我们的客户实体的NamedEntityGraph，该实体应该仅加载客户名称及其订单。订单在子图ordersGraph中有更详细的描述。在这里，我们看到我们只想加载订单的字段ID和CampaignId。

@NamedEntityGraph(name = "CustomersWithOrderId",attributeNodes = {@NamedAttributeNode(value = "name"),@NamedAttributeNode(value = "orders", subgraph = "ordersGraph")},subgraphs = {@NamedSubgraph(name = "ordersGraph",attributeNodes = {@NamedAttributeNode(value = "id"),@NamedAttributeNode(value = "campaignId")})}
)

在通过NameManager通过EntityManager加载JPQL查询后，将其命名为JPQL查询的提示：

EntityGraph entityGraph = entityManager.getEntityGraph("CustomersWithOrderId");
entityManager.createQuery("SELECT c FROM CustomerEntity AS c", CustomerEntity.class).setHint("javax.persistence.fetchgraph", entityGraph).getResultList();

Hibernate从版本4.3.0.CR1开始支持@NamedEntityGraph批注，并为上面显示的JPQL查询创建以下SQL语句：

Hibernate: selectcustomeren0_.id as id1_1_0_,orders1_.id as id1_2_1_,customeren0_.name as name2_1_0_,customeren0_.numberOfPurchases as numberOf3_1_0_,orders1_.campaignId as campaign2_2_1_,orders1_.CUSTOMER_ID as CUSTOMER4_2_1_,orders1_.timestamp as timestam3_2_1_,orders1_.CUSTOMER_ID as CUSTOMER4_1_0__,orders1_.id as id1_2_0__ fromCustomerEntity customeren0_ left outer joinOrderEntity orders1_ on customeren0_.id=orders1_.CUSTOMER_ID

我们看到Hibernate不会发出N + 1查询，而是@NamedEntityGraph注释强制Hibernate为每个左外部联接加载订单。当然，这与FETCH JOIN子句有微妙的区别，在子句中，Hibernate创建了内部联接。与FETCH JOIN子句相反，左外部联接还将加载不存在订单的客户，在FETCH JOIN子句中，我们仅加载具有至少一个订单的客户。

有趣的是，Hibernate加载的负载比表CustomerEntity和OrderEntity的指定属性更多。由于这与@NamedEntityGraph的规范（第3.7.4节）相冲突，因此我为此创建了一个JIRA问题。