我假设我遇到了N + 1选择问题。
我有这个实体：
@Entity
@Table(name = "Devices")
public class Device implements Serializable {
@OneToOne(mappedBy="holdingDevice", fetch=FetchType.LAZY)
@Cascade(CascadeType.ALL)
@PrimaryKeyJoinColumn
private WarrantyEntry warranty;
}
这是另一个实体：
@Entity
@Table(name = "Warranty")
public class WarrantyEntry implements Serializable{
private static final long serialVersionUID = 1L;
@Id
@GeneratedValue(strategy=GenerationType.AUTO)
private Long id;
@OneToOne
@JoinColumn(name = "serial")
@PrimaryKeyJoinColumn
private Device holdingDevice;
现在，当我开始迭代这个循环时：
Set<Device> customerDevices = user.getCustomer().getDevices();
for (Device device : customerDevices) {
...
}
我被困住了，我在日志中看到Hibernate选择：
Hibernate：* /选择warrantyen0_.id为...... Hibernate：/ *加载myclass.Device * /
Hibernate：* /选择warrantyen0_.id为.. Hibernate：/ *加载myclass.Device * / ...
我一遍又一遍地猜测我有n + 1选择问题。
任何建议我如何只用一个来解决和替换所有这些选择？
完成此操作后
String query="from Customer c left join fetch c.devices d \n" +
"left join fetch d.tradeInOldDevice " +
"left join fetch d.tradeInNewDevice "+
"left join fetch d.warranty";
deviceDao.getSessionFactory().openSession().createQuery(query);
for (Device device : customerDevices) {
..
}
我仍然明白：
Hibernate:
devices0_.owningCompany_customerRefId as owningC15_0_1_,
devices0_.serial as serial1_,
devices0_.serial as serial9_0_,
devices0_.blackListed as blackLis2_9_0_,
devices0_.Creation_id as Creation12_9_0_,
devices0_.deactivated as deactiva3_9_0_,
devices0_.deviceComment as deviceCo4_9_0_,
devices0_.deviceName as deviceName9_0_,
devices0_.deviceType as deviceType9_0_,
devices0_.distributor_customerRefId as distrib13_9_0_,
devices0_.endCustomer_customerRefId as endCust14_9_0_,
devices0_.owningCompany_customerRefId as owningC15_9_0_,
devices0_.paChallenge as paChalle7_9_0_,
devices0_.parent_serial as parent16_9_0_,
devices0_.pendingDeactivation as pendingD8_9_0_,
devices0_.safetyStock as safetySt9_9_0_,
devices0_.serialSalt as serialSalt9_0_,
devices0_.signedBlackBerry as signedB11_9_0_,
devices0_.tradeInOldDevice as tradeIn17_9_0_
from
Devices devices0_
where
devices0_.owningCompany_customerRefId=?
Hibernate:
device0_.serial as serial9_0_,
device0_.blackListed as blackLis2_9_0_,
device0_.Creation_id as Creation12_9_0_,
device0_.deactivated as deactiva3_9_0_,
device0_.deviceComment as deviceCo4_9_0_,
device0_.deviceName as deviceName9_0_,
device0_.deviceType as deviceType9_0_,
device0_.distributor_customerRefId as distrib13_9_0_,
device0_.endCustomer_customerRefId as endCust14_9_0_,
device0_.owningCompany_customerRefId as owningC15_9_0_,
device0_.paChallenge as paChalle7_9_0_,
device0_.parent_serial as parent16_9_0_,
device0_.pendingDeactivation as pendingD8_9_0_,
device0_.safetyStock as safetySt9_9_0_,
device0_.serialSalt as serialSalt9_0_,
device0_.signedBlackBerry as signedB11_9_0_,
device0_.tradeInOldDevice as tradeIn17_9_0_
from
Devices device0_
where
device0_.tradeInOldDevice=?
Hibernate:
warrantyen0_.id as id34_2_,
warrantyen0_.createdTime as createdT2_34_2_,
warrantyen0_.deleted as deleted34_2_,
warrantyen0_.expiryDate as expiryDate34_2_,
warrantyen0_.serial as serial34_2_,
warrantyen0_.updateTime as updateTime34_2_,
warrantyen0_.updateUser as updateUser34_2_,
device1_.serial as serial9_0_,
device1_.blackListed as blackLis2_9_0_,
device1_.Creation_id as Creation12_9_0_,
device1_.deactivated as deactiva3_9_0_,
device1_.deviceComment as deviceCo4_9_0_,
device1_.deviceName as deviceName9_0_,
device1_.deviceType as deviceType9_0_,
device1_.distributor_customerRefId as distrib13_9_0_,
device1_.endCustomer_customerRefId as endCust14_9_0_,
device1_.owningCompany_customerRefId as owningC15_9_0_,
device1_.paChallenge as paChalle7_9_0_,
device1_.parent_serial as parent16_9_0_,
device1_.pendingDeactivation as pendingD8_9_0_,
device1_.safetyStock as safetySt9_9_0_,
device1_.serialSalt as serialSalt9_0_,
device1_.signedBlackBerry as signedB11_9_0_,
device1_.tradeInOldDevice as tradeIn17_9_0_,
management2_.id as id22_1_,
management2_1_.deleted as deleted22_1_,
management2_1_.firstName as firstName22_1_,
management2_1_.lastLogin as lastLogin22_1_,
management2_1_.lastName as lastName22_1_,
management2_1_.password as password22_1_,
management2_1_.primaryEmail as primaryE7_22_1_,
management2_1_.userName as userName22_1_,
management2_.authority as authority23_1_,
management2_.isViewer as isViewer23_1_,
management2_3_.distributor as distribu1_25_1_,
management2_4_.umeKeysQuota as umeKeysQ1_27_1_,
case
when management2_2_.id is not null then 2
when management2_3_.id is not null then 3
when management2_4_.id is not null then 5
when management2_5_.id is not null then 6
when management2_6_.id is not null then 7
when management2_.id is not null then 1
end as clazz_1_,
cids3_.Users_id as Users1_22_4_,
cids3_.element as element4_,
emails4_.Users_id as Users1_22_5_,
emails4_.element as element5_,
roles5_.Users_Management_id as Users1_22_6_,
roles5_.element as element6_
from
Warranty warrantyen0_
left outer join
Devices device1_
on warrantyen0_.serial=device1_.serial
left outer join
Users_Management management2_
on warrantyen0_.updateUser=management2_.id
left outer join
Users management2_1_
on management2_.id=management2_1_.id
left outer join
Users_Management_Administrators management2_2_
on management2_.id=management2_2_.id
left outer join
Users_Management_Distributors management2_3_
on management2_.id=management2_3_.id
left outer join
Users_Management_Limited management2_4_
on management2_.id=management2_4_.id
left outer join
Users_Management_Managers management2_5_
on management2_.id=management2_5_.id
left outer join
Users_Management_Workers management2_6_
on management2_.id=management2_6_.id
left outer join
Users_CID cids3_
on management2_.id=cids3_.Users_id
left outer join
Users_Emails emails4_
on management2_.id=emails4_.Users_id
left outer join
Users_Management_roles roles5_
on management2_.id=roles5_.Users_Management_id
where
warrantyen0_.serial=?
谢谢， 射线。
答案 0 :(得分：2)
假设您已在customerDevices类中定义Customer关联，如下所示：
@OneToMany(fetch = FetchType.LAZY, mappedBy = "device")
private Set<Device> customerDevices;
此LAZY映射关联会向您显示n+1选择问题。让我们考虑一个简单的查询，为给定的customer检索customerId：
session().createQuery("from Customer c where c.name=:name").setParameter("name", name);
这将返回customer，其中customerDevices的集合是未初始化的集合包装器。现在当你迭代循环时：
Set<Device> customerDevices = user.getCustomer().getDevices();
for (Device device : customerDevices) {
...
}
并且在访问devices的集合时，Hibernate必须从执行额外select语句的数据库中获取此惰性集合。
此问题的推荐解决方案是在代码中覆盖运行时的默认提取策略，您可以使用以下查询来实现：
session().createQuery(from Customer c left join fetch c.devices d
left join fetch d.warrantyEntry)
这将返回一个Customer以及相关的集合。因此，不是仅检索初始查询中的顶级对象，而是通过准确指定哪些关联将在初始查询中获取所有需要的数据在正在进行的工作单元中访问。
修改
有时，一对一映射会导致HQL急切获取失败。您可以在以下讨论中找到有关此问题的更多信息：HQL eager fetch failure和N+1 selects on left join。
提到的一个解决方案是将OneToOne映射更改为ManyToOne和OneToMany，我尝试过（使用类似类型的模型类），它对我来说非常适合。我通过一个SQL查询选择了所有结果。
例如，在您的Device课程中，您可以将OneToOne映射更改为ManyToOne：
@ManyToOne(cascade = CascadeType.ALL)
@JoinColumn(name = "warrantyFK", unique = true)
private WarrantyEntry warranty;
unique=true属性会使此关联成为一对一关联，因为没有两个Device可以拥有相同的WarrantyEntry。
您可以在Device类中定义WarrantyEntry的集合：
@OneToMany(mappedBy = "warranty")
private Set<Device> holdingDevices;
主要想法是使用OneToMany和ManyToOne代替OneToOne。您只需确保最多将一个项添加到映射的set侧的OneToMany。
答案 1 :(得分：0)
可以＆＃34; n + 1选择问题＆＃34;，您可以确保将这些属性放在persistence.xml中：
<property name="hibernate.show_sql">true</property>
<property name="hibernate.format_sql">true</property>
<property name="hibernate.use_sql_comments">true</property>
我认为你也可以在JoinColumn中使用unique = true，nullable = false。
您可以尝试使用JPQL / HQL进行急切获取并解决此问题（也许使用Criteria也会解决，不确定）。