[英]How to use second level cache for lazy loaded collections in Hibernate?
假设我有两个实体Employee
和Skill
。 每个员工都有一套技能。 现在,当我懒散地通过Employee
实例加载技能时,缓存不用于Employee
不同实例中的技能。
让我们考虑以下数据集。
Employee - 1 : Java, PHP
Employee - 2 : Java, PHP
当我在Employee-1之后加载Employee-2时,我不想让休眠进入数据库以获取技能,而是使用缓存中已经可用的Skill
实例。 这可能吗? 如果可以,怎么办?
休眠配置
<session-factory>
<property name="hibernate.connection.driver_class">com.mysql.jdbc.Driver</property>
<property name="hibernate.connection.password">pass</property>
<property name="hibernate.connection.url">jdbc:mysql://localhost/cache</property>
<property name="hibernate.connection.username">root</property>
<property name="hibernate.dialect">org.hibernate.dialect.MySQLInnoDBDialect</property>
<property name="hibernate.cache.use_second_level_cache">true</property>
<property name="hibernate.cache.use_query_cache">true</property>
<property name="hibernate.cache.provider_class">net.sf.ehcache.hibernate.EhCacheProvider</property>
<property name="hibernate.hbm2ddl.auto">update</property>
<property name="hibernate.show_sql">true</property>
<mapping class="org.cache.models.Employee" />
<mapping class="org.cache.models.Skill" />
</session-factory>
删除了具有导入,获取和设置程序的实体
@Entity
@Table(name = "employee")
@Cache(usage = CacheConcurrencyStrategy.READ_WRITE)
public class Employee {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private int id;
private String name;
public Employee() {
}
@ManyToMany
@JoinTable(name = "employee_skills", joinColumns = @JoinColumn(name = "employee_id"), inverseJoinColumns = @JoinColumn(name = "skill_id"))
@Cache(usage = CacheConcurrencyStrategy.READ_WRITE)
private List<Skill> skills;
}
@Entity
@Table(name = "skill")
@Cache(usage = CacheConcurrencyStrategy.READ_WRITE)
public class Skill {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private int id;
private String name;
}
用于加载第二名员工及其技能的SQL
Hibernate: select employee0_.id as id0_0_, employee0_.name as name0_0_ from employee employee0_ where employee0_.id=?
Hibernate: select skills0_.employee_id as employee1_1_, skills0_.skill_id as skill2_1_, skill1_.id as id1_0_, skill1_.name as name1_0_ from employee_skills skills0_ left outer join skill skill1_ on skills0_.skill_id=skill1_.id where skills0_.employee_id=?
我特别想避免第二个查询,因为无论如何第一个查询是不可避免的。
您需要缓存Employee--<>Skills
关联。 下面的示例来自使用二级缓存来加速您的Hibernate应用程序 :
<hibernate-mapping package="com.wakaleo.articles.caching.businessobjects">
<class name="Employee" table="EMPLOYEE" dynamic-update="true">
<meta attribute="implement-equals">true</meta>
<id name="id" type="long" unsaved-value="null" >
<column name="emp_id" not-null="true"/>
<generator class="increment"/>
</id>
<property column="emp_surname" name="surname" type="string"/>
<property column="emp_firstname" name="firstname" type="string"/>
<many-to-one name="country"
column="cn_id"
class="com.wakaleo.articles.caching.businessobjects.Country"
not-null="true" />
<!-- Lazy-loading is deactivated to demonstrate caching behavior -->
<set name="languages" table="EMPLOYEE_SPEAKS_LANGUAGE" lazy="false">
<cache usage="read-write"/>
<key column="emp_id"/>
<many-to-many column="lan_id" class="Language"/>
</set>
</class>
</hibernate-mapping>
注意语言中的<cache>
元素。
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