如何缓存hibernate集合？

Question

我有两个实体com.gn.entities.Report和com.gn.entities.ReportSummary。

 <?xml version="1.0"?>
 <!DOCTYPE hibernate-mapping PUBLIC  "-//Hibernate/Hibernate Mapping DTD z.0//EN"  "http://hibernate.sourceforge.net/hibernate-mapping-3.0.dtd">
 <hibernate-mapping>
<class name="com.gn.entities.Report" table="REPORT" >
     <id name="id" type="int" column="id" >
             <generator class="native"/>
     </id>
     <property name="name">
           <column name="NAME" />
     </property>
     <set name="reportSummaryList" table="REPORT_SUMMARY" cascade="all" inverse="true" fetch="join" >
         <key column="RPT_ID" not-null="true"></key>
         <one-to-many class="com.gn.entities.ReportSummary"/>
     </set>
 </class>

 <?xml version="1.0"?>
 <!DOCTYPE hibernate-mapping PUBLIC"-//Hibernate/Hibernate Mapping DTD 3.0//EN""http://hibernate.sourceforge.net/hibernate-mapping-3.0.dtd">

<hibernate-mapping>
<class name="com.gn.entities.ReportSummary" table="REPORT_SUMMARY" >
    <composite-id name="id" class="com.gn.entities.ReportSummaryId">
         <key-property name="id" column="ID"></key-property>
         <key-many-to-one name="report" class="com.gn.entities.Report" column="RPT_ID"/>
    </composite-id>
    <property name="name">
         <column name="NAME" />
    </property>
</class>

我想知道它们之间有什么区别

场景1：实体和集合上的缓存标记

<class name="com.gn.entities.Report" table="REPORT" >
<cache usage="read-write" region="report"/>

和

<set name="reportSummaryList" table="REPORT_SUMMARY" cascade="all" inverse="true" fetch="join" >
      <cache usage="read-write" region="report"/>

  <?xml version="1.0"?>
  <!DOCTYPE hibernate-mapping PUBLIC "-//Hibernate/Hibernate Mapping DTD 3.0//EN" "http://hibernate.sourceforge.net/hibernate-mapping-3.0.dtd">
  <hibernate-mapping>
<class name="com.gn.entities.Report" table="REPORT" >
    <cache usage="read-write" region="report"/>
     <id name="id" type="int" column="id" >
             <generator class="native"/>
     </id>
     <property name="name">
           <column name="NAME" />
     </property>
     <set name="reportSummaryList" table="REPORT_SUMMARY" cascade="all" inverse="true" fetch="join" >
         <cache usage="read-write" region="report"/>
         <key column="RPT_ID" not-null="true"></key>
         <one-to-many class="com.gn.entities.ReportSummary"/>
     </set>
 </class>

和

方案2：仅在实体上的缓存标记（<cache usage="read-write" region="report"/>）

 <?xml version="1.0"?>
 <!DOCTYPE hibernate-mapping PUBLIC"-//Hibernate/Hibernate Mapping DTD 3.0//EN" "http://hibernate.sourceforge.net/hibernate-mapping-3.0.dtd">
 <hibernate-mapping>
<class name="com.gn.entities.Report" table="REPORT" >
    <cache usage="read-write" region="report"/>
     <id name="id" type="int" column="id" >
             <generator class="native"/>
     </id>
     <property name="name">
           <column name="NAME" />
     </property>
     <set name="reportSummaryList" table="REPORT_SUMMARY" cascade="all" inverse="true" fetch="join" >


   <!-- <cache usage="read-write" region="report"/>  -->

         <key column="RPT_ID" not-null="true"></key>
         <one-to-many class="com.gn.entities.ReportSummary"/>
     </set>
 </class>

问题1：在两种情况下，Report和ReportSummary都将被缓存？如果是，那么使实体和集合可以缓存的用途是什么？

问题2：此外，是否提取影响缓存（fetch = join或fetch = select）？在两种情况下？

     <set name="reportSummaryList" table="REPORT_SUMMARY" cascade="all" inverse="true" fetch="join" >

     and
     <set name="reportSummaryList" table="REPORT_SUMMARY" cascade="all" inverse="true" fetch="select" >

更新： -

好的，这意味着我必须使ReportSummary也可以缓存。完成后我发现我仍然收到如下日志消息： -

17：23：04,824 TRACE DefaultLoadEventListener：403 - 尝试解决：[com.gn.entities.ReportSummary #component [id，report] {report = com.gn.entities.Report＃1，id = 1}]

17：23：04,824 TRACE DefaultLoadEventListener：427 - 对象未在任何缓存中解析：[com.gn.entities.ReportSummary #component [id，report] {report = com.gn.entities.Report＃1，id = 1}]

17：23：04,824 TRACE AbstractEntityPersister：3923 - 获取实体：[com.gn.entities.ReportSummary #component [id，report] {report = com.gn.entities.Report＃1，id = 1}] 17:23： 04,824 DEBUG Loader：2105 - 正在加载实体：[com.gn.entities.ReportSummary #component [id，report] {report = com.gn.entities.Report＃1，id = 1}] 17：23：04,824 DEBUG SQL： 104 - / * load com.gn.entities.ReportSummary / select reportsumm0_.ID as ID1_1_0_，reportsumm0_.RPT_ID as RPT2_1_0_，reportsumm0_.NAME as NAME3_1_0_ from REPORT_SUMMARY reportsumm0_ where reportsumm0_.ID =？和reportsumm0_.RPT_ID =？ Hibernate：/ load com.gn.entities.ReportSummary * / select reportsumm0_.ID as ID1_1_0_，reportsumm0_.RPT_ID as RPT2_1_0_，reportsumm0_.NAME as NAME3_1_0_ from REPORT_SUMMARY reportsumm0_ where reportsumm0_.ID =？和reportsumm0_.RPT_ID =？

17：23：04,824 TRACE JdbcCoordinatorImpl：319 - 注册语句[com.mysql.jdbc.JDBC4PreparedStatement@e3c0e40：/ * load com.gn.entities.ReportSummary * / select reportsumm0_.ID as ID1_1_0_，reportsumm0_.RPT_ID as RPT2_1_0_，reportsumm0_。名称为NAME3_1_0_，来自REPORT_SUMMARY reportsumm0_，其中reportsumm0_.ID = **未指定**和reportsumm0_.RPT_ID = **未指定] 17：23：04,824 TRACE JdbcCoordinatorImpl：329 - 注册最后一个查询语句[com.mysql.jdbc.JDBC4PreparedStatement @ e3c0e40：/ * load com.gn.entities.ReportSummary * / select reportsumm0_.ID as ID1_1_0_，reportsumm0_.RPT_ID as RPT2_1_0_，reportsumm0_.NAME as NAME3_1_0_ from REPORT_SUMMARY reportsumm0_ where reportsumm0_.ID = NOT SPECIFIED **和reportsumm0_.RPT_ID = * * 未标明 **]

这意味着hibernate找到了ReportSummary id，它将命中数据库以使用该id获取ReportSummary实体？

那么，如果有10000个ReportSummary实体，那么将有10000个数据库命中？

Answer 1

在任何一种情况下都不会缓存ReportSummary实例，因为您没有将该实体类标记为可缓存。

如果Report实体被声明为可缓存，那么显然Hibernate会将其实例缓存在二级缓存中。

如果reportSummaryList不可缓存，则无论Report是否可缓存，都不会缓存该集合。

如果reportSummaryList是可缓存的，那么Hibernate会缓存ReportSummary中存在的reportSummaryList实例的id。 ReportSummary是否可缓存无关紧要; ids缓存在单独的缓存条目中。但是，在这种情况下，您可能还希望使ReportSummary实体可缓存，以便在根据缓存集合条目中包含的ID读取其实例时提高性能。

更多细节here和here。