一?/span>hibernate抓取{略Q单端代理的扚w抓取fetch=select(默认Q?/span>/joinQ?/span>
试用例Q?/span>
Student student = (Student)session.get(Student.class, 1);
System.out.println(student.getName());
System.out.println(student.getClasses().getName());
1Q保?/span>默认Q同fetch="select",如：
<many-to-one name="classes" column="classesid" fetch="select"/>
fetch="select",另外发送一?/span>select语句抓取当前对象兌实体或集?/span>
执行l果Q?/span>2条语?/span>
Hibernate: select student0_.id as id1_0_, student0_.name as name1_0_, student0_.class_id as class3_1_0_ from student_join student0_ where student0_.id=?
学生1
Hibernate: select classes0_.id as id0_0_, classes0_.name as name0_0_ from classes_join classes0_ where classes0_.id=?
高一(1)?/span>
======================================
2Q设|?/span>fetch="join",如：
<many-to-one name="classes" column="classesid" fetch="join"/>
fetch="join",hibernate会通过select语句使用外连接来加蝲其关联实体或集合
此时lazy会失?/span>
执行l果Q一?/span>join语句
Hibernate: select student0_.id as id1_1_, student0_.name as name1_1_, student0_.class_id as class3_1_1_, classes1_.id as id0_0_, classes1_.name as name0_0_ from student_join student0_ left outer join classes_join classes1_ on student0_.class_id=classes1_.id where student0_.id=?
学生1
高一(1)?/span>
======================================================
二?/span>hibernate抓取{略Q集合代理的扚w抓取Q?/span>fetch=selectQ默认）/join/subselectQ?/span>
试用例Q?/span>
Classes c = (Classes) session.load(Classes.class, new Integer(1));
System.out.println("Class.name=" + c.getName());
Set stuSet = c.getStudents();
System.out.println(stuSet.size());
if(stuSet != null && !stuSet.isEmpty()){
for(Iterator it = stuSet.iterator(); it.hasNext();){
Student s = (Student) it.next();
System.out.println("student.name=" + s.getName());
}
}
1Q保持默认，?/span>fetch="select",如：
<set name="students" inverse="true" fetch="select">
fetch="select",另外发送一?/span>select语句抓取当前对象兌实体或集?/span>
试l果Q?/span>2条独立的查询语句
Hibernate: select classes0_.id as id0_0_, classes0_.name as name0_0_ from classes_join classes0_ where classes0_.id=?
Class.name=高一(1)?/span>
Hibernate: select students0_.class_id as class3_1_, students0_.id as id1_, students0_.id as id1_0_, students0_.name as name1_0_, students0_.class_id as class3_1_0_ from student_join students0_ where students0_.class_id=?
9
student.name=学生7
student.name=学生3
student.name=学生1
student.name=学生8
student.name=学生2
student.name=学生4
student.name=学生5
student.name=学生9
student.name=学生6
(2)讄fetch="join",如：
<set name="students" inverse="true" fetch="join">
fetch="join",hibernate会通过select语句使用外连接来加蝲其关联实体或集合
此时lazy会失?/span>
试l果Q?/span>1条独立的join查询语句
Hibernate: select classes0_.id as id0_1_, classes0_.name as name0_1_, students1_.class_id as class3_3_, students1_.id as id3_, students1_.id as id1_0_, students1_.name as name1_0_, students1_.class_id as class3_1_0_ from classes_join classes0_ left outer join student_join students1_ on classes0_.id=students1_.class_id where classes0_.id=?
Class.name=高一(1)?/span>
9
student.name=学生6
student.name=学生4
student.name=学生9
student.name=学生7
student.name=学生2
student.name=学生3
student.name=学生8
student.name=学生1
student.name=学生5
(3)讄fetch="subselect",如：用在查询语句?/span>
<set name="students" inverse="true" fetch="subselect">
fetch="subselect",另外发送一?/span>select语句抓取在前面查询到的所有实体对象的兌集合
试用例Q?/span>
List classList = session.createQuery("from Classes where id in (1,2,3)").list();
for(Iterator iter = classList.iterator(); iter.hasNext();){
Classes c = (Classes)iter.next();
System.out.println("Class.name=" + c.getName());
Set stuSet = c.getStudents();
System.out.println(stuSet.size());
if(stuSet != null && !stuSet.isEmpty()){
for(Iterator it = stuSet.iterator(); it.hasNext();){
Student s = (Student) it.next();
System.out.println("student.name=" + s.getName());
}
}
}
当不?/span>fetch="subselect" ,卻I<set name="students" inverse="true">,l果如下Q?/span>
执行?/span>3条查询语?/span>
Hibernate: select classes0_.id as id0_, classes0_.name as name0_ from classes_join classes0_ where classes0_.id in (1 , 2 , 3)
Class.name=高一(1)?/span>
Hibernate: select students0_.class_id as class3_1_, students0_.id as id1_, students0_.id as id1_0_, students0_.name as name1_0_, students0_.class_id as class3_1_0_ from student_join students0_ where students0_.class_id=?
9
student.name=学生8
student.name=学生5
student.name=学生3
student.name=学生9
student.name=学生7
student.name=学生1
student.name=学生4
student.name=学生6
student.name=学生2
Class.name=高一(2)?/span>
Hibernate: select students0_.class_id as class3_1_, students0_.id as id1_, students0_.id as id1_0_, students0_.name as name1_0_, students0_.class_id as class3_1_0_ from student_join students0_ where students0_.class_id=?
4
student.name=学生3
student.name=学生4
student.name=学生1
student.name=学生2
Class.name=高一(3)?/span>
Hibernate: select students0_.class_id as class3_1_, students0_.id as id1_, students0_.id as id1_0_, students0_.name as name1_0_, students0_.class_id as class3_1_0_ from student_join students0_ where students0_.class_id=?
0
当不?/span>fetch="subselect" ,卻I<set name="students" inverse="true" fetch="subselect">,l果如下Q?/span>
执行?/span>1条查询语句（嵌套子查询）
Hibernate: select classes0_.id as id0_, classes0_.name as name0_ from classes_join classes0_ where classes0_.id in (1 , 2 , 3)
Class.name=高一(1)?/span>
Hibernate: select students0_.class_id as class3_1_, students0_.id as id1_, students0_.id as id1_0_, students0_.name as name1_0_, students0_.class_id as class3_1_0_ from student_join students0_ where students0_.class_id in (select classes0_.id from classes_join classes0_ where classes0_.id in (1 , 2 , 3))
9
student.name=学生8
student.name=学生4
student.name=学生5
student.name=学生9
student.name=学生6
student.name=学生2
student.name=学生3
student.name=学生1
student.name=学生7
Class.name=高一(2)?/span>
4
student.name=学生3
student.name=学生4
student.name=学生2
student.name=学生1
Class.name=高一(3)?/span>
0
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1.量使用many-to-oneQ避免用单one-to-many
2.灉|使用单向one-to-many
3.不用一对一Q用多对一代替一对一
4.配置对象~存Q不使用集合~存
5.一对多使用Bag 多对一使用Set
6.l承使用昄多?nbsp; HQL:from object polymorphism="exlicit" 避免查处所有对?br />7.消除大表Q用二U缓?br /> 对于上面q些QRobbinq行了详l的讲解?br />one-to-manyQ?/strong>
使用inverse=false(default)Q对象的兌关系是由parent对象来维护的
而inverse=true的情况下Q一般用户双向多对多兌Q由子对象维护关联关p，增加子对象的时候需要显C：child.setParent(child)
Z提高性能Q应该尽量用双向one-to-many inverse=trueQ在MVCl构中的DAO接口中应该直接用Session持久化对象，避免通过兌关系Q这句话有点不理解）Q而在单项关系中正用二U缓存，则可以大q提高以查询Z的应用?br /> 多对一性能问题比较，但是要避免经典N+1问题?br /> 通过主键q行兌Q相当于大表拆分表。（q个是区分面向对象设计和面向q程设计的一个关键点Q?br />list、bag、set的正运?/strong>
one-to-manyQ?br /> A、用list 需要维护Index Column字段Q不能被用于双向兌Q而且必须使用inverse=falseQ需要}慎用在某些E有场合（基本上是不予考虑使用Q?br /> B、bag/set在one-to-many中语义基本相同，推荐使用bag
many-to-oneQ?br /> A、bag和set不同Qbag允许重复插入Q徏议用set
在庞大的集合分页中应该用session.createFilter
session.createFilter(parent.getChildren(),""),setFirstResult(0),setMaxResult(10))
避免N+1 参考（http://www.iteye.com/post/266972Q?/strong>
在多对一的情况下Q查询child对象Q当在页面上昄每个子类的父cd象的时候会DN+1ơ查询，需要采用下面的Ҏ避免Qmany-to-one fetch="join|select"Q该Ҏ可能有问题）
inverse=true 无法l护集合~存Q还不是很理解集合缓存和对象~存Q?/strong>
OLTPcd的web应用Q可以群集水qx展，不可避免的出现数据库瓉
框架能降低访问数据库的压力，采用~存是衡量一个框架是否优U的重要标准，从缓存方面看Hibernate
A、对象缓存，l颗_度Q是针对表的U别Q透明化访问，因ؓ有不改变代码的好处，所以是ORM提高性能的法?br /> B、Hibernate是目前ORM框架中缓存性能最好的框架
C、查询缓?br />最后Robbinq针对大家经常出现的Hibernate vs iBatis的讨行了一个ȝQ?br /> 对于OLTP应用Q用ORM框架 而OLEB应用Q不定是什么应用）最好采用JDBC或者其他方法处?br /> Hibernate們于细颗粒度设计，面向对象Q将大表拆分为多个小表，消除冗余字段Q通过二~存提升性能?br /> iBatis們于粗颗粒度设计，面向关系Q尽量把表合qӞ通过Column冗余Q消除关联关p，但是iBatis没有有效的缓存手Dc?/span>
可以说Robbin的性能ȝ对于使用Hibernate的开发h员有着很重要的Ҏ作用。非常感谢他无私奉献自己的经验?/span>
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