类关系:  User N~1 Group

测试代码:

         System.out.println("1");
List stuList = session.createQuery("from User s where s.password ='123'").list(); //(1)
System.out.println("2");
for(Iterator it = stuList.iterator(); it.hasNext();){
User stu = (User)it.next();
System.out.println("3");
System.out.println("[value]"+stu.getName()); //(2)
System.out.println("[value]"+stu.getGroup().getName()); //(3)
}

执行效果:

(1)查询符合条件的User集合. 发出的SQL为 select  * from test_User where password='123'
(2)不再发出SQL. 使用缓存中的User数据.

(3)迭代User时,检查是否加载过对应User的Group,若无则发出一次SQL:  select * from test_group where group_id=?

     如果User有10个.对应在3个group中. 则发出3条select * from test_group where group_id=?

1
Hibernate: 
    select
        user0_.id as id0_,
        user0_.name as name0_,
        user0_.password as password0_,
        user0_.createTime as createTime0_,
        user0_.updateTime as updateTime0_,
        user0_.group_id as group6_0_
    from
        TEST_User user0_
    where
        user0_.password='123'
2
3
[value]jordan
Hibernate:
    select
        group0_.id as id1_0_,
        group0_.name as name1_0_
    from
        TEST_Group group0_
    where
        group0_.id=?
[value]集团2 
  
3
[value]kobe
[value]集团2
3
[value]james
[value]集团2
//以下继续开始循环遍历
3
[value]张三111
Hibernate:
    select
        group0_.id as id1_0_,
        group0_.name as name1_0_
    from
        TEST_Group group0_
    where
        group0_.id=?
[value]集团1
3
[value]李四111
[value]集团1
3
[value]李四111
[value]集团1
3
[value]AAA
Hibernate:
    select
        group0_.id as id1_0_,
        group0_.name as name1_0_
    from
        TEST_Group group0_
    where
        group0_.id=?
[value]集团3
3
[value]BBB
[value]集团3
3
[value]CCC
[value]集团3
3
[value]a1
Hibernate:
    select
        group0_.id as id1_0_,
        group0_.name as name1_0_
    from
        TEST_Group group0_
    where
        group0_.id=?
[value]集团4
3
[value]a11
[value]集团4
3
[value]a111
[value]集团4
3
[value]b1
Hibernate:
    select
        group0_.id as id1_0_,
        group0_.name as name1_0_
    from
        TEST_Group group0_
    where
        group0_.id=?
[value]集团5
3
[value]b11
[value]集团5
3
[value]b111
[value]集团5

 配置batch-size

<class name="com.test.entity.Group" table="TEST_Group" batch-size="2">

 执行效果

 (1)(2)同上

 (3)在遍历User,并且获取对应的Group时.检测是否加载过对应的数据.若无根据设置的batch-size=2

     发出SQL语句: select * from test_group where group_id in (?,?)  一次查询2个Group放到内存中.

     在此遍历User及其Group时,若内存中无加载过对应数据,则再次发出SQL  select * from test_group where group_id in (?,?)  一次查询2个Group放到内存中.

     如Group的数据最后不足.则使用select * from test_group where group_id =? 进行加载

     即如Group 有5个. 则通过2个2个1个的方式进行查询获取到.

1
Hibernate:
select
user0_.id as id0_,
user0_.name as name0_,
user0_.password as password0_,
user0_.createTime as createTime0_,
user0_.updateTime as updateTime0_,
user0_.group_id as group6_0_
from
TEST_User user0_
where
user0_.password='123'
2
3
[value]jordan
Hibernate:
select
group0_.id as id1_0_,
group0_.name as name1_0_
from
TEST_Group group0_
where
group0_.id in (
?, ?
)
[value]集团2
3
[value]kobe
[value]集团2
3
[value]james
[value]集团2
3
[value]张三111
[value]集团1
3
[value]李四111
[value]集团1
3
[value]李四111
[value]集团1
3
[value]AAA
Hibernate:
select
group0_.id as id1_0_,
group0_.name as name1_0_
from
TEST_Group group0_
where
group0_.id in (
?, ?
)
[value]集团3
3
[value]BBB
[value]集团3
3
[value]CCC
[value]集团3
3
[value]a1
[value]集团4
3
[value]a11
[value]集团4
3
[value]a111
[value]集团4
3
[value]b1
Hibernate:
select
group0_.id as id1_0_,
group0_.name as name1_0_
from
TEST_Group group0_
where
group0_.id=?
[value]集团5
3
[value]b11
[value]集团5
3
[value]b111
[value]集团5

P.S 注意如果使用iterate进行迭代而非List ,则Batch-size无效.

如下

        System.out.println("1");
Iterator iter = session.createQuery(" from Group g where g.name like'集团%'").iterate(); //(1)
System.out.println("2");
while (iter.hasNext()) {
System.out.println("3");
Group g = (Group) iter.next();
System.out.println("4");
System.out.println("[value]" + g.getName()); //(2)
System.out.println("5");
System.out.println("[value]" + g.getUsers().size()); //(3)
}

执行效果

原因的出现是因为

(1)处使用了iterate返回集合.iterate默认会开始发出一条语句:查询所有记录符合条件的user的ID. select id from user where password =?

因为(1)只查询出id , 所以在(2)时. 需要再发出一条SQL获取user的属性数据. select * from user where id=?

这就是iterate很出名的N+1问题.

至于(3)处, 如果group没加载过,每次都会发出一条SQL. select * from group where id=?

所以此情况下, batch-size已无法达到预期效果.

1
Hibernate:
select
user0_.id as col_0_0_
from
TEST_User user0_
where
user0_.password='123'
2
3
4
Hibernate:
select
user0_.id as id0_0_,
user0_.name as name0_0_,
user0_.password as password0_0_,
user0_.createTime as createTime0_0_,
user0_.updateTime as updateTime0_0_,
user0_.group_id as group6_0_0_
from
TEST_User user0_
where
user0_.id=?
[value]jordan
5
Hibernate:
select
group0_.id as id1_0_,
group0_.name as name1_0_
from
TEST_Group group0_
where
group0_.id=?
[value]集团2
3
4
Hibernate:
select
user0_.id as id0_0_,
user0_.name as name0_0_,
user0_.password as password0_0_,
user0_.createTime as createTime0_0_,
user0_.updateTime as updateTime0_0_,
user0_.group_id as group6_0_0_
from
TEST_User user0_
where
user0_.id=?
[value]kobe
5
[value]集团2
3
4
Hibernate:
select
user0_.id as id0_0_,
user0_.name as name0_0_,
user0_.password as password0_0_,
user0_.createTime as createTime0_0_,
user0_.updateTime as updateTime0_0_,
user0_.group_id as group6_0_0_
from
TEST_User user0_
where
user0_.id=?
[value]james
5
[value]集团2
3
4
Hibernate:
select
user0_.id as id0_0_,
user0_.name as name0_0_,
user0_.password as password0_0_,
user0_.createTime as createTime0_0_,
user0_.updateTime as updateTime0_0_,
user0_.group_id as group6_0_0_
from
TEST_User user0_
where
user0_.id=?
[value]张三111
5
Hibernate:
select
group0_.id as id1_0_,
group0_.name as name1_0_
from
TEST_Group group0_
where
group0_.id=?
[value]集团1
3
4
Hibernate:
select
user0_.id as id0_0_,
user0_.name as name0_0_,
user0_.password as password0_0_,
user0_.createTime as createTime0_0_,
user0_.updateTime as updateTime0_0_,
user0_.group_id as group6_0_0_
from
TEST_User user0_
where
user0_.id=?
[value]李四111
5
[value]集团1
3
4
Hibernate:
select
user0_.id as id0_0_,
user0_.name as name0_0_,
user0_.password as password0_0_,
user0_.createTime as createTime0_0_,
user0_.updateTime as updateTime0_0_,
user0_.group_id as group6_0_0_
from
TEST_User user0_
where
user0_.id=?
[value]李四111
5
[value]集团1
3
4
Hibernate:
select
user0_.id as id0_0_,
user0_.name as name0_0_,
user0_.password as password0_0_,
user0_.createTime as createTime0_0_,
user0_.updateTime as updateTime0_0_,
user0_.group_id as group6_0_0_
from
TEST_User user0_
where
user0_.id=?
[value]AAA
5
Hibernate:
select
group0_.id as id1_0_,
group0_.name as name1_0_
from
TEST_Group group0_
where
group0_.id=?
[value]集团3
3
4
Hibernate:
select
user0_.id as id0_0_,
user0_.name as name0_0_,
user0_.password as password0_0_,
user0_.createTime as createTime0_0_,
user0_.updateTime as updateTime0_0_,
user0_.group_id as group6_0_0_
from
TEST_User user0_
where
user0_.id=?
[value]BBB
5
[value]集团3
3
4
Hibernate:
select
user0_.id as id0_0_,
user0_.name as name0_0_,
user0_.password as password0_0_,
user0_.createTime as createTime0_0_,
user0_.updateTime as updateTime0_0_,
user0_.group_id as group6_0_0_
from
TEST_User user0_
where
user0_.id=?
[value]CCC
5
[value]集团3
3
4
Hibernate:
select
user0_.id as id0_0_,
user0_.name as name0_0_,
user0_.password as password0_0_,
user0_.createTime as createTime0_0_,
user0_.updateTime as updateTime0_0_,
user0_.group_id as group6_0_0_
from
TEST_User user0_
where
user0_.id=?
[value]a1
5
Hibernate:
select
group0_.id as id1_0_,
group0_.name as name1_0_
from
TEST_Group group0_
where
group0_.id=?
[value]集团4
3
4
Hibernate:
select
user0_.id as id0_0_,
user0_.name as name0_0_,
user0_.password as password0_0_,
user0_.createTime as createTime0_0_,
user0_.updateTime as updateTime0_0_,
user0_.group_id as group6_0_0_
from
TEST_User user0_
where
user0_.id=?
[value]a11
5
[value]集团4
3
4
Hibernate:
select
user0_.id as id0_0_,
user0_.name as name0_0_,
user0_.password as password0_0_,
user0_.createTime as createTime0_0_,
user0_.updateTime as updateTime0_0_,
user0_.group_id as group6_0_0_
from
TEST_User user0_
where
user0_.id=?
[value]a111
5
[value]集团4
3
4
Hibernate:
select
user0_.id as id0_0_,
user0_.name as name0_0_,
user0_.password as password0_0_,
user0_.createTime as createTime0_0_,
user0_.updateTime as updateTime0_0_,
user0_.group_id as group6_0_0_
from
TEST_User user0_
where
user0_.id=?
[value]b1
5
Hibernate:
select
group0_.id as id1_0_,
group0_.name as name1_0_
from
TEST_Group group0_
where
group0_.id=?
[value]集团5
3
4
Hibernate:
select
user0_.id as id0_0_,
user0_.name as name0_0_,
user0_.password as password0_0_,
user0_.createTime as createTime0_0_,
user0_.updateTime as updateTime0_0_,
user0_.group_id as group6_0_0_
from
TEST_User user0_
where
user0_.id=?
[value]b11
5
[value]集团5
3
4
Hibernate:
select
user0_.id as id0_0_,
user0_.name as name0_0_,
user0_.password as password0_0_,
user0_.createTime as createTime0_0_,
user0_.updateTime as updateTime0_0_,
user0_.group_id as group6_0_0_
from
TEST_User user0_
where
user0_.id=?
[value]b111
5
[value]集团5

hibernate.jdbc.fetch_size 和 hibernate.jdbc.batch_size

摘自:http://dongruan00.iteye.com/blog/1835081

hibernate.jdbc.fetch_size 50 //读

hibernate.jdbc.batch_size 30 //写

hiberante.cfg.xml(Oracle ,sql server 支持,mysql不支持)

配置文件参考如下:

<property name="hibernate.jdbc.fetch_size">50</property>

<property name="hibernate.jdbc.batch_size">30</property>

这两个选项非常非常非常重要!!!将严重影响Hibernate的CRUD性能!

C = create, R = read, U = update, D = delete

Fetch Size 是设定JDBC的Statement读取数据的时候每次从数据库中取出的记录条数。

例如一次查询1万条记录,对于Oracle的JDBC驱动来说,是不会1次性把1万条取出来的,而只会取出Fetch Size条数,当纪录集遍历完了这些记录以后,再去数据库取Fetch Size条数据。因此大大节省了无谓的内存消耗。当然Fetch Size设的越大,读数据库的次数越少,速度越快;Fetch Size越小,读数据库的次数越多,速度越慢。这有点像平时我们写程序写硬盘文件一样,设立一个Buffer,每次写入Buffer,等Buffer满了以后,一次写入硬盘,道理相同。

Oracle数据库的JDBC驱动默认的Fetch Size=10,是一个非常保守的设定,根据我的测试,当Fetch Size=50的时候,性能会提升1倍之多,当Fetch Size=100,性能还能继续提升20%,Fetch Size继续增大,性能提升的就不显著了。

因此我建议使用Oracle的一定要将Fetch Size设到50。

不过并不是所有的数据库都支持Fetch Size特性,例如MySQL就不支持。

MySQL就像我上面说的那种最坏的情况,他总是一下就把1万条记录完全取出来,内存消耗会非常非常惊人!这个情况就没有什么好办法了 :(

Batch Size是设定对数据库进行批量删除,批量更新和批量插入的时候的批次大小,有点相当于设置Buffer缓冲区大小的意思。

Batch Size越大,批量操作的向数据库发送sql的次数越少,速度就越快。我做的一个测试结果是当Batch Size=0的时候,使用Hibernate对Oracle数据库删除1万条记录需要25秒,Batch Size = 50的时候,删除仅仅需要5秒!!!

我们通常不会直接操作一个对象的标识符(identifier), 因此标识符的setter方法应该被声明为私有的(private)。这样当一个对象被保存的时候,只有Hibernate可以为它分配标识符。 你会发现Hibernate可以直接访问被声明为public,private和protected等不同级别访问控制的方法(accessor method)和字段(field)。 所以选择哪种方式来访问属性是完全取决于你,你可以使你的选择与你的程序设计相吻合。

所有的持久类(persistent classes)都要求有无参的构造器(no-argument constructor); 因为Hibernate必须要使用Java反射机制(Reflection)来实例化对象。构造器(constructor)的访问控制可以是私有的(private), 然而当生成运行时代理(runtime proxy)的时候将要求使用至少是package级别的访问控制,这样在没有字节码编入 (bytecode instrumentation)的情况下,从持久化类里获取数据会更有效率一些。

hibernate.max_fetch_depth 设置外连接抓取树的最大深度

取值. 建议设置为03之间

就是每次你在查询时,会级联查询的深度,譬如你对关联vo设置了eager的话,如果fetch_depth值太小的话,会发多很多条sql

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