Hibernate06_配置文件详解

dynamic-insert=“true” 动态添加

dynamic-update=“true” 动态更新

where=“id = 1” 附加查询条件

inverse=“true” 放弃关系维护权

cascade=“true” 设置级联操作

dynamic-insert=“true” 动态添加

测试：

People people  =new People();
people.setName("李四");
session.save(people);
session.beginTransaction().commit();
session.close();

结果：

Hibernate: 
    insert 
    into
        people
        (name, money) 
    values
        (?, ?)

修改People.hbm.xml，设置 dynamic-insert="true"

<?xml version="1.0" encoding="UTF-8"?>
<!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.blu.entity.People" table="people" dynamic-insert="true">
		<id name="id" type="java.lang.Integer">
			<column name="id"></column>
			<generator class="identity"></generator>
		</id>
		<property name="name" type="java.lang.String">
			<column name="name"></column>
		</property>
		<property name="money" type="java.lang.Double">
			<column name="money"></column>
		</property>
	</class>

</hibernate-mapping>

再次测试的结果：

Hibernate: 
    insert 
    into
        people
        (name) 
    values
        (?)

dynamic-update=“true” 动态更新

测试：

People people = session.get(People.class, 2);
people.setMoney(2000.0);
session.update(people);
session.beginTransaction().commit();
session.close();

结果：

Hibernate: 
    select
        people0_.id as id1_3_0_,
        people0_.name as name2_3_0_,
        people0_.money as money3_3_0_ 
    from
        people people0_ 
    where
        people0_.id=?
Hibernate: 
    update
        people 
    set
        name=?,
        money=? 
    where
        id=?

修改People.hbm.xml，设置 dynamic-update="true"

配置文件略

再次测试：

People people = session.get(People.class, 3);
people.setMoney(3000.0);
session.update(people);
session.beginTransaction().commit();
session.close();

结果：

Hibernate: 
    select
        people0_.id as id1_3_0_,
        people0_.name as name2_3_0_,
        people0_.money as money3_3_0_ 
    from
        people people0_ 
    where
        people0_.id=?
Hibernate: 
    update
        people 
    set
        money=? 
    where
        id=?

总结：使用 dynamic-insert=“true” 和 dynamic-update=“true” 后，SQL语句被优化，执行效率提高，提升了系统性能！

where=“id = 1” 附加查询条件

测试

String hql = "from People";
Query query = session.createQuery(hql);
List<People> list = query.list();
for(People people : list) {
	System.out.println(people);
}
session.beginTransaction().commit();
session.close();

结果：

Hibernate: 
    select
        people0_.id as id1_3_,
        people0_.name as name2_3_,
        people0_.money as money3_3_ 
    from
        people people0_
People(id=1, name=张三, money=1000.0)
People(id=2, name=李四, money=2000.0)
People(id=3, name=王五, money=3000.0)

修改People.hbm.xml，设置 where="id = 1"

配置文件略

再次测试的结果：

Hibernate: 
    select
        people0_.id as id1_3_,
        people0_.name as name2_3_,
        people0_.money as money3_3_ 
    from
        people people0_ 
    where
        (
            people0_.id = 1
        )
People(id=1, name=张三, money=1000.0)

inverse="true" 放弃关系维护权

测试：

Customer customer = new Customer();
customer.setName("李四");
		
Orders order1 = new Orders();
order1.setName("订单a");
order1.setCustomer(customer);
		
Orders order2 = new Orders();
order2.setName("订单b");
order2.setCustomer(customer);
		
Set<Orders> orders = new HashSet<Orders>();
orders.add(order1);
orders.add(order2);
customer.setOrders(orders);

session.save(customer);	
session.save(order1);
session.save(order2);
session.beginTransaction().commit();
session.close();

结果执行了5条SQL（如果先保存订单后保存客户，则会执行七条SQL）：

Hibernate: 
    insert 
    into
        customer
        (name) 
    values
        (?)
Hibernate: 
    insert 
    into
        orders
        (name, cid) 
    values
        (?, ?)
Hibernate: 
    insert 
    into
        orders
        (name, cid) 
    values
        (?, ?)
Hibernate: 
    update
        orders 
    set
        cid=? 
    where
        id=?
Hibernate: 
    update
        orders 
    set
        cid=? 
    where
        id=?

原因是Customer和Orders都在维护一对多关系，所以会重复设置主外键约束关系。

解决方法1：去掉一方维护关系的代码

Customer customer = new Customer();
customer.setName("李四");
		
Orders order1 = new Orders();
order1.setName("订单a");
order1.setCustomer(customer);
		
Orders order2 = new Orders();
order2.setName("订单b");
order2.setCustomer(customer);

session.save(customer);
session.save(order1);
session.save(order2);
session.beginTransaction().commit();
session.close();

结果：

Hibernate: 
    insert 
    into
        customer
        (name) 
    values
        (?)
Hibernate: 
    insert 
    into
        orders
        (name, cid) 
    values
        (?, ?)
Hibernate: 
    insert 
    into
        orders
        (name, cid) 
    values
        (?, ?)

解决方法2：通过配置 inverse="true" 让Customer放弃关系维护权

<set name="orders" table="orders" lazy="extra" inverse="true">
	<key column="cid"></key>
	<one-to-many class="com.blu.entity.Orders"></one-to-many>
</set>

cascade="true" 设置级联操作

测试：

Customer customer = session.get(Customer.class, 10);
session.delete(customer);
session.beginTransaction().commit();
session.close();

删除报错，原因：外键约束
解决方法1：先删除所有的orders，再删除customer

Customer customer = session.get(Customer.class, 10);
Iterator<Orders> iterator = customer.getOrders().iterator();
while(iterator.hasNext()) {
	session.delete(iterator.next());
}
session.delete(customer);
session.beginTransaction().commit();
session.close();

结果：

Hibernate: 
    select
        customer0_.id as id1_1_0_,
        customer0_.name as name2_1_0_ 
    from
        customer customer0_ 
    where
        customer0_.id=?
Hibernate: 
    select
        orders0_.cid as cid3_2_0_,
        orders0_.id as id1_2_0_,
        orders0_.id as id1_2_1_,
        orders0_.name as name2_2_1_,
        orders0_.cid as cid3_2_1_ 
    from
        orders orders0_ 
    where
        orders0_.cid=?
Hibernate: 
    delete 
    from
        orders 
    where
        id=?
Hibernate: 
    delete 
    from
        orders 
    where
        id=?
Hibernate: 
    delete 
    from
        customer 
    where
        id=?

解决方法2： 通过配置 cascade="delete" 开启级联删除操作

<set name="orders" table="orders" lazy="extra" inverse="true" cascade="delete">
	<key column="cid"></key>
	<one-to-many class="com.blu.entity.Orders"></one-to-many>
</set>

测试：

Customer customer = session.get(Customer.class, 11);
session.delete(customer);
session.beginTransaction().commit();
session.close();

结果：

Hibernate: 
    select
        customer0_.id as id1_1_0_,
        customer0_.name as name2_1_0_ 
    from
        customer customer0_ 
    where
        customer0_.id=?
Hibernate: 
    select
        orders0_.cid as cid3_2_0_,
        orders0_.id as id1_2_0_,
        orders0_.id as id1_2_1_,
        orders0_.name as name2_2_1_,
        orders0_.cid as cid3_2_1_ 
    from
        orders orders0_ 
    where
        orders0_.cid=?
Hibernate: 
    delete 
    from
        orders 
    where
        id=?
Hibernate: 
    delete 
    from
        customer 
    where
        id=?