Hibernate06_配置文件详解

比眉伴天荒 2023-03-06 05:05 20阅读 0赞

**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
            (?, ?)

![在这里插入图片描述][20200805114401887.png]  
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
            (?)

![在这里插入图片描述][20200805114804606.png]

--------------------

##### 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=?

![在这里插入图片描述][20200805115656396.png]

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：

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=?

![在这里插入图片描述][20200805153441427.png]  
![在这里插入图片描述][20200805153501584.png]  
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

--------------------

##### **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

测试：

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

结果：

[20200805114401887.png]: /images/20230208/ceda4e950294448fb689bf8ab7828900.png
[20200805114804606.png]: /images/20230208/546894f8bb1d4aa4b255fe2cfb884042.png
[20200805115656396.png]: /images/20230208/7133c0e8374947d7a6bc4c20e07700b6.png
[20200805153441427.png]: /images/20230208/4cd3fdb037ba454d85bc0067b8da0e32.png
[20200805153501584.png]: /images/20230208/c0225a06448044ca876b4c21f8b1eab1.png