深入理解Hibernate中Iterate方法

本是古典何须时尚 2023-10-17 11:31 14阅读 0赞

**深入理解Hibernate中Iterate方法**

### **一、数据准备（参考前面一篇博文）** ###

[数据准备参考博文（这是链接）][Link 1]  
**在实际需求中有时需要对一些业务数据进行频繁的查询处理，可能会这样处理：**  
**二、list查询剖析**

public void testQuerySingleTable_HQL_List() {
            Session session = HbnUtils.getSession();
            try {
                session.beginTransaction();
                String hql = "from Student where age > 17";
                // 执行第一次查操作
                List<Student> list = session.createQuery(hql).list();
                for (Student stu : list) {
                System.out.println(stu);
                }
                System.out.println("\t-------第一次查询与第二次查询的分割线--------");
                // 执行第二次查操作
                List<Student> list_1 = session.createQuery(hql).list();
                for (Student stu : list) {
                    System.out.println(stu);
                }
                session.getTransaction().commit();
            } catch (Exception e) {
                e.printStackTrace();
                session.getTransaction().rollback();
            }
        }

**下面是控制台打印的SQL语句和提示语句**

Hibernate: 
        select
            student0_.t_id as t_id1_0_,
            student0_.t_name as t_name2_0_,
            student0_.t_age as t_age3_0_,
            student0_.t_score as t_score4_0_ 
        from
            t_student student0_ 
        where
            student0_.t_age>17
    Student [id=8, name=张8三, age=18, score=88.0]
    Student [id=9, name=张9三, age=19, score=89.0]
    Student [id=10, name=张10三, age=20, score=90.0]
        -------第一次查询与第二次查询的分割线--------
    Hibernate: 
        select
            student0_.t_id as t_id1_0_,
            student0_.t_name as t_name2_0_,
            student0_.t_age as t_age3_0_,
            student0_.t_score as t_score4_0_ 
        from
            t_student student0_ 
        where
            student0_.t_age>17
    Student [id=8, name=张8三, age=18, score=88.0]
    Student [id=9, name=张9三, age=19, score=89.0]
    Student [id=10, name=张10三, age=20, score=90.0]

**从上面的输出信息可以看出，用List方法Hibernate查询了几次就访问了几次数据库，为了提高效率，本应该在第二次查询的时候可以从缓存中获取，因为第一次的结果已经保存在Session缓存中了！**

### **三、iterate查询剖析** ###

**代码**

/**
    * iterate
     *  */
        @Test
        public void testQuerySingleTable_HQL_Iterate() {
            Session session = HbnUtils.getSession();
            try {
                session.beginTransaction();
                String hql = "from Student where age>17";
                // 执行(第一次查询)
                Iterator<Student> iterator = session.createQuery(hql).iterate();
                while(iterator.hasNext()){
                    System.out.println(iterator.next());
                }
                System.out.println("\t-------第一次查询与第二次查询的分割线--------");
                //执行第二次查询操作
                List<Student> list_1 = (List<Student>) session.createQuery(hql).iterate();
                while(iterator.hasNext()){
                    System.out.println(iterator.next());
                }
                session.getTransaction().commit();
            } catch (Exception e) {
                e.printStackTrace();
                session.getTransaction().rollback();
            }
        }

**执行后控制台输出信息**

Hibernate: 
        select
            student0_.t_id as col_0_0_ 
        from
            t_student student0_ 
        where
            student0_.t_age>17
    Hibernate: 
        select
            student0_.t_id as t_id1_0_0_,
            student0_.t_name as t_name2_0_0_,
            student0_.t_age as t_age3_0_0_,
            student0_.t_score as t_score4_0_0_ 
        from
            t_student student0_ 
        where
            student0_.t_id=?
    Student [id=8, name=张8三, age=18, score=88.0]
    Hibernate: 
        select
            student0_.t_id as t_id1_0_0_,
            student0_.t_name as t_name2_0_0_,
            student0_.t_age as t_age3_0_0_,
            student0_.t_score as t_score4_0_0_ 
        from
            t_student student0_ 
        where
            student0_.t_id=?
    Student [id=9, name=张9三, age=19, score=89.0]
    Hibernate: 
        select
            student0_.t_id as t_id1_0_0_,
            student0_.t_name as t_name2_0_0_,
            student0_.t_age as t_age3_0_0_,
            student0_.t_score as t_score4_0_0_ 
        from
            t_student student0_ 
        where
            student0_.t_id=?
    Student [id=10, name=张10三, age=20, score=90.0]
        -------第一次查询与第二次查询的分割线--------
    Hibernate: 
        select
            student0_.t_id as col_0_0_ 
        from
            t_student student0_ 
        where
            student0_.t_age>17

**分析输出信息可以看出：Hibernate的iterate执行过程**  
1）：  
`Hibernate: select student0_.t_id as col_0_0_ from t_student student0_ where student0_.t_age>17 Student [id=8, name=张8三, age=18, score=88.0] Student [id=9, name=张9三, age=19, score=89.0] Student [id=10, name=张10三, age=20, score=90.0]`  
1）：先把所有满足条件的记录的id从数据库查询出来  
2）：然后根据查询出来的id逐个去查询每一条记录的详细信息  
3）：先根据id在Session缓存中查找  
4）：如果没有，就需要去数据库查找，有几个满足条件的id，就需要访问几次数据库  
3）：当进行第二次查询时，先到数据库进行了一次主键查询，查查满足条件的id，然后直接根据id从Session缓存中读出满足条件的记录的详细信息

### **四、Query接口的List方法与Iterate方法比较** ###

List方法不会从Session缓存中读取数据

Iterate方法会从Session缓存中读取数据

List方法第一次查询效率高

Iterate方法第一次查询效率很低（2n+1次）

所以为了避免n+1问题，我们取长补短，第一次查询用List,第二次查询用Iterate

### **五、Query接口的List方法与Iterate方法结合** ###

public void testQuerySingleTable_HQL_IterateAndList() {
            Session session = HbnUtils.getSession();
            try {
                session.beginTransaction();
                String hql = "from Student where age>17";
                // 执行(第一次查询)
                List<Student> list = session.createQuery(hql).list();
                for (Student stu : list) {
                System.out.println(stu);
                }
                System.out.println("\t-------第一次查询与第二次查询的分割线--------");
                //执行第二次查询操作
                Iterator<Student> iterator_1 = session.createQuery(hql).iterate();
                while(iterator_1.hasNext()){
                    System.out.println(iterator_1.next());
                }
                session.getTransaction().commit();
            } catch (Exception e) {
                e.printStackTrace();
                session.getTransaction().rollback();
            }
        }

**看结果：———**

Hibernate: 
        select
            student0_.t_id as t_id1_0_,
            student0_.t_name as t_name2_0_,
            student0_.t_age as t_age3_0_,
            student0_.t_score as t_score4_0_ 
        from
            t_student student0_ 
        where
            student0_.t_age>17
    Student [id=8, name=张8三, age=18, score=88.0]
    Student [id=9, name=张9三, age=19, score=89.0]
    Student [id=10, name=张10三, age=20, score=90.0]
        -------第一次查询与第二次查询的分割线--------
    Hibernate: 
        select
            student0_.t_id as col_0_0_ 
        from
            t_student student0_ 
        where
            student0_.t_age>17
    Student [id=8, name=张8三, age=18, score=88.0]
    Student [id=9, name=张9三, age=19, score=89.0]
    Student [id=10, name=张10三, age=20, score=90.0]

**从结果分析效率确实提高了不少！！！**

[Link 1]: http://blog.csdn.net/leisure_life/article/details/61192386