默认情况下lazy=true,fetch=select
Session session = HibernateUtil.getSessionFactory().getCurrentSession();
session.beginTransaction();
Person p = (Person)session.get(Person.class, 1);
session.getTransaction().commit();
结果：
Hibernate: select person0_.id as id0_0_, person0_.name as name0_0_, person0_.age as age0_0_ from Person person0_ where person0_.id=?
如果改为：
虽然lazy=true，但是因为fetch是join，运行上面的代码，sql有变化，直接一个sql 通过 left join加载出所有phone
Hibernate: select person0_.id as id0_1_, person0_.name as name0_1_, person0_.age as age0_1_, phones1_.person_id as person3_3_, phones1_.id as id3_, phones1_.id as id1_0_, phones1_.num as num1_0_ from Person person0_ left outer join Phone phones1_ on person0_.id=phones1_.person_id where person0_.id=?
以上只是针对根据ID查询时候的情况，当用list()方法查询多个记录的时候，虽然配置了join，但是lazy依然有效
并没有通过join的方式查询phone
Hibernate: select person0_.id as id0_, person0_.name as name0_, person0_.age as age0_ from Person person0_
再来，lazy=false,fetch="join"，但是这次我们修改查询语句
Session session = HibernateUtil.getSessionFactory().getCurrentSession();
session.beginTransaction();
Person p = (Person)session.createQuery("from Person").list().get(0);
session.getTransaction().commit();
出现了著名的N+1问题，先查出所有person，再挨个person查phone
Hibernate: select person0_.id as id0_, person0_.name as name0_, person0_.age as age0_ from Person person0_ Hibernate: select phones0_.person_id as person3_1_, phones0_.id as id1_, phones0_.id as id1_0_, phones0_.num as num1_0_ from Phone phones0_ where phones0_.person_id=? Hibernate: select phones0_.person_id as person3_1_, phones0_.id as id1_, phones0_.id as id1_0_, phones0_.num as num1_0_ from Phone phones0_ where phones0_.person_id=?
总结：
1、在get和load查询的时候，不管lazy是什么值，只要fetch=join，就会用left join的方式查询子表