我有一个用户和一个消息表。用户到消息是一对多关系,消息到用户是多对一关系。我已将多对一之一标记为获取连接。当我“获取”单个消息时,Hibernate 运行联接查询,但是当我获取所有消息时,Hibernate 运行选择查询而不是联接。可能是什么原因?详情如下:
它们之间的关系:
用户
<set name="messagesForFromUserUid" lazy="true" table="message" inverse="true" cascade="save-update">
<key>
<column name="from_user_uid" not-null="true" />
</key>
<one-to-many class="repository.Message" />
</set>
<set name="messagesForToUserUid" lazy="true" table="message" fetch="select">
<key>
<column name="to_user_uid" not-null="true" />
</key>
<one-to-many class="repository.Message" />
</set>
留言
<many-to-one name="userByFromUserUid" class="repository.User" fetch="join" lazy="false">
<column name="from_user_uid" not-null="true" />
</many-to-one>
<many-to-one name="userByToUserUid" class="repository.User" fetch="select" lazy="proxy">
<column name="to_user_uid" not-null="true" />
</many-to-one>
当我获取单个消息对象时,Hibernate 按预期运行一个连接查询:
Message m = (Message) s.get(Message.class, 2);
Hibernate:
select
message0_.message_uid as message1_1_1_,
message0_.from_user_uid as from2_1_1_,
message0_.to_user_uid as to3_1_1_,
message0_.message_text as message4_1_1_,
message0_.created_dt as created5_1_1_,
user1_.user_uid as user1_0_0_,
user1_.user_name as user2_0_0_,
user1_.user_password as user3_0_0_,
user1_.email as email0_0_,
user1_.first_name as first5_0_0_,
user1_.last_name as last6_0_0_,
user1_.created_dt as created7_0_0_
from
hello.message message0_
inner join
hello.user user1_
on message0_.from_user_uid=user1_.user_uid
where
message0_.message_uid=?
但是当我一次性获取所有消息时,Hibernate 会运行选择查询:
List<Message> l = s.createQuery("from Message").list();
Hibernate:
select
message0_.message_uid as message1_1_,
message0_.from_user_uid as from2_1_,
message0_.to_user_uid as to3_1_,
message0_.message_text as message4_1_,
message0_.created_dt as created5_1_
from
hello.message message0_
Hibernate:
select
user0_.user_uid as user1_0_0_,
user0_.user_name as user2_0_0_,
user0_.user_password as user3_0_0_,
user0_.email as email0_0_,
user0_.first_name as first5_0_0_,
user0_.last_name as last6_0_0_,
user0_.created_dt as created7_0_0_
from
hello.user user0_
where
user0_.user_uid=?
Hibernate:
select
user0_.user_uid as user1_0_0_,
user0_.user_name as user2_0_0_,
user0_.user_password as user3_0_0_,
user0_.email as email0_0_,
user0_.first_name as first5_0_0_,
user0_.last_name as last6_0_0_,
user0_.created_dt as created7_0_0_
from
hello.user user0_
where
user0_.user_uid=?
最佳答案
看起来 Hibernate 并不总是使用映射中定义的获取策略来进行 HQL 或 Criteria 查询。它们通常用于获取/加载。在这里找到引用:https://forum.hibernate.org/viewtopic.php?f=1&t=957561
关于java - Hibernate 'join' 获取奇怪的行为,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/9695597/