更新：这似乎只在使用PostgreSQL时发生。我在这里提交了一个项目：https://github.com/jbwiv/testfetch有两个分支。 ＆＃34;主＆＃34;使用h2和＆＃34; postgres＆＃34;使用PostgreSQL。
尝试每个分支，然后转到localhost：8080 / testfetch / console。
在该控制台中，键入：
import testfetch.Employee
import testfetch.User
Employee.list()
在H2，我得到了我的期望：
hibernate.SQL select this_.id as id1_1_1_, this_.version as version2_1_1_, user2_.id as id1_0_0_, user2_.version as version2_0_0_, user2_.employee_id as employee3_0_0_ from employee this_ left outer join app_user user2_ on this_.id=user2_.employee_id
然而，在postgres分支中，我得到：
hibernate.SQL /* criteria query */ select this_.id as id1_1_0_, this_.version as version2_1_0_ from employee this_
hibernate.SQL /* load testfetch.User */ select user0_.id as id1_0_0_, user0_.version as version2_0_0_, user0_.employee_id as employee3_0_0_ from app_user user0_ where user0_.employee_id=?
hibernate.SQL /* load testfetch.User */ select user0_.id as id1_0_0_, user0_.version as version2_0_0_, user0_.employee_id as employee3_0_0_ from app_user user0_ where user0_.employee_id=?
hibernate.SQL /* load testfetch.User */ select user0_.id as id1_0_0_, user0_.version as version2_0_0_, user0_.employee_id as employee3_0_0_ from app_user user0_ where user0_.employee_id=?
hibernate.SQL /* load testfetch.User */ select user0_.id as id1_0_0_, user0_.version as version2_0_0_, user0_.employee_id as employee3_0_0_ from app_user user0_ where user0_.employee_id=?
hibernate.SQL /* load testfetch.User */ select user0_.id as id1_0_0_, user0_.version as version2_0_0_, user0_.employee_id as employee3_0_0_ from app_user user0_ where user0_.employee_id=?
hibernate.SQL /* load testfetch.User */ select user0_.id as id1_0_0_, user0_.version as version2_0_0_, user0_.employee_id as employee3_0_0_ from app_user user0_ where user0_.employee_id=?
hibernate.SQL /* load testfetch.User */ select user0_.id as id1_0_0_, user0_.version as version2_0_0_, user0_.employee_id as employee3_0_0_ from app_user user0_ where user0_.employee_id=?
hibernate.SQL /* load testfetch.User */ select user0_.id as id1_0_0_, user0_.version as version2_0_0_, user0_.employee_id as employee3_0_0_ from app_user user0_ where user0_.employee_id=?
hibernate.SQL /* load testfetch.User */ select user0_.id as id1_0_0_, user0_.version as version2_0_0_, user0_.employee_id as employee3_0_0_ from app_user user0_ where user0_.employee_id=?
hibernate.SQL /* load testfetch.User */ select user0_.id as id1_0_0_, user0_.version as version2_0_0_, user0_.employee_id as employee3_0_0_ from app_user user0_ where user0_.employee_id=?
hibernate.SQL /* load testfetch.User */ select user0_.id as id1_0_0_, user0_.version as version2_0_0_, user0_.employee_id as employee3_0_0_ from app_user user0_ where user0_.employee_id=?
hibernate.SQL /* load testfetch.User */ select user0_.id as id1_0_0_, user0_.version as version2_0_0_, user0_.employee_id as employee3_0_0_ from app_user user0_ where user0_.employee_id=?
hibernate.SQL /* load testfetch.User */ select user0_.id as id1_0_0_, user0_.version as version2_0_0_, user0_.employee_id as employee3_0_0_ from app_user user0_ where user0_.employee_id=?
hibernate.SQL /* load testfetch.User */ select user0_.id as id1_0_0_, user0_.version as version2_0_0_, user0_.employee_id as employee3_0_0_ from app_user user0_ where user0_.employee_id=?
hibernate.SQL /* load testfetch.User */ select user0_.id as id1_0_0_, user0_.version as version2_0_0_, user0_.employee_id as employee3_0_0_ from app_user user0_ where user0_.employee_id=?
hibernate.SQL /* load testfetch.User */ select user0_.id as id1_0_0_, user0_.version as version2_0_0_, user0_.employee_id as employee3_0_0_ from app_user user0_ where user0_.employee_id=?
hibernate.SQL /* load testfetch.User */ select user0_.id as id1_0_0_, user0_.version as version2_0_0_, user0_.employee_id as employee3_0_0_ from app_user user0_ where user0_.employee_id=?
hibernate.SQL /* load testfetch.User */ select user0_.id as id1_0_0_, user0_.version as version2_0_0_, user0_.employee_id as employee3_0_0_ from app_user user0_ where user0_.employee_id=?
hibernate.SQL /* load testfetch.User */ select user0_.id as id1_0_0_, user0_.version as version2_0_0_, user0_.employee_id as employee3_0_0_ from app_user user0_ where user0_.employee_id=?
hibernate.SQL /* load testfetch.User */ select user0_.id as id1_0_0_, user0_.version as version2_0_0_, user0_.employee_id as employee3_0_0_ from app_user user0_ where user0_.employee_id=?
这对我们来说是一个巨大的阻碍。我非常感谢您提供的任何反馈。
我正在实时使用Hibernate4在Grails 2.4.4中正常工作。
例如，我有两个类，Employee和User。
class Employee {
User user
static mapping = {
user fetch:'join'
}
static belongsTo = User
}
class User {
Employee employee
static mapping = {
employee fetch:'join'
table 'app_user'
}
}
但是，当我尝试使用此设置查询这些对象时，我仍然最终出现N + 1问题。例如
Employee.list()
Employee.executeQuery("select e from Employee e")
所有这些导致对Employee表的一次查询，该表返回21名员工，然后对app_user表（用户映射到的）进行21次查询。
然而，
Employee.withCriteria {fetchMode 'user', FetchMode.JOIN)
的工作原理。在这种情况下，我只执行了一个查询。
我做错了什么？
蒂亚。
答案 0 :(得分：0)
zyro在JIRA问题中正确地指出我们的PostgreSQL配置中有“max_fetch_depth = 0”。
引用他：
postgres分支有dataSource.hibernate.max_fetch_depth = 0，而hibernate docs说A 0禁用默认的外连接提取。这可能是原因吗？
确实，在删除它之后，它以预期的方式执行。