Hibernate加入

Question

如何在Hibernate查询或标准中编写这个简单的工作选择

SELECT * FROM `call` AS c JOIN `event_customer` AS e ON c.customer_id = e.customer_id WHERE e.dgdelete = FALSE AND c.eventruletype_id = 1 AND e.event_id = 74 AND c.callactivity_id = 3 AND e.calltype = 2

我发现了很多建议，但没有任何帮助。

它可能是hibernate映射的东西吗？

谢谢。

PS： 这不起作用

query = session.createQuery("select c from " + Call.class.getSimpleName()+ " as c " + EventCustomer.class.getSimpleName()+ " as e where e.customer_id = c.customer_id and e.event_id = 74"); 

这也不起作用

 query = session.createQuery("select c from " + Call.class.getSimpleName()+ " as c join " + EventCustomer.class.getSimpleName()+ " as e where e.customer_id = c.customer_id and e.event_id = 74");

还有这个

 query = session.createQuery("select c from " + Call.class.getSimpleName()+ " as c join " + EventCustomer.class.getSimpleName()+ " as e on e.customer_id = c.customer_id  where  e.event_id = 74"); 

还有这个

query = session.createQuery("select c from " + Call.class.getSimpleName()+ " as c left join " + EventCustomer.class.getSimpleName()+ " as e on e.customer_id = c.customer_id  where  e.event_id = 74");

并且在某些情况下我收到此错误

Answer 1

不要连接，而是使用标准：

Criteria crit = session.createCriteria(Call.class, "c")
    .createCriteria("eventCustomerDto","e")
    .add(Restrictions.eq("event_id",74));
List<Call> list=crit.list();

代码可能有拼写错误，但这就是想法。另外，对于我的例子，我认为Call和EventCustomer之间的Call hibernate映射存在一对多的关系。

Answer 2

这应该有效。 （用逗号替换JOIN）

 Query query = session.createQuery("select c from " + Call.class.getSimpleName() + " as c , " + EventCustomer.class.getSimpleName() + " as e where e.customer.id = c.customer.id and c.event.id = 74");