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Question

我收到以下异常：

HTTP Status 500 - Request processing failed;
nested exception is org.springframework.orm.hibernate3.HibernateQueryException: expecting OPEN, found '.' near line 1, column 35 [from in.bets.crm.model.User u1, in.bets.crm.model.User u2 where u1.manager is not null and u2.manager.id = u1.id];
nested exception is org.hibernate.hql.ast.QuerySyntaxException: expecting OPEN, found '.' near line 1, column 35 [from in.bets.crm.model.User u1, in.bets.crm.model.User u2 where u1.manager is not null and u2.manager.id = u1.id]

我编写的查询如下：

users=hibernateTemplate.find("from User u1, User u2 where u1.manager is not null and u2.manager.id = u1.id");

请帮助解决此错误。 谢谢

Answer 1

试试这个:(只需添加选择和加入关键字！）

users = hibernateTemplate.find（“select u from User u inner join User m，其中u.manager不为null，u.manager.id = m.id”）;