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我有两个对象

Person(long id, String name, PersonInfo info) 

PersonInfo(long id, String email, String lastname, in age)

现在我尝试创建一个javax.persistence.Query

我尝试创造select to person where personinfo.email="somemail"

我试试

String queryString="select Person from Person p where p.PersonInfo._email='somemail'";
Query query=emf.createQuery(queryString);
List resultList = query.getResultList();

但我得到了这个例外:

java.lang.IllegalArgumentException: org.hibernate.QueryException: could not resolve property: PersonInfo of: Entities.Person [select Person from Entities.Person p where p.PersonInfo._email='somemail'] [Ljava.lang.StackTraceElement;@748e6591

我错了什么?

4

1 回答 1

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下面的查询字符串呢:

String queryString="select p from Person p where p.personInfo.email='somemail'";
于 2012-05-07T19:40:22.787 回答