2

嗨，我有 3 个表的数据库员工：

mysql> describe person;
+------------+--------------+------+-----+---------+----------------+
| Field      | Type         | Null | Key | Default | Extra          |
+------------+--------------+------+-----+---------+----------------+
| person_id  | bigint(20)   | NO   | PRI | NULL    | auto_increment |
| date       | datetime     | YES  |     | NULL    |                |
| first_name | varchar(255) | YES  |     | NULL    |                |
| last_name  | varchar(255) | YES  |     | NULL    |                |
| position   | varchar(255) | YES  |     | NULL    |                |
| salary     | double       | YES  |     | NULL    |                |
+------------+--------------+------+-----+---------+----------------+

mysql> describe department;
+---------------+--------------+------+-----+---------+----------------+
| Field         | Type         | Null | Key | Default | Extra          |
+---------------+--------------+------+-----+---------+----------------+
| department_id | bigint(20)   | NO   | PRI | NULL    | auto_increment |
| dept_name     | varchar(255) | YES  |     | NULL    |                |
+---------------+--------------+------+-----+---------+----------------+

mysql> describe department_person;
+---------------+------------+------+-----+---------+-------+
| Field         | Type       | Null | Key | Default | Extra |
+---------------+------------+------+-----+---------+-------+
| department_id | bigint(20) | NO   | PRI | NULL    |       |
| person_id     | bigint(20) | NO   | PRI | NULL    |       |
+---------------+------------+------+-----+---------+-------+

因此，我想编写 HQL 语句以使用 person.lastName 从 person.position 获取数据，并且我使用以下代码来提取这些数据，但没有运气：

public List<Person> findPosition(){
        Session session =
                HiberUtil.getSessionFactory().getCurrentSession();
        session.beginTransaction();
        List<Person> result = session.createQuery("select position from Person p where p.lastName= 'Anderson'").list();
        for(Person a : result) {
            Hibernate.initialize(a.getDepartmentList());
        }
        session.getTransaction().commit();
        return result;

    }

在这就是它向我展示了这个例外：

Hibernate: select person0_.position as col_0_0_ from person person0_ where person0_.last_name='Anderson'
Exception in thread "main" java.lang.ClassCastException: java.lang.String cannot be cast to main.Person
    at main.StaffDAO.findPosition(StaffDAO.java:100)
    at main.Main.main(Main.java:37)

您能帮我编写正确的 HQL 语句吗？

4

3 回答 3

1 
List<Person> result = session.createQuery("select position from Person p where p.lastName= 'Anderson'").list();

您正在选择一个包含 VARCHAR(255) 但希望它以List<Person>. 那根本不计算。

你可以做

List<Person> result = session.createQuery("select p from Person p where p.lastName= 'Anderson'").list();
于 2013-04-15T20:04:01.070 回答
1

好吧，您正在position查询中选择 a，并尝试将其放入 Person 列表中。相反，您应该select p from Person p where p.lastName= 'Anderson'.

于 2013-04-15T20:05:46.277 回答
0

从人 p 中选择人，其中 p.lastName='Anderson'

于 2013-04-15T20:10:30.333 回答