Question

您好我使用spring mvc 3和hibernate 4开发了一个Web应用程序。我正在为所有配置使用注释。

当我尝试选择实体时，我收到错误：

DaoException: unexpected token: MEMBER near line 1, column 37 [from clanwar.model.ClanMember where MEMBER = :member]
at clanwar.dao.impl.ClanMemberDao.findByPlayerId(ClanMemberDao.java:71)

如果我使用.createSQLQuery() hibernate执行查询，请返回Object[]实体ClanMember实例。

DAO中的方法：

    @Repository
    @Transactional

    ...

    @Override
    public ClanMember findByPlayerId(int id) throws DaoException {

        ClanMember foundMember;

        try {
            foundMember = (ClanMember) getSession()
                .createQuery("from ClanMember where MEMBER = :member")
                .setParameter("member", id)
                .uniqueResult();
        } catch (Exception e) {
            throw new DaoException(e.getMessage());
        }

        return foundMember;
    }

    ...

实体：

@Entity
@Table(name = "CLAN_MEMBERS")
public class ClanMember implements Serializable {

    private static final long serialVersionUID = 1L;

    @Id
    @ManyToOne
    @JoinColumn(name = "CLAN")
    private Clan clan;

    @Id
    @ManyToOne
    @JoinColumn(name = "MEMBER")
    private Player member;

    @ManyToOne
    @JoinColumn(name = "ROLE")
    private ClanRole role;

    public ClanMember() {}

    // Setters and getters

}

表CLAN_MEMBERS：

+--------+---------+------+-----+---------+-------+
| Field  | Type    | Null | Key | Default | Extra |
+--------+---------+------+-----+---------+-------+
| CLAN   | int(11) | NO   | PRI | NULL    |       |
| MEMBER | int(11) | NO   | PRI | NULL    |       |
| ROLE   | int(11) | NO   |     | 1       |       |
+--------+---------+------+-----+---------+-------+

我做错了什么？提前谢谢。

Answer 1

您的查询应如下所示：

from ClanMember where member = :member

因此，获取成员的代码应如下所示：

foundMember = (ClanMember) getSession()
                .createQuery("from ClanMember where member = :member")
                .setParameter("member", id)
                .uniqueResult();

另请注意，您传递的id应该是Player类的实例。

Answer 2

member是HQL中的保留字。通过添加表别名来逃避它。

此外，由于member的类型为Player，因此当您的where clause包含member时，您需要提供Player类型的参数。或者，如果您想提供您正在执行的ID，则需要明确搜索id：

所以：

 foundMember = (ClanMember) getSession()
            .createQuery("from ClanMember c where c.member = :member")
            .setParameter("member", someInstanceOfPlayer);
            .uniqueResult();

或

 foundMember = (ClanMember) getSession()
            .createQuery("from ClanMember c where c.member = :member")
            .setParameter("member", idOfAPlayer);
            .uniqueResult();

在Hibernate 4中获取ERROR意外令牌

2 个答案: