Question

我有以下实体：

1.User entity。

 @Entity
    @Table(name = "USER")
    public class User {

        @Id
        @GeneratedValue
        private int userId;

        @Column
        private String username;

        @Column
        private String password;

        @ManyToMany(fetch=FetchType.EAGER)
        @JoinTable(name = "USER_ROLE", joinColumns={@JoinColumn(name="userId")}, inverseJoinColumns={@JoinColumn(name="roleId")})
        private List<Role> roles;

//+setters and getters}

角色实体。

@Entity
@Table(name="ROLE")
public class Role {

    @Id
    @GeneratedValue
    private long roleId;

    @Column
    private String name;

    @ManyToMany(mappedBy="roles", fetch=FetchType.EAGER)
    private List<User> users;
//+setters and getters
}

现在，我正在尝试在存储库中提取具有其角色（role_name）的用户（用户名）。我希望为Spring Security Auth做这件事。这是我的HQL代码：

@Query("select u.username, r.name from User u"+
            " join User_Role ur on u.userId = ur.userId"+
            " join Role r on ur.roleId = r.roleId"+
            " where u.username =:username")

......我收到了这个错误：

Caused by: org.hibernate.hql.internal.ast.QuerySyntaxException: Path expected for join! [select u.username, r.name from ro.app.mmapp.persistence.entity.User u join User_Role ur on u.userId = ur.userId join Role r on ur.roleId = r.roleId where u.username =:username]

我尝试了多种类型的连接，但得到了相同的结果。谁可以帮我提一些建议？

谢谢！

Answer 1

任何基本的JPQL教程都会告诉您加入一个关系。

select u.username, r.name from User u JOIN u.roles r WHERE u.username =:username

ManyToMany HQL Join - 期望加入的路径

1 个答案: