Java 如何在HQL中编写联接查询？

我正在尝试运行一个HQL查询，它给我的错误是：

org.hibernate.QueryException: could not resolve property:
UserType of: EntityPack.UserTypeMenu 
[from EntityPack.UserTypeMenu as utm ,EntityPack.UserType as ut 
 where utm.UserType.userTypeId=ut.userTypeId and ut.userTypeDesc like ' %ad%' ]

这是我编写查询的函数：

    public ObservableList PostTableusertypemenu(String search, int q) {
            ObservableList data;
            data = FXCollections.observableArrayList();
            List ll=null;
                 ll = pop.UrviewTable("from UserTypeMenu as utm ,UserType as ut "+
                                      "where utm.UserType.userTypeId=ut.userTypeId"+
                                    " and ut.userTypeDesc like ' %"+ search +"%' ");
            Iterator ite = ll.iterator();
            while (ite.hasNext()) {
                UserTypeMenu obj = (UserTypeMenu) ite.next();
                data.add(obj);
            }  
            return data;
        }

我的UserType实体类：

    @Entity
    @Table(name = "user_type")
    public class UserType {
            @Id
        @GeneratedValue(strategy = GenerationType.AUTO)
        @Basic(optional = false)
        @Column(name = "User_Type_Id")
        private Integer userTypeId;
        @Basic(optional = false)
        @Column(name = "User_Type")
        private String userType;
        @Basic(optional = false)
        @Column(name = "User_Type_Desc")
        private String userTypeDesc;
        @Basic(optional = false)
        @Column(name = "Status")
        private String status;
          @Basic(optional = false)
        @Column(name = "Markers")
        private String markers;
    }

    @Entity
    @Table(name = "user_type_menu")
    @NamedQueries({
        @NamedQuery(name = "UserTypeMenu.findAll", query = "SELECT u FROM UserTypeMenu u")})
    public class UserTypeMenu implements Serializable {
        private static final long serialVersionUID = 1L;
        @Id
        @GeneratedValue(strategy = GenerationType.AUTO)
        @Basic(optional = false)
        @Column(name = "User_Type_Menu_Id")
        private Integer userTypeMenuId;
        @Basic(optional = false)
        @Column(name = "Status")
        private String status;
        @Basic(optional = false)
        @Column(name = "Markers")
        private String markers;
        @ManyToOne(optional = false)
        private UserType userType;
        @ManyToOne(optional = false)
        private UserMenuMaster userMenuMaster;
        @ManyToOne(optional = false)
        private UserMenuBar userMenuBar;
    }

UserTypeMenu实体类：

    @Entity
    @Table(name = "user_type")
    public class UserType {
            @Id
        @GeneratedValue(strategy = GenerationType.AUTO)
        @Basic(optional = false)
        @Column(name = "User_Type_Id")
        private Integer userTypeId;
        @Basic(optional = false)
        @Column(name = "User_Type")
        private String userType;
        @Basic(optional = false)
        @Column(name = "User_Type_Desc")
        private String userTypeDesc;
        @Basic(optional = false)
        @Column(name = "Status")
        private String status;
          @Basic(optional = false)
        @Column(name = "Markers")
        private String markers;
    }

    @Entity
    @Table(name = "user_type_menu")
    @NamedQueries({
        @NamedQuery(name = "UserTypeMenu.findAll", query = "SELECT u FROM UserTypeMenu u")})
    public class UserTypeMenu implements Serializable {
        private static final long serialVersionUID = 1L;
        @Id
        @GeneratedValue(strategy = GenerationType.AUTO)
        @Basic(optional = false)
        @Column(name = "User_Type_Menu_Id")
        private Integer userTypeMenuId;
        @Basic(optional = false)
        @Column(name = "Status")
        private String status;
        @Basic(optional = false)
        @Column(name = "Markers")
        private String markers;
        @ManyToOne(optional = false)
        private UserType userType;
        @ManyToOne(optional = false)
        private UserMenuMaster userMenuMaster;
        @ManyToOne(optional = false)
        private UserMenuBar userMenuBar;
    }

我想要的是根据Usertype描述从UsertypeMenu获取数据

请帮帮我。。

---

谢谢。：）

不像在SQL中那样在HQL中编写联接，而是使用

符号在对象图中导航。试试这个

"from UserTypeMenu as utm where utm.userType.userTypeDesc like ' %"+ search +"%' "

实际上，您可以直接在关联上使用联接。下面是相同的查询，但使用联接语法

"from UserTypeMenu as utm join utm.userType ut where ut.userTypeDesc like ' %"+ search +"%' "

---

这里使用连接的好处是，例如，如果关系不是强制性的，并且您不想因为使用

时隐式使用的

内部连接而丢失任何结果，您可以指定
左连接
，
您能给我推荐任何网站，帮助我更详细地了解hql吗当然，最好的地方是。当然，由于HQL是JPQL的扩展，您也可以检查。它是givimg me java.lang.ClassCastException:[Ljava.lang.Object；无法强制转换为EntityPack.UserTypeMenu此错误。在UserTypeMenu obj=（UserTypeMenu）ite.next（）；data.add（obj）；您必须发布
pop.UrviewTable（）的代码
“公共列表UrviewTable（字符串sQuery）{session=sf.sesFactory.openSession（）；session.beginTransaction（）；Query qee1=session.createQuery（sQuery）；List ll=qee1.List（）；session.getTransaction（）.commit（）；session.close（）；}”