Question

我遇到问题hql
我有3张桌子 1.用户（user_id = pk，用户名）
2. groups（group_id = pk，group_name）
3. usergroups（user_id = fk，group_id = fk）

<hibernate-mapping>
<class name="com.model.authRole.UserGroups" table="user_groups" catalog="db_realm" optimistic-lock="version">
    <composite-id name="id" class="com.model.authRole.UserGroupsId">
        <key-property name="userId" type="int">
            <column name="user_id" />
        </key-property>
        <key-property name="groupId" type="int">
            <column name="group_id" />
        </key-property>
    </composite-id>
    <many-to-one name="groups" class="com.model.authRole.Groups" update="false" insert="false" fetch="select">
        <column name="group_id" not-null="true" />
    </many-to-one>
    <many-to-one name="users" class="com.model.authRole.Users" update="false" insert="false" fetch="select">
        <column name="user_id" not-null="true" />
    </many-to-one>
    <property name="aktiv" type="boolean">
        <column name="aktiv" not-null="true" />
    </property>
</class>

我想显示用户名和group_name，但是hql

的错误

from UserGroups  ug 
inner  join  Users  u on u.user = ug.user_id
inner  join  Groups g on g.group = ug.group_id

并记录错误

org.hibernate.hql.internal.ast.QuerySyntaxException：预期的路径加入！ [来自com.model.authRole.UserGroups ug内部联接用户u on u.user = ug.user_id inner join g g.group = ug.group_id]

如何解决？

Answer 1

我假设类UserGroups包含来自User的对象和来自Group类的对象

session.createQuery("from UserGroups ug where ug.User = :user and ug.Group =:group").setParameter("user", (Users)session.get(Users.class,user_id)).setParameter("group", (Groups)session.get(Groups.class,group_id));

org.hibernate.hql.internal.ast.QuerySyntaxException：连接所需的路径

1 个答案: