Question

<hibernate-mapping package="entity.associative">
  <class name="UserGroup" table="user_group">
    <meta attribute="class-description">
    This class contains user group information. 
    </meta>
    <composite-id>
      <key-property name="userID" column="u_id"/>
      <key-property name="groupID" column="g_id"/>
    </composite-id>
    <property name="hours" type="double">
      <column name="hours"/>
    </property>
  </class>
</hibernate-mapping>

<hibernate-mapping package="entity.user">
  <class name="User" table="user">
    <meta attribute="class-description">
        This class contains user information. 
    </meta>
    <id name="userID" type="string">
        <column name="u_id" length="50"/>
    </id>

    <property name="fName" type="string">
        <column name="f_name" length="100"/>
    </property>
.
.
.
</hibernate-mapping>

我收到了错误

org.hibernate.hql.internal.ast.QuerySyntaxException: 
Path expected for join! 
[FROM entity.user.User u INNER JOIN UserGroup ug 
    where u.userID = ug.userID and ug.groupID = 'EORG']

最有可能的原因是它们位于不同的包中 - 一个位于entity.user，另一个位于entity.associative。有没有办法改变我的查询？

我目前的查询是：

tx = session.beginTransaction();
String queryStr = "FROM User u INNER JOIN UserGroup ug where u.userID = ug.userID and ug.groupID = 'EORG'";
Query query = session.createQuery(queryStr);
userList = (ArrayList<User>) query.list();

Answer 1

在您的查询中，您必须使用以下显式包：

tx = session.beginTransaction();
String queryStr = "FROM entity.user.User u, 
    entity.associative.UserGroup ug 
    where u.userID = ug.userID and ug.groupID = 'EORG'";
Query query = session.createQuery(queryStr);
userList = (ArrayList) query.list();

通常为了防止这种行为，我写这样的查询：

String myQuery = " FROM " + User.class.getName() + " u, " + 
    UserGroup.class.getName() + " ug 
    where u.userID = ug.userID and ug.groupID = 'EORG'";

使用getName（）可以获得类的完整路径

Answer 2

要加入User和UserGroup，您需要在User至UserGroup

中建立关联

class User {

  @OneToOne
  private UserGroup userGroup;

}

FROM User u INNER JOIN u.userGroup ug where ug.groupID = 'EORG'

此查询只是一个示例。它可以用更简单的形式重写

FROM User u where u.userGroup.groupID = 'EORG'

Hibernate的最新版本能够使用没有这种关联的连接，但需要使用on子句。

HQL内部加入不同的包

2 个答案: