hibernate关联查询方法
今天做了一个两张表的查询 要求查询文章分类及该分类文章的数目
hibernate刚学，我的解决方案如下，不知道大家有什么好的方法，请赐教

public List<Sort> getAllSort() {//String hql = "from Sort sort_ order by sort_.id desc";//return (List<Sort>)this.getHibernateTemplate().find(hql);Session session =  HibernateUtil.getSession();Transaction tx = session.beginTransaction();Connection conn = session.connection();PreparedStatement ps = null;ResultSet rs = null;try {ps = conn.prepareStatement("SELECT sort_.id_,sort_.name_,sort_.tag_,COUNT(sort_id) as counts FROM note_,sort_ WHERE sort_id=sort_.id_ GROUP BY sort_id ORDER BY counts desc");rs = ps.executeQuery();} catch (SQLException e) {e.printStackTrace();} tx.commit();List<Sort> list = new ArrayList<Sort>();Sort sort = null;try {while(rs.next()){sort = new Sort();sort.setId(rs.getString(1));sort.setName(rs.getString(2));sort.setTag(rs.getInt(3));sort.setCounts(rs.getInt(4));list.add(sort);SysLog.loger(rs.getString(1));SysLog.loger(rs.getString(2));SysLog.loger(rs.getInt(3));SysLog.loger(rs.getInt(4));}} catch (SQLException e) {e.printStackTrace();}return list;}

======================================================
table:note_ (表示文章列表)
+-------------+--------------+------+-----+---------+-------+
| Field | Type | Null | Key | Default | Extra |
+-------------+--------------+------+-----+---------+-------+
| id_ | varchar(255) | NO | PRI | | |
| title_ | varchar(50) | YES | | NULL | |
| content_ | text | YES | | NULL | |
| createTime_ | datetime | YES | | NULL | |
| visitCount_ | int(11) | YES | | NULL | |
| sort_id | varchar(255) | YES | MUL | NULL | |
+-------------+--------------+------+-----+---------+-------+

table:sort_(表示文章的类别)
+-------+--------------+------+-----+---------+-------+
| Field | Type | Null | Key | Default | Extra |
+-------+--------------+------+-----+---------+-------+
| id_ | varchar(255) | NO | PRI | | |
| name_ | varchar(50) | YES | | NULL | |
| tag_ | int(11) | YES | | NULL | |
+-------+--------------+------+-----+---------+-------+

Note.java
private String id;
private String title;
private Sort sort;
private String content;
private Date createTime;
private int visitCount;
...getter,setter

Sort.java
private String id;
private String name;
private Set Note;
private int tag;
private int counts;
...getter,setter

Note.hbm.xml
<id name="id" type="string" column="id_">
<generator type="string" column="title_" length="50" />
<property name="content" type="string" column="content_" length="1024" />
<property name="createTime" type="java.util.Date" column="createTime_" length="50" />
<property name="visitCount" type="int" column="visitCount_" length="50" />
<many-to-one name="sort" column="sort_id"/>

Sort.hbm.xml
<id name="id" type="string" column="id_">
<generator type="string" column="name_" length="50" />
<property name="tag" column="tag_" length="50" />
<set name="note" cascade="save-update">
<key column="sort_id"></key>
<one-to-many class="Note"/>
</set> 1 楼 zuzg 2009-08-06 这个做法和hibernate有关系吗？ 2 楼 xiaoban0514 2009-08-06 没有关系，，就是因为不会用oo的思想去查询所有旧出此下策了。。。