Question

我只需要在hibernate（将歌曲与播放列表相关联）的多个表/对象上添加一个名为SongsPlaylist的条目。

表的结构如此定义：Songs---->SongPlaylist<----Playlists

映射文件（省略不是文件的重要部分）：

<hibernate-mapping>
    <class name="model.pojo.Songs" table="songs" catalog="dbname" lazy="false">
        <id name="id" type="java.lang.Integer">
            <column name="id" />
            <generator class="identity" />
        </id>
        <set name="songsPlaylists" table="songs_playlist" inverse="true" lazy="true" fetch="select">
            <key>
                <column name="song" not-null="true" unique="true" />
            </key>
            <one-to-many class="model.pojo.SongsPlaylist" />
        </set>
    </class>
</hibernate-mapping>
<!-- Generated 24-set-2014 8.40.14 by Hibernate Tools 3.6.0 -->
<hibernate-mapping>
    <class name="model.pojo.SongsPlaylist" table="songs_playlist" catalog="dbname" lazy="false">
        <id name="id" type="java.lang.Integer">
            <column name="id" />
            <generator class="identity" />
        </id>
        <many-to-one name="songs" class="model.pojo.Songs" fetch="select">
            <column name="song" not-null="true" unique="true" />
        </many-to-one>
        <many-to-one name="playlists" class="model.pojo.Playlists" fetch="select">
            <column name="playlist" not-null="true" unique="true" />
        </many-to-one>
    </class>
</hibernate-mapping>

<!-- Generated 24-set-2014 8.40.14 by Hibernate Tools 3.6.0 -->
<hibernate-mapping>
    <class name="model.pojo.Playlists" table="playlists" catalog="dbname" lazy="false">
        <id name="id" type="java.lang.Integer">
            <column name="id" />
            <generator class="identity" />
        </id>
        <set name="songsPlaylists" table="songs_playlist" inverse="true" lazy="true" fetch="select">
            <key>
                <column name="playlist" not-null="true" unique="true" />
            </key>
            <one-to-many class="model.pojo.SongsPlaylist" />
        </set>
    </class>
</hibernate-mapping>

pojo课程：

public class SongsPlaylist  implements java.io.Serializable {
     private Integer id;
     private Songs songs;
     private Playlists playlists;
     (...)
}

public class Songs  implements java.io.Serializable {
     private Integer id;
     private Set songsPlaylists = new HashSet(0);
     (...)
}

public class Playlists  implements java.io.Serializable {
     private Integer id;
     private Set songsPlaylists = new HashSet(0);
     (...)
}

这是我的功能

            s = new HibernateUtil().getSessionFactory().openSession();
            transaction = s.beginTransaction();
            (...)
                q = s.createQuery("select sp from SongsPlaylist sp where sp.playlists = :p");
                q.setParameter("p", p);
                sp = q.list();
                //already exists a song binded to that playlist
            if(sp.size()>0){
                Set<Playlists> sP = sp.get(0).getPlaylists().getSongsPlaylists();
                sP.add(p);
                song.setSongsPlaylists(sP);
                s.save(s);
            }else{
                //or it is in another playlist...
                q = s.createQuery("select sp from SongsPlaylist sp where sp.songs = :s");
                q.setParameter("s", song);
                sp = q.list();
                if(sp.size()>0){
                    Set<Songs> sP = sp.get(0).getSongs().getSongsPlaylists();
                    sP.add(song);
                    p.setSongsPlaylists(sP);
                    s.save(p);
                }else{
                    s.save(new SongsPlaylist(song, p));
                }
            }
            }
            transaction.commit();
            s.close();

并且它与

一起停留在transaction.commit();（最后一行）

org.hibernate.HibernateException: Found shared references to a collection: 
model.pojo.Songs.songsPlaylists

在持久对象上持久化Set<Playlists（当我尝试用Set<Songs>执行此操作时相同）时，我似乎做错了。为了插入“SongPlaylist”并避免提到的hibernate异常，这个案例的最佳做法是什么？

Answer 1

找到了解决方案......

在else块（而不是我写的）中，现在是：

Set<Playlists> sP = song.getPlaylistses(); // terrific netbeans naming when automatic pojo mapping :-)
sP.add(p);
song.setPlaylistses(sP);
s.save(song);

只是一个与这个没有关系的漏洞，但是

SongsPlaylist不应该像hbm文件一样。 hibernate映射应该是

<hibernate-mapping>
    <class name="model.pojo.Songs" table="songs" catalog="dbname" lazy="false">
        <id name="id" type="java.lang.Integer">
            <column name="id" />
            <generator class="identity" />
        </id>
        (...)
        <set name="playlistses" table="songs_playlist" inverse="false" lazy="false" fetch="select">
            <key>
                <column name="song" not-null="true" />
            </key>
            <many-to-many entity-name="model.pojo.Playlists">
                <column name="playlist" not-null="true" />
            </many-to-many>
        </set>
    </class>
</hibernate-mapping>


<hibernate-mapping>
    <class name="model.pojo.Playlists" table="playlists" catalog="dbname" lazy="false">
        <id name="id" type="java.lang.Integer">
            <column name="id" />
            <generator class="identity" />
        </id>
        <set name="songses" table="songs_playlist" inverse="false" lazy="false" fetch="select">
            <key>
                <column name="playlist" not-null="true" />
            </key>
            <many-to-many entity-name="model.pojo.Songs">
                <column name="song" not-null="true" />
            </many-to-many>
        </set>
    </class>
</hibernate-mapping>

多对多设置 - 添加条目 - org.hibernate.HibernateException：找到对集合的共享引用：

1 个答案: