续集和嵌套结果？

Sequelize允许数据关系，但我似乎无法看到是否有办法让它返回由关系组成的查询结果？ 我可以做这两个查询，但我很好奇Sequelize是否提供这个？ 例如播放列表和播放列表条目：

Playlist: sequelize.define('playlist', { name: Sequelize.STRING, description: Sequelize.STRING }), PlaylistEntry: sequelize.define('playlist_entry', { playlist: Sequelize.INTEGER //track: Sequelize.INTEGER }) PlaylistEntry.belongsTo( Playlist, { as: 'Playlist', foreignKey: { name: 'fk_playlist' }}); 

我会希望的（伪代码）：

查询：

 Playlist.find(where: {id: playlistId}, nestedResult: true); 

结果：

 [{ id: 123, name: 'abc' playlistEntries: [{ id: 321, track: 431 }] }] 

Answer 1

 Playlist.find({ where: {}, include: [ {model: PlaylistEntry} //this should be PlaylistEntry model ] }) 

http://docs.sequelizejs.com/en/latest/docs/models-usage/#eager-loading