我有以下查询来获取一部名为“I.D”的电影中的所有学分 -
SELECT title, platform, platform_id, type, credit, role
FROM main_iteminstance i inner join main_credit c on c.item_instance_id=i.id inner join main_creditperson p on c.person_id=p.id
WHERE i.id = 196365940 order by role_type_id, p.name
这给了我以下数据:
+---------------------- main table --------------------+ +--------- joined table --------+
name platform platform_id type credit role
I.D. iTunes 1168241742 Movie Geetanjali Actor
I.D. iTunes 1168241742 Movie Murari Kumar Actor
I.D. iTunes 1168241742 Movie Kamal K.M. Director
I.D. iTunes 1168241742 Movie Geetanjali Producer
我想将其展平为具有最少重复数据的对象,以将其放入json对象中。例如,下面的对象如下所示:
{
"platform": "iTunes",
"platform_id": "1168241742",
"type": "Movie",
"name": "I.D.",
"credits": {
"actors": ["Geetanjali", "Murari Kumar"],
"directors": ["Kamal K.M."],
"producers": ["Geetanjali"]
},
}
我们可以看到,除了信用属性之外,一切都是“平坦的”。有没有办法进行查询,我们基本上'json-ify'是一个连接表的数据?建议的方法是什么?基本上,有没有办法从sql到mongodb / json而不使用中间的代码?