[英]SQLAlchemy join through JSON sub field
I have an existing SQLAlchemy model that is supposed to have a relation to another model, however the connecting point there is sadly a (sub)field of a JSON PostgreSQL column.
因此,假设一个表“类别”是:
id(bigint) | name(string)
-------------------------
5 | Comics
...第二个表“英雄”是:
id(bigint) | name(string) | info(JSON)
-------------------------------------------------------------
2 | Tranquility | {"category_id":5, "genre_id": 17}
我将如何在 python SQLAlchemy&PostgreSQL9.5 中通过hero
的info["category_id"]
category
hero
上的类别以获得以下结果?
id(bigint) | name(string) | category_name | info(JSON)
-------------------------------------------------------------
2 | Tranquility | Comics | {"category_id":5, "genre_id": 17}
根据这个答案,我知道 PostgreSQL 本身有可能发生这样的事情
https://dba.stackexchange.com/a/83935 ( http://sqlfiddle.com/#!15/226c33/1 )
并且可能需要一些技巧来铸造或绕过默认方式
如json 字段的 sqlalchemy 过滤器中所述
注意:请注意,我不能更改 DB 结构,也不能更改 model 的现有部分,而且我也不是在寻找我已经知道的双重查询绕过。
您可以使用json_to_record
创建一个内存表,然后加入该表。
https://www.postgresql.org/docs/9.4/functions-json.html
另一个问题已经回答了这个问题,但使用json_to_recordset
作为数据是对象列表https://stackoverflow.com/a/31333794/3358570
这里是 SQL 您可以根据需要转换为 sqlalchemy,
SELECT hero.id as hero_id, hero.name as name, d.name as category_name, info
FROM hero,
json_to_record(hero.info) AS x(category_id int)
JOIN category d on d.id = category_id
这里 db fiddle 为你https://www.db-fiddle.com/f/7ops4KXVDF6KpY5JZau7vG/1
另一种更简单的方法是在 join 中访问 JSON 值并输入种姓category_id
SELECT hero.id AS hero_id,
hero.name AS hero_name,
category.name AS category_name,
category.id AS category_id,
hero.info AS hero_info
FROM hero
JOIN category ON category.id = CAST(hero.info ->> 'category_id' AS INTEGER)
等效的 qalchemy 会变成,
category_id = cast(Hero.info.op("->>")("category_id"), Integer)
query = (db.session.query(Hero.id, Hero.name, Category.name, Category.id, Hero.info)
.join(Category, Category.id == category_id))
data = query.all()
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.