如何将其他表中的选择分组为sql中的一列？

here is two tables:

这是两个表：

a:

A：

+-----+------------------------+
| id  | conten                 | 
+-----+------------------------+
| 1   | q.                     |
| 2   | q.                     |
| 3   | s.                     |
| 4   | g                      |
| 1   | a                      |
| 2   | a                      |
+-----+------------------------+

b:

b：

+-----+------------------------+
| id  | type                   | 
+-----+------------------------+
| 1   | I.                     |
| 2   | II.                    |
| 3   | III.                   |
| 4   | IV                     |
| 5   | V                      |
| 6   | VI                     |
+-----+------------------------+

Is there a way to select from a and b so that for one id 2, there will be one additional field that groups all content from that id? the select result should be something like this:

有没有办法从a和b中进行选择，以便对于一个id 2，会有一个附加字段对该ID中的所有内容进行分组？选择结果应该是这样的：

+-----+------------------------+-----------+
| id  | type                   | contents  |
+-----+------------------------+-----------+
| 2   |I.                      | q,a       |
+-----+------------------------+-----------+

Edited btw, if there is a way to do it by sqlahcmey, that would be sweet.

编辑顺便说一句，如果有一种方法可以通过sqlahcmey来做，那将是甜蜜的。

1 个解决方案

#1

SELECT b.id, b.type, IFNULL(GROUP_CONCAT(a.conten), '') AS contents
FROM b
LEFT JOIN a ON a.id = b.id
GROUP BY b.id

See How do I write a group_concat function in sqlalchemy? for how to translate GROUP_CONCAT to SQLAlchemy.

请参阅如何在sqlalchemy中编写group_concat函数？如何将GROUP_CONCAT转换为SQLAlchemy。