Mysql 如何在组中使用sum()?

Mysql 如何在组中使用sum()?,mysql,sum,group-concat,Mysql,Sum,Group Concat,修订问题 我真的想要一组和。。。 表:商店 +---------+--------+--------+ | shop_id | name | state | +---------+--------+--------+ | 0 | shop 0 | 5 | | 1 | shop 1 | 5 | | 2 | shop 2 | 5 | | 3 | shop 3 | 2 | +---------+-------

修订问题

我真的想要一组和。。。 表:商店

+---------+--------+--------+
| shop_id | name   | state  |
+---------+--------+--------+
|    0    | shop 0 |    5   |
|    1    | shop 1 |    5   |
|    2    | shop 2 |    5   |
|    3    | shop 3 |    2   |
+---------+--------+--------+
表:项目

+------------+--------------+
|   shop  | item | quantity | 
+------------+--------------+
|    0    |  0   |    1     |
|    0    |  1   |    2     |
|    0    |  2   |    3     |
|    1    |  0   |    1     |
|    1    |  1   |    2     |
|    1    |  2   |    3     |
|    2    |  0   |    1     |
|    2    |  1   |    2     |
|    2    |  2   |    3     |
|    3    |  0   |    1     |
|    3    |  1   |    2     |
|    3    |  2   |    3     |
+------------+--------------+

    SELECT state,SUM(i.quantity) total
    FROM shops s2
    LEFT JOIN items i ON i.shop=s2.shopid
    WHERE state=5
    GROUP by item

result #1:

+--------+---------+
| state  |  total  |
+--------+---------+
|    5   |    3    |
+--------+---------+
|    5   |    6    |
+--------+---------+
|    5   |    9    |
+--------+---------+

But I would like the totals, like this:
result #2:
+--------+---------+---------+----------+
| state  | total 0 | total 1 |  total 2 |
+--------+---------+---------+----------+
|    5   |    3    |     6   |    9     |
+--------+---------+---------+----------+

or using group_concat()
result #3

+--------+---------+
| state  | totals  |
+--------+---------+
|    5   |  3,6,9  |
+--------+---------+
我似乎无法让group_concat获取结果1中的总计列

提前感谢更改:

group_concat(CAST(quantity AS CHAR))

--


据我所知,你不能在MySQL中做到这一点。仅在组_contcat()的范围内支持动态列,该组仍然将多个结果行聚合到单个列中


只有当您有固定/有限数量的
Total X
-s时,才可以在查询中明确声明它们本身。

找到了一种方法:

SELECT state,GROUP_CONCAT(cast(total as char))
FROM
(
    SELECT state,SUM(i.quantity) total
    FROM shops s
    LEFT JOIN items i ON i.shop=s.shopid
    WHERE state=5
    GROUP by item
) s

谢谢你的快速回答,我最初问的是正确的,但我问的不是我想要的…@mahks-我想你不可能只用一个问题就得到你想要的
SELECT s.`state`, i.`item`, SUM(i.`quantity`) AS quantities
FROM `shops` AS s
    LEFT JOIN `items` AS i ON i.`shop` = s.`shopid`
WHERE s.`state` = 5
GROUP BY i.`item`
SELECT state,GROUP_CONCAT(cast(total as char))
FROM
(
    SELECT state,SUM(i.quantity) total
    FROM shops s
    LEFT JOIN items i ON i.shop=s.shopid
    WHERE state=5
    GROUP by item
) s