我如何按 book_id 和 order_id 对金额组求和

Question

我有一张桌子：

book_id  order_id  amount
-------- -------   -------
   a      d1         10
   a      d1         10
   a      d1         10
   a      d2         20
   a      d2         20
   a      d3         30

我试试这个

SELECT book_id,sum(amount) as sum FROM table WHERE book_id= 'a' GROUP BY book_id;

结果

book_id      sum
--------   -------
   a         100

但我想像这样按 book_id 和总金额对它们进行分组

book_id      sum
--------   -------
   a         60     ==> d1+d2+d3

如何将 mySQL 写入 SUM 行，例如 d1+d2+d3？

Answer 1

给定当前数据，您将基于 DISTINCT 行 SUM：

Working Test Case - Updated

所有的结果：

book_id SUM(amount)
-------------------
a       60
-------------------

WITH cte AS (
       SELECT DISTINCT book_id,  order_id,  amount FROM tbl
     )
SELECT book_id
     , SUM(amount)
  FROM cte
 GROUP BY book_id
;

根据需要为cte 术语或最终查询表达式添加过滤。

WITH cte AS (
       SELECT DISTINCT book_id,  order_id,  amount
         FROM tbl
        WHERE book_id = 'a'
     )
SELECT book_id
     , SUM(amount)
  FROM cte
 GROUP BY book_id
;

对于旧版本的 MySQL，您可以使用 derived table，而不是 CTE 术语：

SELECT book_id
     , SUM(amount)
  FROM (
         SELECT DISTINCT book_id,  order_id,  amount
           FROM tbl
          WHERE book_id = 'a'
       ) AS cte
 GROUP BY book_id
;

尝试SUM(DISTINCT amount) 是不正确的，因为一些不同的订单可能具有相同的amount，并且它们会被压缩为一个值。

Answer 2

有两种方法。第一个是Distinct：

Select t.book_id, sum(t.amount)
from
(
   SELECT DISTINCT book_id,  order_id,  amount 
   FROM book
) t
group by t.book_id

第二个是With:

WITH w AS (
       SELECT DISTINCT book_id,  order_id,  amount FROM book
          )
SELECT book_id, SUM(amount)
  FROM w
 GROUP BY book_id;

Answer 3

试试这个嵌套的SELECT。

SELECT book_id, SUM(amount) AS `sum` FROM (SELECT DISTINCT book_id, order_id, amount FROM book GROUP BY book_id, order_id) AS a;

结果

book_id  sum     
-------  --------
a        60      

Answer 4

   SELECT DISTINCT(book_id), SUM(amount) FROM table GROUP BY book_id WHERE book_id ='a';