MySQL如何先将子组加总，然后加总

Question

我在table_1和table_2中的数据。

TABLE_1

id  id_1 num   ids_2
1   3    33    666,777,888
2   3    333   6666,7777,8888
3   4    44    111,222,333
4   4    444   1111,2222,3333

TABLE_2

id_2   num
111    1
222    2
333    3
1111   1
2222   2
3333   3
666    6
777    7
888    8
6666   6
7777   7
8888   8

我只知道如何通过两个步骤做我想做的事情：

首先获得LEFT JOIN：

SELECT t1.id_1, sum(t2.num) 
FROM table_1 AS t1 
LEFT JOIN table_2 AS t2 
ON FIND_IN_SET(t2.id_2, t1.ids_2) 
GROUP BY t1.id_1;

id_1 sum(t2.num)
3    6+7+8+6+7+8
4    1+2+3+1+2+3

然后再次与table_1联接到sum（table_1.num）+ sum（table_2.num）：

id_1 sum(table_1.num)+sum(table_2.num)
3    6+7+8+6+7+8+33+333
4    1+2+3+1+2+3+44+444

我可以只在一个SQL中执行吗？

Answer 1

<强> Here is the SQLFIddel Demo

以下是您可以尝试的查询

SELECT A.id_1, sum(B.num)+sum(distinct A.num)
  FROM table_1 AS A 
  LEFT JOIN table_2 AS B
    on FIND_IN_SET(B.id_2, A.ids_2) 
 GROUP BY A.id_1;

Answer 2

从How to count items in comma separated list MySQL

获得灵感

SELECT t1.id_1,
  SUM(t1.num / (LENGTH(t1.ids_2) - LENGTH(REPLACE(t1.ids_2, ',', '')) + 1) + t2.num)
    AS total
  FROM table_1 AS t1
  LEFT JOIN table_2 AS t2
    ON FIND_IN_SET(t2.id_2, t1.ids_2) 
  GROUP BY t1.id_1;

如果没有上述分部，结果将是：

id_1 total
3    6+7+8 + 6+7+8 + 33+33+33 + 333+333+333
4    1+2+3 + 1+2+3 + 44+44+44 + 444+444+444

通过划分，结果将是我想要的结果：

id_1 total
3    6+7+8 + 6+7+8 + 33/3+33/3+33/3 + 333/3+333/3+333/3
4    1+2+3 + 1+2+3 + 44/3+44/3+44/3 + 444/3+444/3+444/3