从带有联合部分2的多个表中选择计数相关的行

Question

在第2部分中，我有两个表

来自tb1的第一个表，来自tb2的第二个表。

+------+----------+------+
| kd   | skl      | name |
+------+----------+------+
| 001  | database | a    |
| 001  | web      | b    |
| 001  | web      | c    |
| 002  | app      | d    |
| 002  | web      | e    |
| 002  | app      | f    |
| 003  | json     | g    |
| 003  | -        | h    |
| 003  | -        | i    |
| 004  | ruby     | j    |
| 004  | database | k    |
| 004  | web      | l    |
| 005  | -        | m    |
| 005  | -        | n    |
| 005  | -        | o    |
+------+----------+------+

+------+----------+------+
| kd   | skl      | name |
+------+----------+------+
| 001  | database | p    |
| 001  | web      | q    |
| 001  | web      | r    |
| 002  | app      | s    |
| 002  | web      | t    |
| 002  | app      | u    |
| 003  | json     | v    |
| 003  | web      | w    |
| 003  | app      | x    |
| 004  | ruby     | y    |
| 004  | database | z    |
| 004  | web      | d    |
| 005  | -        | c    |
| 005  | web      | b    |
| 005  | app      | a    |
+------+----------+------+

看起来就像我从带有联合的多个表中选择计数相关行中的第一个问题一样

我在下面接受这个查询。

    select kd,skl,name,sum(row) brs from
(select a.kd,skl,name,count(a.kd) row
from tb1 a where skl in('web','app')
group by a.kd,skl,name
union all
select b.kd,skl,name,count(b.kd)
from tb2 b where skl in('web','app')
group by b.kd,skl,name)t
group by kd,name,skl;

选择查询给出结果

+------+-----+------+------+
| kd   | skl | name | brs  |
+------+-----+------+------+
| 001  | web | b    |    1 |
| 001  | web | c    |    1 |
| 001  | web | q    |    1 |
| 001  | web | r    |    1 |
| 002  | app | d    |    1 |
| 002  | web | e    |    1 |
| 002  | app | f    |    1 |
| 002  | app | s    |    1 |
| 002  | web | t    |    1 |
| 002  | app | u    |    1 |
| 003  | web | w    |    1 |
| 003  | app | x    |    1 |
| 004  | web | d    |    1 |
| 004  | web | l    |    1 |
| 005  | app | a    |    1 |
| 005  | web | b    |    1 |
+------+-----+------+------+

问题是brs列不计数取决于kd列数据。

我需要下面的结果

+------+-----+------+------+
| kd   | skl | name | brs  |
+------+-----+------+------+
| 001  | web | b    |    4 |
| 001  | web | c    |    4 |
| 001  | web | q    |    4 |
| 001  | web | r    |    4 |
| 002  | app | d    |    6 |
| 002  | web | e    |    6 |
| 002  | app | f    |    6 |
| 002  | app | s    |    6 |
| 002  | web | t    |    6 |
| 002  | app | u    |    6 |
| 003  | web | w    |    2 |
| 003  | app | x    |    2 |
| 004  | web | l    |    2 |
| 004  | web | d    |    2 |
| 005  | web | b    |    2 |
| 005  | app | a    |    2 |
+------+-----+------+------+

请提供选择查询示例或该主题的线索。

非常感谢您的所有建议。

Answer 1

这不是最美观的查询，但它应该可以工作：

SELECT t1.kd, t1.sk1, t1.name, t2.brs
FROM
(
    SELECT a.kd, skl, name
    FROM tb1 a WHERE skl IN ('web','app')
    UNION ALL
    SELECT b.kd, skl, name,
    FROM tb2 b WHERE skl IN ('web','app')
) t1
INNER JOIN
(
    SELECT r.kd, SUM(r.row) brs
    FROM
    (
        SELECT a.kd, COUNT(a.kd) row
        FROM tb1 a WHERE skl IN ('web','app')
        GROUP BY a.kd
        UNION ALL
        SELECT b.kd, COUNT(b.kd)
        FROM tb2 b WHERE skl IN ('web','app')
        GROUP BY b.kd
    ) r
    GROUP BY r.kd
) t2
ON t1.kd = t2.kd