如何找到具有特定值的按列计数

Question

我有一张桌子attendance

+-----------------------------------------------+
+ p_id  | day_1 | day_2 | day_3 | day_4 | day_5 |
+-----------------------------------------------+
+  1    | P     |  P    | P     | A     | A     |
+  2    | P     |  A    | A     | P     | P     |
+  3    | A     |  A    | P     | P     | P     |
+-----------------------------------------------+

这里p_id = people.id外键

还有一个表people

+------------+
+ id   | gend|
+------------+
| 1    | M   |
| 2    | F   |
| 3    | M   |
+------------+

我想要得到这样的最终输出，其中包含缺席总人数及其各自的男性/女性人数：

+---------------------------------------+
+  day     | total_a | males  | females +
+---------------------------------------+
+  day_1   |  1      | 1      | 0       +
+  day_2   |  2      | 1      | 1       +
+  day_3   |  1      | 0      | 1       +

以此类推

我不知道从哪里开始这样的查询。任何帮助表示赞赏。

Answer 1

这听起来像您需要先取消attendance表的透视，然后join并重新聚合：

select a.day, sum(a.pa = 'P') as total,
       sum(a.pa = 'P' and p.gend = 'M') as num_males,
       sum(a.pa = 'P' and p.gend = 'F') as num_females
from ((select p_id, day_1 as pa, 'day_1' as day
       from attendance a
      ) union all
      (select p_id, day_2, 'day_2' as day
       from attendance a
      ) union all
      (select p_id, day_3, 'day_3' as day
       from attendance a
      ) union all
      (select p_id, day_4, 'day_4' as day
       from attendance a
      ) union all
      (select p_id, day_5, 'day_5' as day
       from attendance a
      )
     ) a join 
     people p
     on p.id = a.p_id
group by day
order by day;