sql: подсчитать, сколько ум_но по разным разделам и накопить за час

Question

как посчитать, сколько ум_но по разным разделам и кумулятивно по часам без конкретной функции

оригинал:

+-------+------+------+-----+
| um_no | nums | hour | day |
+-------+------+------+-----+
| mary  |    3 |    8 |   1 |
| john  |    6 |    8 |   1 |
| peter |    2 |    8 |   1 |
| jason |    1 |    8 |   1 |
| mary  |    5 |    9 |   1 |
| john  |    6 |    9 |   1 |
| peter |    6 |    9 |   1 |
| jason |    1 |    9 |   1 |
| mary  |    5 |   10 |   1 |
| john  |    4 |   10 |   1 |
| peter |    2 |   10 |   1 |
| jason |    4 |   10 |   1 |
+-------+------+------+-----+

хочу:

+-----+------+---------+----------+-----------+----------+
| day | hour | nums0_5 | nums5_10 | nums10_15 | nums15up |
+-----+------+---------+----------+-----------+----------+
|   1 |    8 |       3 |        1 |         0 |        0 |
|   1 |    9 |       1 |        2 |         1 |        0 |
|   1 |   10 |       0 |        2 |         1 |        1 |
+-----+------+---------+----------+-----------+----------+

Answer 1

Вы можете попробовать .. Используйте с предложением Гордона, что действительно здорово ..

select
  day,
  hour,
  count(*) filter (where sum_nums >= 0 and sum_nums <= 5) as nums0_5,
  count(*) filter (where sum_nums >= 5 and sum_nums <= 10) as nums5_10,
  count(*) filter (where sum_nums >= 10 and sum_nums <= 15) as nums10_15,
  count(*) filter (where sum_nums >= 15) as nums15up
from
  (select 
      *,
      sum(nums) over (partition by um_no, day order by hour) as sum_nums
   from 
      tbl) t
group by
  day,
  hour

Вы можете увидеть здесь DEMO

Answer 2

select tt.day, tt.hour,
       sum(tt.nums_0_5) as nums_0_5 ,
       sum(tt.nums_5_10) as nums_5_10 ,
       sum(tt.nums_10_15) as nums_10_15, 
       sum(tt.nums15up) as nums15up
from
(select sum(case when nums >= 0 and nums < 5 then 1 else 0 end ) as nums_0_5,   
        sum( case when nums >= 5 and nums < 10 then 1 else 0 end ) as nums_5_10,
        sum( case when nums >= 10 and nums < 15 then 1 else 0 end ) as nums_10_15,
        sum( case when nums >= 15 then 1 else 0 end ) as nums15up,
        t.day,
        t.hour, 
        t.um_no
        from (select tbl.day, 
                     tbl.hour,
                     tbl.um_no,
                     tbl.nums,
                     Row_Number() over(partition by hour order by hour asc) as row_nums
              from tblTest tbl group by hour, um_no, nums, day) t
              group by t.day, t.hour, t.um_no
) tt group by  tt.hour, tt.day

Answer 3

Использовать условное агрегирование:

select day, hour,
       sum( case when nums >= 0 and nums < 5 then 1 else 0 end ) as nums_0_5,
       sum( case when nums >= 5 and nums < 10 then 1 else 0 end ) as nums_5_10,
       . . .
from t
group by day, hour;

В Postgres суммы могут быть упрощены до:

       count(*) filter (where nums >= 0 and nums < 5) as nums_0_5,

В MySQL:

       sum( nums >= 0 and nums < 5 ) as nums_0_5,

РЕДАКТИРОВАТЬ:

Вы можете получить кумулятивные числа с помощью оконных функций и затем агрегировать:

select day, hour,
       sum( case when cume_nums >= 0 and cume_nums < 5 then 1 else 0 end ) as nums_0_5,
       sum( case when cume_nums >= 5 and cume_nums < 10 then 1 else 0 end ) as nums_5_10,
       . . .
from (select t.*,
             sum(nums) over (partition by um_no, date order by hour) as cume_nums
      from t
     ) t
group by day, hour;