Как ограничить количество строк до 3 и суммировать суммы за каждый день недели?

Question

Мне нужно объединить результат в 3 строки за апрель, май и июнь соответственно, в то время как я получаю сумму за каждый день недели (с понедельника по воскресенье) для этих 3 строк. Я пытался использовать разные для удаления избыточных названий месяцев но это не сработало.

select distinct t.calendar_month_name as "SALES_MONTH", 
case when t.day_number_in_week = 1 then sum(s.amount_sold) else 0 end as MONDAY,
case when t.day_number_in_week = 2 then sum(s.amount_sold) else 0 end as TUESDAY,
case when t.day_number_in_week = 3 then sum(s.amount_sold) else 0 end as WEDNESDAY,
case when t.day_number_in_week = 4 then sum(s.amount_sold) else 0 end as THURSDAY,
case when t.day_number_in_week = 5 then sum(s.amount_sold) else 0 end as FRIDAY,
case when t.day_number_in_week = 6 then sum(s.amount_sold) else 0 end as SATURDAY,
case when t.day_number_in_week = 7 then sum(s.amount_sold) else 0 end as SUNDAY
from products p
join sales s on p.prod_id = s.prod_id
join times t on t.time_id = s.TIME_ID
where
p.prod_id = 5  and 
t.calendar_year = 2000 and
t.calendar_quarter_number = 2
group by t.calendar_month_name, t.day_number_in_week
having sum(s.amount_sold) > 1
order by 1;

Answer 1

Вам не нужна таблица PROD здесь. В моей схеме SH у меня нет PROD_ID = 5, поэтому я использовал 13.

select * from (
  select t.calendar_month_name, t.day_number_in_week,
    sum(s.amount_sold) amount_sold
  from sales s
  join times t on t.time_id = s.TIME_ID
  where
  s.prod_id = 13  and 
  t.calendar_year = 2000 and
  t.calendar_quarter_number = 2
  group by t.calendar_month_name, t.day_number_in_week
  having sum(s.amount_sold) > 1
)
pivot(sum(amount_sold) for day_number_in_week in (
  1 as MONDAY,
  2 as TUESDAY,
  3 as WEDNESDAY,
  4 as THURSDAY,
  5 as FRIDAY,
  6 as SATURDAY,
  7 as SUNDAY
))
order by 1;

CALENDAR_     MONDAY    TUESDAY  WEDNESDAY   THURSDAY     FRIDAY   SATURDAY     SUNDAY
--------- ---------- ---------- ---------- ---------- ---------- ---------- ----------
April       20792,32   35724,12   43189,77   16688,08    11436,9              49960,37
June        27037,04   16809,54   11504,03              37709,26   23923,83   35908,33
May                   123982,43   18773,74               5283,56              21167,23

Если бы не было фильтра HAVING, вам бы не понадобилось GROUP BY во встроенном представлении, потому что неявной группировки предложения PIVOT будет достаточно:

select * from (
  select t.calendar_month_name, t.day_number_in_week,
    s.amount_sold
  from sales s
  join times t on t.time_id = s.TIME_ID
  where
  s.prod_id = 13  and 
  t.calendar_year = 2000 and
  t.calendar_quarter_number = 2
)
pivot(sum(amount_sold) for day_number_in_week in (
  1 as MONDAY,
  2 as TUESDAY,
  3 as WEDNESDAY,
  4 as THURSDAY,
  5 as FRIDAY,
  6 as SATURDAY,
  7 as SUNDAY
))
order by 1;

С уважением, Рагу Эштон