あなたはこれを試すことができます:
select max(date_field_two) as date_field_two
from
(
select date'2018-08-30'+
cast(case when to_char(date'2018-08-30'+level,'D','NLS_DATE_LANGUAGE=ENGLISH')
in ('6','7') then
0
else
level
end as int) as date_field_two,
sum(cast(case when to_char(date'2018-08-30'+level,'D','NLS_DATE_LANGUAGE=ENGLISH')
in ('6','7') then
0
else
1
end as int)) over (order by level) as next_day
from dual
connect by level <= 20*1.5
-- 20 is the day to be added, every time 5(#of business days)*1.5 > 7(#of week days)
-- 7=5+2<5+(5/2)=5*(1+1/2)=5*1.5 [where 1.5 is just a coefficient might be replaced a greater one like 2]
-- so 4*5*1.5=20*1.5 > 4*7
)
where next_day = 20;
DATE_FIELD_TWO
-----------------
27.09.2018
connect by dual
を使用する 条項。
追伸週末のみに関連する質問に応じて、文化ごとに異なる祝日の場合は無視しました。
編集: '2018-09-25'と'2018-09-26'(この一連の日)に国民の祝日があると仮定して、次のことを考慮してください:
select max(date_field_two) as date_field_two
from
(
select date'2018-08-30'+
(case when to_char(date'2018-08-30'+level,'D','NLS_DATE_LANGUAGE=ENGLISH')
in ('6','7') then
0
when date'2018-08-30'+level in (date'2018-09-25',date'2018-09-26') then
0
else
level
end) as date_field_two,
sum(cast(case when to_char(date'2018-08-30'+level,'D','NLS_DATE_LANGUAGE=ENGLISH')
in ('6','7') then
0
when date'2018-08-30'+level in (date'2018-09-25',date'2018-09-26') then
0
else
1
end as int)) over (order by level) as next_day
from dual
connect by level <= 20*2
)
where next_day = 20;
DATE_FIELD_TWO
-----------------
01.10.2018
この休日が週末と一致しない限り、この場合のように、次の日に繰り返されます。