Du kan prøve dette :
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
ved at bruge connect by dual
klausul.
P.S. Ignorerede tilfældet for helligdage, som adskiller sig fra den ene kultur til den anden, afhængigt af, at spørgsmålet kun er relateret til weekender.
Rediger: Antag, at du har en national helligdag på "2018-09-25" og "2018-09-26" (i dette sæt af dage), så overvej følgende :
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
som gentages den næste dag, som i dette tilfælde, medmindre denne ferie falder sammen med weekenden.