Puoi provare questo:
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
utilizzando connect by dual clausola.
PS Ignorato il caso dei giorni festivi, che differiscono da una cultura all'altra, a seconda che la domanda sia correlata solo ai fine settimana.
Modifica : Supponi di avere festività nazionali il "25-09-2018" e il "26-09-2018" (in questo insieme di giorni), quindi considera quanto segue:
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
che itera un giorno successivo, come in questo caso, a meno che questa festività non coincida con il fine settimana.
