declare @T table (Address_col varchar(20))
insert into @T values
('Nevada,USA'),
('Tokyo,Japan'),
('Hanoi,Vietnam')
select left(Address_col, charindex(',', Address_col)-1) as Address_col,
stuff(Address_col, 1, charindex(',', Address_col), '') as Country
from @T
Aktualizace:
Rozdělení řetězce na tři části může vypadat takto:deklarace @T tabulka (Address_col varchar(20))
insert into @T values
('Nevada,USA,World'),
('Tokyo,Japan,World'),
('Hanoi,Vietnam,World')
select parsename(C, 3),
parsename(C, 2),
parsename(C, 1)
from @T
cross apply (select replace(Address_col, ',', '.')) as T(C)