awk or sed to print text between the first occurance of a pattern and the last occurance of a pattern

Question

i have seen a couple of solutions to print between patterns but am unable to put things together to cater to my problem. I have a text file that contains a view definition as shown below and need to extract the definition between the first and last braces. Please note: there are open and closed braces in between which should be printed. any solution in awk or sed?

create view view_name 
as(select column1 as someDATE,
    column2,
    column3,
    substring(convert(char(19),(DATEadd(hh,8,column4)),121),12) as someTIME,
    from table_name
    where NAME in('test')
    and column5='something')

Output reqired:

select column1 as someDATE,
column2,
column3,
substring(convert(char(19),(DATEadd(hh,8,column4)),121),12) as someTIME,
from table_name
where NAME in('test')
and column5='something'

Answer 1

Crude but effective:

sed -n -e '/(/,/)/p' filename | sed '1s/[^(]*(//;$s/)[^)]*$//'

EDIT:

(Corrected first solution.)

More sophisticated, but pulls the whole file into the hold space at once:

sed -n 'H;${x;s/^[^(]*(//;s/)[^)]*$//;p;}' filename