It is easy enough to extract a week value from a date in hiveql:
select date_format('2020-10-18','w');
Is there a function for reversing this process: extracting the end date when provided a year and week number?
To get accurate date you need to provide also week day along with year and week number in a year.
select date_format('2020-10-18','w'), from_unixtime(unix_timestamp('2020, 43, 7', 'yyyy, w, u'), 'yyyy-MM-dd');
Returns:
43 2020-10-18
It looks like week number in a year counted from Sundays and day number in a week is counted from Mondays because Monday is 19th:
select date_format('2020-10-18','w u'), from_unixtime(unix_timestamp('2020, 43, 1', 'yyyy, w, u'), 'yyyy-MM-dd');
returns
43 2020-10-19
If that is true, you can fix it by subtracting 60*60*24 from unix_timestamp:
select date_format('2020-10-18','w'), from_unixtime(unix_timestamp('2020, 43, 1', 'yyyy, w, u')-60*60*24, 'yyyy-MM-dd');
Returns:
43 2020-10-18
UPDATE: Surprisingly, if not providing day in a week, only year and week number, it works also counting Sunday as a week day by default but it will be not correct for other dates for example 2020-01-20, it will return the same Sunday 2020-01-18, check it yourself:
select date_format('2020-10-18','w'), from_unixtime(unix_timestamp('2020, 43', 'yyyy, w'), 'yyyy-MM-dd');
returns:
43 2020-10-18
So, if you do not have day in a week and do not need absolutely accurate date, then use
from_unixtime(unix_timestamp('2020, 43', 'yyyy, w'), 'yyyy-MM-dd');
Or like this (year and week number are selected from the table):
select from_unixtime(unix_timestamp(concat(col_year, ', ', col_week), 'yyyy, w'), 'yyyy-MM-dd') from your_table;