Incremental business day column that resets each month - sql

Question

I need to create a table that contains records with 1) all 365 days of the year and 2) a counter representing which business day of the month the day is. Non-business days should be represented with a 0. For example:
Date | Business Day
2019-10-01 1
2019-10-02 2
2019-10-03 3
2019-10-04 4
2019-10-05 0 // Saturday
2019-10-06 0 // Sunday
2019-10-07 5
....
2019-11-01 1
2019-11-02 0 // Saturday
2019-11-03 0 // Sunday
2019-11-04 2
So far, I've been able to create a table that contains all dates of the year.
CREATE TABLE ${TMPID}_days_of_the_year
(
`theDate` STRING
);
INSERT OVERWRITE TABLE ${TMPID}_days_of_the_year
select
dt_set.theDate
from
(
-- last 0~99 months
select date_sub('2019-12-31', a.s + 10*b.s + 100*c.s) as theDate
from
(
select 0 as s union all select 1 union all select 2 union all select 3 union all select 4 union all select 5 union all select 6 union all select 7 union all select 8 union all select 9
) a
cross join
(
select 0 as s union all select 1 union all select 2 union all select 3 union all select 4 union all select 5 union all select 6 union all select 7 union all select 8 union all select 9
) b
cross join
(
select 0 as s union all select 1 union all select 2 union all select 3
) c
) dt_set
where dt_set.theDate between '2019-01-01' and '2019-12-31'
order by dt_set.theDate DESC;
And I also have a table that contains all of the weekend days and holidays (this data is loaded from a file, and the date format is YYYY-MM-DD)
CREATE TABLE ${TMPID}_company_holiday
(
`holidayDate` STRING
)
;
LOAD DATA LOCAL INPATH '${FILE}' INTO TABLE ${TMPID}_company_holiday;
My question is.... how do I join these tables together while creating the business day counter column shown as in the sample data above?

Answer 1

You can use row_number() for the enumeration. This is a little tricky, because it needs to be conditional, but the information you need is provided by a left join:
select dy.*,
(case when ch.holiday_date is null
then row_number() over (partition by trunc(dy.date, 'MONTH'), ch.holiday_date
order by dy.date
)
else 0
end) as business_day
from days_of_the_year dy left join
company_holiday ch
on dy.date = ch.holiday_date;