Using TIME_DIFF with multiple conditions in Google BigQuery - sql

Question

I am trying to calculate the worked hours for specific days in Google BigQuery (SQL).
The pay wage is $10 when you work on a day time but $15 when you work on a night time.
Day time is defined as 6am to 10pm whereas night time is defined as 10pm to 6am.
Employees can work flexibly as they are limousine drivers.
The following is an example of my table:
id
start_at
end_at
date
abc123
04:00:00
07:00:00
2020-01-05
abc123
09:00:00
15:32:00
2020-01-05
abc123
23:00:00
23:35:00
2020-01-05
abc123
23:40:00
23:59:00
2020-01-05
abc123
23:59:00
01:35:00
2020-01-05
abc123
02:02:00
04:35:00
2020-01-06
abc123
05:40:00
06:59:00
2020-01-06
So the actual work hours is calculated by taking the difference between start_at and end_at but the day time and night time conditions are becoming a hassle in my query..
*the date column is based on start_at. Even when you start at 11:59pm and end at the next day 12:05am, the date follows the date of the start_at instead of end_at.
Any ideas? Thanks in advance!

Answer 1

Consider below solution
create temp function night_day_split(start_at time, end_at time, date date) as (array(
select as struct
extract(date from time_point) day,
if(extract(hour from time_point) between 6 and 22, 'day', 'night') day_night,
count(1) minutes
from unnest(generate_timestamp_array(
timestamp(datetime(date, start_at)),
timestamp(datetime(if(start_at < end_at, date, date + 1), end_at)),
interval 1 minute
)) time_point
group by 1, 2
));
select id, day,
sum(if(day_night = 'day', minutes, null)) day_minutes,
sum(if(day_night = 'night', minutes, null)) night_minutes
from yourtable,
unnest(night_day_split(start_at, end_at, date)) v
group by id, day
if applied to sample data in your question - output is

Answer 2

You can try following code :-
with mytable as (
select 'abc123' id, cast( '04:00:00' as time) start_dt, cast( '07:00:00' as time) end_dt, date('2020-01-05' ) date union all
select 'abc123', cast( '09:00:00' as time), cast( '15:32:00' as time), date('2020-01-05') union all
select 'abc123', cast( '23:00:00' as time), cast( '23:35:00' as time), date('2020-01-05' ) union all
select 'abc123', cast('23:40:00' as time), cast( '23:59:00' as time), date('2020-01-05') union all
select 'abc123', cast ('23:59:00' as time), cast( '01:35:00' as time), date('2020-01-05') union all
select 'abc123', cast('02:02:00' as time), cast( '04:35:00' as time), date('2020-01-06') union all
select 'abc123', cast('05:40:00' as time), cast( '06:59:00' as time), date('2020-01-06')
)
select id, date, sum (value) as sal from(
select id, date,
case when start_dt > cast( '06:00:00' as time) and end_dt < cast( '22:00:00' as time) and start_dt < end_dt then (time_diff(end_dt, start_dt, Minute)/60) * 10
when start_dt < cast( '06:00:00' as time) and end_dt < cast( '06:00:00' as time) then (time_diff(end_dt, start_dt, Minute)/60) * 15
when start_dt < cast( '06:00:00' as time) and end_dt < cast( '22:00:00' as time) then (time_diff(cast( '06:00:00' as time), start_dt, Minute)/60) * 15 + (time_diff( end_dt,cast( '06:00:00' as time), Minute)/60) * 10
when start_dt > cast( '22:00:00' as time) and end_dt < cast( '06:00:00' as time) then (time_diff(cast( '23:59:00' as time), start_dt, Minute)/60) * 15 + (time_diff( end_dt,cast( '00:00:00' as time), Minute)/60) * 15
when start_dt > cast( '22:00:00' as time) and end_dt > cast( '22:00:00' as time) then (time_diff(end_dt, start_dt, Minute)/60) * 15
else 0
end as value
from mytable) group by id, date
Output :-
You can further group by on month for monthly salary.