Redshift SQL: Get date difference based on start and end dates - sql

Question

My table has start_date and end_date from which I need to find the hour difference. The issue is that both of these date times are not on the same day.
user start_date end_date difference
Alex 7/25/2016 16:00 7/26/2016 0:30 8.5
Alex 7/24/2016 16:00 7/25/2016 0:30 8.5
Alex 7/21/2016 16:00 7/22/2016 0:30 8.5
Alex 7/20/2016 16:00 7/21/2016 0:30 8.5
Alex 7/19/2016 16:00 7/20/2016 0:30 8.5
Alex 7/18/2016 16:00 7/19/2016 0:30 8.5
Alex 7/17/2016 16:00 7/18/2016 0:30 8.5
Alex 7/14/2016 16:00 7/15/2016 0:30 8.5
Alex 7/13/2016 16:00 7/14/2016 0:30 8.5
Alex 7/12/2016 16:00 7/13/2016 0:30 8.5
Alex 7/11/2016 16:00 7/12/2016 0:30 8.5
Alex 7/10/2016 16:00 7/11/2016 0:30 8.5
Usually it is 5 working days and I get the answer if I group them by start_date. But I need an new date column where I need the output as below. Please note that 15/7/2016 and 22/7/2016 was not present in the above table. I need the additional 0.5 hour & date for the 6th day to be included to my derived table.
User Date difference
Alex 7/25/2016 8.5
Alex 7/24/2016 8.5
Alex 7/22/2016 0.5
Alex 7/21/2016 8.0
Alex 7/20/2016 8.5
Alex 7/19/2016 8.5
Alex 7/18/2016 8.5
Alex 7/17/2016 8.5
Alex 7/15/2016 0.5
Alex 7/14/2016 8.0
Alex 7/13/2016 8.5
Alex 7/12/2016 8.5
Alex 7/11/2016 8.5
Alex 7/10/2016 8.5
I calculate the difference as
round(cast(datediff(seconds, start_date, end_date) as decimal)/3600,2)

Answer 1

Whenever there is sophisticated logic, I'd suggest to use union queries and split the logic into a select query (or even table) each. Then you'd be able to calculate this in two steps. The main difference seems to be whether the 0.5 between 00:00:00 and 00:30:00 should be counted to the previous workday or whether it should stand alone. The latter seems to be determined based on whether the end_date is also a workday itself. I see three cases:
Next day is a workday:
Report all hours on start_date
Next day is not a workday:
Report hours from start_date to midnight on start_date
Report hours from midnight to end_date on end_date
I used the following example data based on your description:
create temporary table _test (user varchar(20), start_date timestamp, end_date timestamp);
insert into _test values ('Alex', '7/25/2016 16:00', '7/26/2016 0:30'), ('Alex', '7/24/2016 16:00', '7/25/2016 0:30'), ('Alex', '7/21/2016 16:00', '7/22/2016 0:30'), ('Alex', '7/20/2016 16:00', '7/21/2016 0:30'), ('Alex', '7/19/2016 16:00', '7/20/2016 0:30'), ('Alex', '7/18/2016 16:00', '7/19/2016 0:30'), ('Alex', '7/17/2016 16:00', '7/18/2016 0:30'), ('Alex', '7/14/2016 16:00', '7/15/2016 0:30'), ('Alex', '7/13/2016 16:00', '7/14/2016 0:30'), ('Alex', '7/12/2016 16:00', '7/13/2016 0:30'), ('Alex', '7/11/2016 16:00', '7/12/2016 0:30'), ('Alex', '7/10/2016 16:00', '7/11/2016 0:30');
We will need to know whether the next day is a workday, so I suggest using the lead() window function (see documentation) which will give you the start_date from the next row.
create temporary table _differences as (
select
user_name
, start_date::date as start_date
, end_date::date as end_date
/**
* Calculate difference in hours between start_date and end_date: */
, round(cast(datediff(seconds, start_date, end_date) as decimal)/3600,2) as hours_start_to_end
/**
* Calculate difference in hours between start_date and midnight: */
, round(cast(datediff(seconds, start_date, dateadd(day, 1, start_date::date)) as decimal)/3600,2) as hours_start_to_midnight
/**
* Calculate difference between midnight on end_date and end_date: */
, round(cast(datediff(seconds, end_date::date, end_date) as decimal)/3600,2) as hours_midnight_to_end
/**
* Calculate number of days from end_date until next start_date: */
, datediff(day, end_date::date, lead(start_date::date) over(partition by user_name order by start_date::date)) as days_until_next_workday
from
_test
);
Then the following query:
select
user_name as user_name
, start_date as ref_date
, hours_start_to_end as difference
from
_differences
where
days_until_next_workday = 0 -- report all work hours on start_date
union
select
user_name as user_name
, start_date as ref_date
, hours_start_to_midnight as difference
from
_differences
where
days_until_next_workday > 0 -- report partial work hours on start_date
union
select
user_name as user_name
, end_date as ref_date
, hours_midnight_to_end as difference
from
_differences
where
days_until_next_workday > 0 -- report partial work hours on end_date
order by
user_name
, ref_date desc
;
Would yield the following result:
user_name | ref_date | difference
-----------+------------+------------
Alex | 2016-07-24 | 8.50
Alex | 2016-07-22 | 0.50
Alex | 2016-07-21 | 8.00
Alex | 2016-07-20 | 8.50
Alex | 2016-07-19 | 8.50
Alex | 2016-07-18 | 8.50
Alex | 2016-07-17 | 8.50
Alex | 2016-07-15 | 0.50
Alex | 2016-07-14 | 8.00
Alex | 2016-07-13 | 8.50
Alex | 2016-07-12 | 8.50
Alex | 2016-07-11 | 8.50
Alex | 2016-07-10 | 8.50
(13 rows)
You can see that 7/25/2016 is missing because there is no start_date on or after 7/26/2016, so you'll need to figure out how to account for that special case.

Answer 2

here is how I have done the calc and it works perfectly
select user, trunc(start_time) as date1,
SUM(case when id = 1 then round(cast(datediff(seconds, start_time, st_t1) as decimal)/3600,2) end) as SCHEDULE
from
(
select user, start_time,
case when trunc(start_time) <> trunc(end_time) then cast(to_char(start_time,'yyyy-mm-dd 23:59:59') as timestamp) else cast(to_char(end_time,'yyyy-mm-dd hh24:mi:ss') as timestamp) end as st_t1
from table1 a
where id = 1
group by user_name, trunc(start_time)
union
select user_name, trunc(end_time) as date1,
SUM(case when id = 1 then round(cast(datediff(seconds, st_t2, end_time) as decimal)/3600,2) end) as SCHEDULE
from
(
select user_name, end_time,
case when trunc(start_time) <> trunc(end_time) then cast(to_char(end_time,'yyyy-mm-dd 00:00:00') as timestamp) else cast(to_char(end_time,'yyyy-mm-dd hh24:mi:ss') as timestamp) end as st_t2
from table1 a
where id = 1
)
group by user, trunc(end_time)