Need to split datetime ranges by each day - sql

Question

I have a table that needs to be split on the basis of datetime
Input Table
ID| Start | End
--------------------------------------------
A | 2019-03-04 23:18:04| 2019-03-04 23:21:25
--------------------------------------------
A | 2019-03-04 23:45:05| 2019-03-05 00:15:14
--------------------------------------------
Required Output
ID| Start | End
--------------------------------------------
A | 2019-03-04 23:18:04| 2019-03-04 23:21:25
--------------------------------------------
A | 2019-03-04 23:45:05| 2019-03-04 23:59:59
--------------------------------------------
A | 2019-03-05 00:00:00| 2019-03-05 00:15:14
--------------------------------------------
Thanks!!

Answer 1

Try this below code. This will only work if the start and end date fall in two consecutive day. Not if the start and end date difference is more than 1 day.
MSSQL:
SELECT ID,[Start],[End]
FROM Input_Table A
WHERE DATEDIFF(DD,[Start],[End]) = 0
UNION ALL
SELECT ID,[Start], CAST(CAST(CAST([Start] AS DATE) AS VARCHAR(MAX)) +' 23:59:59' AS DATETIME)
FROM Input_Table A
WHERE DATEDIFF(DD,[Start],[End]) > 0
UNION ALL
SELECT ID,CAST(CAST([End] AS DATE) AS DATETIME),[End]
FROM Input_Table A
WHERE DATEDIFF(DD,[Start],[End]) > 0
ORDER BY 1,2,3
PostgreSQL:
SELECT ID,
TO_TIMESTAMP(startDate,'YYYY-MM-DD HH24:MI:SS'),
TO_TIMESTAMP(endDate, 'YYYY-MM-DD HH24:MI:SS')
FROM mytemp A
WHERE DATE_PART('day', endDate::date) -
DATE_PART('day',startDate::date) = 0
UNION ALL
SELECT ID,
TO_TIMESTAMP(startDate,'YYYY-MM-DD HH24:MI:SS'),
TO_TIMESTAMP(CONCAT(CAST(CAST (startDate AS DATE) AS VARCHAR) ,
' 23:59:59') , 'YYYY-MM-DD HH24:MI:SS')
FROM mytemp A
WHERE DATE_PART('day', endDate::date) -
DATE_PART('day',startDate::date) > 0
UNION ALL
SELECT ID,
TO_TIMESTAMP(CAST(CAST (endDate AS DATE) AS VARCHAR) ,
'YYYY-MM-DD HH24:MI:SS') ,
TO_TIMESTAMP(endDate,'YYYY-MM-DD HH24:MI:SS')
FROM mytemp A
WHERE DATE_PART('day', endDate::date) -
DATE_PART('day',startDate::date) > 0;
PostgreSQL Demo Here

Answer 2

demo:db<>fiddle
This works even when range crosses more than one day
WITH cte AS (
SELECT
id,
start_time,
end_time,
gs,
lag(gs) over (PARTITION BY id ORDER BY gs) -- 2
FROM
a
LEFT JOIN LATERAL
generate_series(start_time::date + 1, end_time::date, interval '1 day') gs --1
ON TRUE
)
SELECT -- 3
id,
COALESCE(lag, start_time) AS start_time,
gs - interval '1 second'
FROM
cte
WHERE gs IS NOT NULL
UNION
SELECT DISTINCT ON (id) -- 4
id,
CASE WHEN start_time::date = end_time::date THEN start_time ELSE end_time::date END, -- 5
end_time
FROM
cte
CTE: the generate_series function generates one row per day new day. So, there is no value if there is no date change
CTE: the lag() window function allows to move the current date value into the next row (the current end is the next start)
With this data set you can calculate the new start and end values. If there is no gs value: There is no date change. This ignored at this point. For all cases with date changes: If there is no lag value, it is the beginning (so it cannot got a previous value). In this case, the normal start_time is taken, otherwise it is a new day which takes the date break time. The end_time is taken with the last second of the day (interval - '1 second')
The second part: Because of the date breaks there is always one additional record which need to be unioned. The last record is from the beginning of the end_time (so cast to date). The CASE clause combines this step with the case of no date change which has been ignored so far. So if start_time and end_time are at the same date, here the original start_time is taken.

Answer 3

Unfortunately, Redshift doesn't have a convenient way to generate a series of numbers. If you table is big enough, you can use it to generate numbers. "Big enough" means that the number of rows is greater than the longest span. Perhaps another table would work, if not this one.
Once you have that, you can use this logic:
with n as (
select row_number() over () - 1 as n
from t
)
select t.id,
greatest(t.s, date_trunc('day', t.s) + n.n * interval '1 day') as s,
least(t.e, date_trunc('day', t.s) + (n.n + 1) * interval '1 day' - interval '1 second') as e
from t join
n
on t.e >= date_trunc('day', t.s) + n.n * interval '1 day';
Here is a db<>fiddle. It uses an old version of Postgres, but not quite old enough for Redshift.

Answer 4

Simulate loop for interval generation using recursive CTE, i.e. take range from start to midnight in seed row, take another day in subsequent rows etc.
with recursive input as (
select 'A' as id, timestamp '2019-03-04 23:18:04' as s, timestamp '2019-03-04 23:21:25' as e union
select 'A' as id, timestamp '2019-03-04 23:45:05' as s, timestamp '2019-03-05 00:15:14' as e union
select 'B' as id, timestamp '2019-03-06 23:45:05' as s, timestamp '2019-03-08 00:15:14' as e union
select 'C' as id, timestamp '2019-03-10 23:45:05' as s, timestamp '2019-03-15 00:15:14' as e
), generate_id as (
select row_number() over () as unique_id, * from input
), rec (unique_id, id, s, e) as (
select unique_id, id, s, least(e, s::date::timestamp + interval '1 day')
from generate_id seed
union
select remaining.unique_id, remaining.id, previous.e, least(remaining.e, previous.e::date::timestamp + interval '1 day')
from rec as previous
join generate_id remaining on previous.unique_id = remaining.unique_id and previous.e < remaining.e
)
select id, s, e from rec
order by id,s,e
Note:
your id column appears not to be unique, so I added custom unique_id column. If id was unique, CTE generate_id was unnecessary. Uniqueness is unavoidable for recursive query to work.
close-open range is better for representation of such data, rather than close-close range. So end time in my query returns 00:00:00, not 23:59:59. If it's not suitable for you, modify query as an exercise.
UPDATE: query works on Postgres. OP originally tagged question postgres, then changed tag to redshift.