T-SQL, list of DATETIME, create <from> - <to> from it - sql

Question

What would I need to do to achieve the following? Somehow I can't seem to find a good solution.
I have a few CTEs, and the last one is producing just a list of DATETIME values, with a row number column, and those are ordered by the DATETIME.
For example
rn datetime
---------------------------
1 2023-01-07 01:00:00.000
2 2023-01-08 05:30:00.000
3 2023-01-08 08:00:00.000
4 2023-01-09 21:30:00.000
How do I have to join this CTE with each other in order to get the following result:
from to
---------------------------------------------------
2023-01-07 01:00:00.000 2023-01-08 05:30:00.000
2023-01-08 08:00:00.000 2023-01-09 21:30:00.000
Doing a regular inner join (with t1.rn = t2.rn - 1) gives me one row too much (the one from 05:30 to 08:00). So basically each date can only be "used" once.
Hope that makes sense... thanks!
Inner joining the CTE with itself, which didn't return the wanted result.

Answer 1

You can pivot the outcome of your CTE and distribute rows using arithmetics : modulo 2 comes to mind.
Assuming that your CTE returns columns dt (a datetime field) and rn (an integer row number) :
select min(dt) dt_from, max(dt) dt_to
from cte
group by ( rn - 1 ) % 2
On T-SQL we could also leverage integer division to express the pair grouping:
group by ( rn - 1 ) / 2

Answer 2

You can avoid both the JOIN and the GROUP BY by using LAG to retrieve a previous column value in the result set. The server may be able to generate an execution plan that iterates over the data just once instead of joining or grouping :
with pairs as (
SELECT rn, lag(datetime) OVER(ORDER BY rn) as dt_from, datetime as dt_to
from another_cte
)
select dt_from,dt_to
from pairs
ORDER BY rn
where rn%2=0
The row number itself can be calculated from datetime :
with pairs as (
SELECT
ROW_NUMBER() OVER(ORDER BY datetime) AS rn,
lag(datetime) OVER(ORDER BY datetime) as dt_from,
datetime as dt_to
from another_cte
)
select dt_from,dt_to
from pairs
ORDER BY dt_to
where rn%2=0