Get Start and End date from multiple rows of dates, excluding weekends - sql

Question

I'm trying figure out how to return Start Date and End date based on data like in the below table:
Name
Date From
Date To
A
2022-01-03
2022-01-03
A
2021-12-29
2021-12-31
A
2021-12-28
2021-12-28
A
2021-12-27
2021-12-27
A
2021-12-23
2021-12-24
A
2021-11-08
2021-11-09
The result I am after would show like this:
Name
Date From
Date To
A
2021-12-23
2022-01-03
A
2021-11-08
2021-11-09
The dates in first table will sometimes go over weekends with the Date From and Date To, but in cases where the row ends on a Friday and next row starts on following Monday it will need to be classified as the same "block", as presented in the second table. I was hoping to use DATEFIRST setting to cater for the weekends to avoid using a calendar table, as per How do I exclude Weekend days in a SQL Server query?, but if calendar table ends up being the easiest way out I'm happy to look into creating one.
In above example I only have 1 Name, but the table will have multiple names and it will need to be grouped by that.
The only examples of this I am seeing are using only 1 date column for records and I struggled changing their code around to cater for my example. The closest example I found doesn't work for me as it is based on datetime fields and the time differences - find start and stop date for contiguous dates in multiple rows

Answer 1

This is a Gaps & Island problem with the twist that you need to consider weekend continuity.
You can do:
select max(name) as name, min(date_from) as date_from, max(date_to) as date_to
from (
select *, sum(inc) over(order by date_to) as grp
from (
select *,
case when lag(ext_to) over(order by date_to) = date_from
then 0 else 1 end as inc
from (
select *,
case when (datepart(weekday, date_to) = 6)
then dateadd(day, 3, date_to)
else dateadd(day, 1, date_to) end as ext_to
from t
) x
) y
) z
group by grp
Result:
name date_from date_to
---- ---------- ----------
A 2021-11-08 2021-11-09
A 2021-12-23 2022-01-03
See running example at db<>fiddle #1.
Note: Your question doesn't mention it, but you probably want to segment per person. I didn't do it.
EDIT: Adding partition by name
Partitioning by name is quite easy actually. The following query does it:
select name, min(date_from) as date_from, max(date_to) as date_to
from (
select *, sum(inc) over(partition by name order by date_to) as grp
from (
select *,
case when lag(ext_to) over(partition by name order by date_to) = date_from
then 0 else 1 end as inc
from (
select *,
case when (datepart(weekday, date_to) = 6)
then dateadd(day, 3, date_to)
else dateadd(day, 1, date_to) end as ext_to
from t
) x
) y
) z
group by name, grp
order by name, grp
See running query at db<>fiddle #2.

Answer 2

with extended as (
select name,
date_from,
case when datepart(weekday, date_to) = 6
then dateadd(day, 2, date_to) else date_to end as date_to
from t
), adjacent as (
select *,
case when dateadd(day, 1,
lag(date_to) over (partition by name order by date_from)) = date_from
then 0 else 1 end as brk
from extended
), blocked as (
select *, sum(brk) over (partition by name order by date_from) as grp
from adjacent
)
select name, min(date_from), max(date_to) from blocked
group by name, grp;
I'm assuming that ranges do no overlap and that all input dates do fall on weekdays. While hammering this out on my cellphone I originally made two mistakes. For some reason I got to and from dates reversed in my head and then I was thinking that Friday is 5 (as with ##datefirst) rather than 6. (Of course this could otherwise vary with the regional setting anyway.) One advantage of using table expressions is to modularize and bury certain details in lower levels of the logic. In this case it would be very easy to adjust dates should some of these assumptions prove to be wrong.
https://dbfiddle.uk/?rdbms=sqlserver_2019&fiddle=42e0c452d57d474232bcf991d6d3c43c