Order By One One Column in MSSQL - sql

Question

I Have the following SQL Tables:
[Calendar]
[CalendarId]
[Name]
SAMPLE DATA:
CalendarId ResourceKey Name
1 1 tk1-Room1
2 2 tk1-Room2
3 3 tk1-noentries
[CalendarEntry]
[CalendarId]
[CalendarEntryId]
[Start]
[End]
SAMPLE DATA:
CalendarId Start End
1 2019-11-18 16:00:00.0000000 2019-11-18 17:00:00.0000000
1 2019-11-19 16:00:00.0000000 2019-11-19 17:00:00.0000000
2 2019-11-25 16:00:00.0000000 2019-11-25 17:00:00.0000000
1 2019-11-25 17:00:00.0000000 2019-11-25 18:00:00.0000000
Expected output:
Name StartDate EndDate ResourceKey
tk1-Room1 2019-11-25 17:00:00 2019-11-25 17:00:00 1
tk1-Room2 2019-11-25 16:00:00 2019-11-25 17:00:00 2
tk1-noentries NULL NULL 3
I am trying to list all Calendar entries, with their corresponding Start(Most Recent) and End Times.
I have the following code which is working partially:
SELECT Name,StartDate,ResourceKey FROM [Calendar].[dbo].[Calendar] CAL
LEFT JOIN(
SELECT
CalendarId,
MAX(ENT.[Start]) as StartDate
FROM [CalendarEntry] ENT
GROUP BY CalendarId
)
AS ST on CAL.CalendarId = ST.CalendarId
However, If i was to include that column, In my sub SELECT, EG:
SELECT
CalendarId,
MAX(ENT.[Start]) as StartDate,
ENT.[End] as endDate
I get the following error:
Column 'CalendarEntry.End' is invalid in the select list because it is not contained in either an aggregate function or the GROUP BY clause.
However, including it in the GROUP BY now causes Multiple CalendarEntry Rows to come back for each Calendar..
What is the best way for me to grab the most recent row out of CalendarEntry which allows me access to all the columns?
Thanks!

Answer 1

This is a typical top 1 per group question.
You can either use row_number():
select *
from (
select
c.*,
e.*,
row_number() over(partition by c.CalendarId order by e.Start desc) rn
from [Calendar].[dbo].[Calendar] c
left join [CalendarEntry] e ON c.CalendarId = e.CalendarId
) t
where rn = 1
Or you can filter with a correlated subquery:
select c.*, e.*
from [Calendar].[dbo].[Calendar] c
left join [CalendarEntry] e
on c.CalendarId = e.CalendarId
and c.Start = (
select max(e1.Start) from [CalendarEntry] e where c.CalendarId = e1.CalendarId
)

Answer 2

I am trying to list all Calendar entries, with their corresponding Start(Most Recent) and End Times.
I interpret this as the most recent record from CalendarEntry for each CalendarId:
select ce.*
from CalendarEntry ce
where ce.StartDate = (select max(ce2.StartDate)
from CalendarEntry ce2
where ce2.CalendarId = ce.CalendarId
);

Answer 3

You can try OUTER APPLY too, however #GMB's answer is a better approach from performance prospective
SELECT Name,
StartDate,
EndDate,
ResourceKey
FROM dbo.Calendar AS C
OUTER APPLY
(
SELECT TOP 1 *
FROM dbo.CalendarEntry
WHERE CalendarId = C.CalendarId
ORDER BY StartDate DESC,
EndDate DESC
) AS K;
You can also try LAST_VALUE/FIRST_VALUE(available in SQL Server 2012 and later) functions too, as below, , however again #GMB's answer is a better approach from performance prospective:
SELECT DISTINCT
Name,
LAST_VALUE(StartDate) OVER (PARTITION BY C.CalendarId
ORDER BY StartDate,EndDate
ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING),
LAST_VALUE(EndDate) OVER (PARTITION BY C.CalendarId
ORDER BY StartDate,EndDate
ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING),
ResourceKey
FROM dbo.Calendar AS C
LEFT JOIN dbo.CalendarEntry
ON CalendarEntry.CalendarId = C.CalendarId;
If you want to use FIRST_VALUE function, then you should rewrite the order by as below:
ORDER BY StartDate DESC,EndDate DESC
And you also you will not need to specify the ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING section
SELECT DISTINCT
Name,
FIRST_VALUE(StartDate) OVER (PARTITION BY C.CalendarId
ORDER BY StartDate DESC,EndDate DESC),
FIRST_VALUE(EndDate) OVER (PARTITION BY C.CalendarId
ORDER BY StartDate DESC,EndDate DESC),
ResourceKey
FROM dbo.Calendar AS C
LEFT JOIN dbo.CalendarEntry
ON CalendarEntry.CalendarId = C.CalendarId;