Query for Min, Max, Avg, and Last Value in TSQL - sql-server

Question

I am trying to find Max, min, avg and last value of a column in single query.
Platform: SQL Server 2012
Sample Table:
SN Month Acc Bal
------------------------
1 7 101 1,000/-
2 7 101 1,500/-
3 7 101 1,700/-
4 8 101 1,200/-
5 8 101 900/-
6 9 101 2,500/-
Query I wrote:
select
[Month], [Acc],
min(Bal) as MinBal,
avg(Bal) as AvgBal,
max(Bal) as MaxBal
--, ??? for as LastBal
from
MyTable
Group By
[Month], [Acc]
Query with Last_Value returns all records instead of aggregated records
select
[Month], [Acc],
min(Bal) as MinBal,
avg(Bal) as AvgBal,
max(Bal) as MaxBal,
LAST_VALUE(Bal) OVER (partition by [Acc] order by [Month]) as LastBal
from
MyTable
Group By
[Month], [Acc], Bal
Also including last_value(bal) is generating an error with bal required on group by list
Column 'Bal' is invalid in the select list because it is not contained in either an aggregate function or the GROUP BY clause.

Answer 1

Please try this solution-
DATA Generation
CREATE TABLE Alls
(
SN INT
,[Month] INT
,Acc INT
,Bal INT
)
GO
INSERT INTO Alls VALUES
(1, 7, 101, 1000),
(2, 7, 101, 1500),
(3, 7, 101, 1700),
(4, 8, 101, 1200),
(5, 8, 101, 900),
(6, 9, 101, 2500)
GO
SOLUTION
SELECT sn,Acc,[Month] ,Bal
, MIN(Bal) OVER(PARTITION BY Acc,[Month]) MinBal
, AVG(Bal*1.) OVER(PARTITION BY Acc,[Month]) AvgBal
, MAX(Bal) OVER(PARTITION BY Acc,[Month]) MaxBal
, FIRST_VALUE(Bal) OVER(PARTITION BY Acc,[Month] ORDER BY SN DESC) lastVal
FROM Alls
ORDER By SN
OUTPUT
sn Acc Month Bal MinBal AvgBal MaxBal lastVal
----------- ----------- ----------- ----------- ----------- ---------------- ----------- -----------
1 101 7 1000 1000 1400.000000 1700 1700
2 101 7 1500 1000 1400.000000 1700 1700
3 101 7 1700 1000 1400.000000 1700 1700
4 101 8 1200 900 1050.000000 1200 900
5 101 8 900 900 1050.000000 1200 900
6 101 9 2500 2500 2500.000000 2500 2500
(6 rows affected)
IF you only need acc,month and other aggregate columns then use below-
SOLUTION
SELECT Acc,[Month],MAX(MinBal)MinBal,MAX(AvgBal)AvgBal,MAX(MaxBal)MaxBal,MAX(lastVal)lastVal
FROM
(
SELECT sn,Acc,[Month] ,Bal
, MIN(Bal) OVER(PARTITION BY Acc,[Month]) MinBal
, AVG(Bal*1.) OVER(PARTITION BY Acc,[Month]) AvgBal
, MAX(Bal) OVER(PARTITION BY Acc,[Month]) MaxBal
, FIRST_VALUE(Bal) OVER(PARTITION BY Acc,[Month] ORDER BY SN DESC) lastVal
FROM Alls
)u GROUP BY Acc,[Month]
OUTPUT
Acc Month MinBal AvgBal MaxBal lastVal
----------- ----------- ----------- ---------------- ----------- -----------
101 7 1000 1400.000000 1700 1700
101 8 900 1050.000000 1200 900
101 9 2500 2500.000000 2500 2500
(3 rows affected)

Answer 2

select *
from
( SELECT sn, Acc, [Month], Bal
, MIN(Bal) OVER(PARTITION BY Acc, [Month]) MinBal
, AVG(Bal) OVER(PARTITION BY Acc, [Month]) AvgBal
, MAX(Bal) OVER(PARTITION BY Acc, [Month]) MaxBal
, row_number() OVER(PARTITION BY Acc, [Month] ORDER BY SN DESC) as rn
) tt
where rn = 1
ORDER By sn

Answer 3

You can achieve as below:
select
tt.Month
, tt.Acc
, min(Bal) as MinBal
, avg(Bal) as AvgBal
, max(Bal) as MaxBal
, latest.balance
FROM #tbl1 as tt
JOIN (
SELECT
id
,month
,acc
,bal as balance
FROM #tbl1 AS t1
WHERE id = (SELECT MAX(id)
FROM #tbl1 AS t2
WHERE t1.month = t2.month
AND t1.acc = t2.acc
GROUP BY month, acc)
) as latest
on tt.month = latest.month
AND tt.acc = latest.acc
Group By tt.Month, tt.Acc, latest.balance
DROP TABLE #tbl1