How to use MAX() to select a single record but prevent GROUP BY from selecting all distinct values and multiple records

Question

I have some Line items that have version numbers:

[Doc No]    [Version No]    [Line No]    [Data-a]    [Data-b]
1           1               1000         abc         abc
1           2               1000         xyz         abc
1           1               2000         null        lmnop
1           2               2000         ggg         lmnop

I want to only select the highest version number per [Line No]:

[Doc No]    [Version No]    [Line No]    [Data-a]    [Data-b]
1           2               1000         xyz         abc
1           2               2000         ggg        lmnop

I am doing this using this tsql:

SELECT [Doc No]
      ,max([Version No])
      ,[Line No]
      ,[Data-a]
      ,[Data-b]
FROM [table]
GROUP BY [Doc No], [Line No], [Data-a], [Data-b]

The problem obviously is ... if you notice in the first table the [Data-a] column has changing values (hence the need for version numbers. So I want to see the latest version and the value contained in [Data-a] for that record. But since I am using max() to get the latest version, I have to put [Data-a] in the GROUP BY clause - which causes each two distinct values to be pulled and thus two records showing both version numbers. How can I solve this problem?

Answer 1

You can accomplish this with a subquery that finds the maximum Version No by Doc No and Line No.

CREATE TABLE #temp ([Doc No]  INT, [Version No] INT, [Line No]  INT, [Data-a] VARCHAR(5), [Data-b] VARCHAR(5));

INSERT INTO #temp VALUES (1,1,1000,'abc','abc')
INSERT INTO #temp VALUES(1,2,1000,'xyz','abc')
INSERT INTO #temp VALUES(1,1,2000,null,'lmnop')
INSERT INTO #temp VALUES(1,2,2000,'ggg','lmnop')

SELECT a.*
FROM #temp as a
JOIN
    ( 
SELECT [Doc No] 
        ,[Line No] 
      ,max([Version No]) as MaxVersion
FROM #temp
GROUP BY [Doc No] , [Line No]
) as b on a.[Doc No]  = b.[Doc No]  and a.[Line No] = b.[Line No] and a.[Version No] = b.MaxVersion

DROP TABLE #temp

Answer 2

Just incase you need another solution. Below uses row_number() over (partition by .. order by)

CREATE TABLE dbo.docStore
    ([Doc No] int, 
     [Version No] int,
     [Line No] int ,
     [Data-a] varchar(10),
     [Data-b] varchar(10))
;

INSERT INTO dbo.docStore
    ([Doc No],[Version No],[Line No],[Data-a],[Data-b])
VALUES
    (1,1,1000,'abc','abc'    ),
    (1,2,1000,'xyz','abc'    ),
    (1,1,2000,null,'lmnop'   ),
    (1,2,2000,'ggg','lmnop'  )
;

with cte as 
(
select [Doc No], [Version No], [Line No], [Data-a], [Data-b], 
        row_number() over (partition by [Doc No], [Line No] 
                           order by [Version No] desc) 
        as rn 
from dbo.docStore
)
select [Doc No], [Version No], [Line No], [Data-a], [Data-b] 
from cte
where rn = 1
;

-- clean up 
drop table dbo.docStore 
;

Answer 3

This get you what you need?

;with cteMaxVers ([Line No], [Version No] 
AS (SELECT [Line No], MAX([Version No])
FROM [table]
GROUP BY [Line No]  
)  
SELECT t.[Doc No]
      ,t.[Version No]
      ,t.[Line No]
      ,t.[Data-a]
      ,t.[Data-b]
FROM [table] t 
JOIN cteMaxVers c ON t.[Line No] = c.[Line No] AND t.[Version No] = c.[Version No];

Answer 4

Borrowing @Kin's code but adding one more record to show the correct answer.

use tempdb

CREATE TABLE dbo.docStore
    ([Doc No] int, [Version No] int,[Line No] int ,[Data-a] varchar(10),[Data-b] varchar(10))
;

INSERT INTO dbo.docStore
    ([Doc No],[Version No],[Line No],[Data-a],[Data-b])
VALUES
    (1,1,1000,'abc','abc'    ),
    (1,2,1000,'xyz','abc'    ),
    (1,1,2000,null,'lmnop'   ),
    (1,2,2000,'ggg','lmnop'  ),
    (1,3,2000,'ddd','test')
;

with cte as 
(
select [Doc No],[Version No],[Line No],[Data-a],[Data-b], row_number() over (partition by [Doc No],[Line No] order by [Version No] desc) as rn from dbo.docStore
)
select [Doc No],[Version No],[Line No],[Data-a],[Data-b] from cte
where rn =1


-- clean up 
drop table dbo.docStore

This will ensure that we will get the largest [Version No] per group of [Doc No] and [Line No]

The result is as follows: