Select all second highest values only from temp table - sql-server

Question

Using T-SQL (SQL Server 2008 R2), I'm trying to list only the rows with the second highest value in a particular column from a temp table and then place the results into a new temp table. The PK is the ID, which can have increasing version numbers and then unique codes.
Example:
ID | Name| Version | Code
------------------------
1 | A | 1 | 10
1 | A | 2 | 20
1 | A | 3 | NULL
2 | B | 1 | 40
2 | B | 2 | 50
2 | C | 1 | 60
The desired outcome of the query is
ID | Version | Code
------------------------
1 | 2 | 20
2 | 1 | 40
To achieve this I need the below query to be adapted to pull the second highest value as long as the result gives a version number greater than 1. These results come from a temp table and will then be placed into a final results temp table. EDIT: Please note this will be applied over 33000 rows of data so I would prefer something neater than INSERT VALUES. Thanks.
Current query:
SELECT
ID
,Version
,Code
INTO
#table2
FROM
#table1
SELECT *
FROM #table2
WHERE Version > 1
ORDER BY ID asc
DROP TABLE #table1
DROP TABLE #table2
I have tried running the where clause WHERE Version < (SELECT MAX(VERSION) FROM #TABLE 2) but this has no effect, presumably due to the unique code values and in any case wouldn't work where I have more than 3 Versions.
Ideas would be gratefully received.
Thanks in advance.

Answer 1

i HAVE TEST THE BELOW CODE AND IT IS GIVING OUTPUT AS PER The YOUR desired outcome of the query is
SELECT ID,Name,[Version],Code
FROM (
SELECT ROW_NUMBER() OVER (PARTITION BY NAME ORDER BY [Version] DESC) AS RNK,*
FROM
(
SELECT 1 ID, 'A' Name ,1 [Version] ,10 Code
UNION ALL
SELECT 1, 'A', 2 ,20
UNION ALL
SELECT 1, 'A', 3 ,30
UNION ALL
SELECT 1, 'A', 4 ,NULL
UNION ALL
SELECT 2, 'B', 1 ,40
UNION ALL
SELECT 2, 'B', 2 ,50
UNION ALL
SELECT 2, 'C', 1 ,60
)B
)BASE
WHERE RNK =2

Answer 2

If your primary key is only ID, you have duplicate rows. So I assume your primary key is something else, for example ID, Version, Name. You have two rows with the same ID and same Version, what kind of rule do you want to apply on this ? Lowest number ?
I made an example that does kind of what you want:
First declare the necessary tables:
declare #table1 table (
Id int,
Name nvarchar(20),
[Version] int,
Code int
)
insert into #table1 values (1,'A',1,10),(1,'A',2,20),(1,'A',3,30),(1,'A',4,NULL)
,(2,'B',1,40),(2,'B',2,50),(2,'C',1,60);
And then the query to get the results:
with HighestVersions (Id, MaxVersion) As
(
select Id, max(version) from #table1 group by Id
)
select
t1.Id,
t1.[Version],
min(t1.Code) as Code
from
#table1 t1
inner join
HighestVersions hv
on
hv.Id = t1.Id
and (hv.MaxVersion-1) = t1.[Version]
group by
t1.Id
,t1.[Version]
I had to do a little dirty trick with the outermost select, this is because of the duplicate 'Id' and 'Version'. Else you would have gotten two rows with ID = 2, Version = 1
If you want to remove the NULL value you can change the WITH part (according to your last edit):
with HighestVersions (Id, MaxVersion) As
(
select Id, max(version) from #table1 where Code is not null group by Id
)

Answer 3

Try this:
DECLARE #List TABLE (ID int, Name char(1), Version int, Code int NULL)
INSERT INTO #List
VALUES
(1, 'A', 1, 10),
(1, 'A', 2, 20),
(1, 'A', 3, 30),
(1, 'A', 4, NULL),
(2, 'B', 1, 40),
(2, 'B', 2, 50),
(2, 'C', 1, 60)
SELECT
ID, Name, Version, Code
FROM
(
SELECT
*,
ROW_NUMBER() OVER (PARTITION BY ID, Name ORDER BY Version DESC) Rn
FROM #List
) a
WHERE
a.Rn = 2