Is it possible to add a identity column to a GROUP BY so that each duplicate has a identity number?
My original data looks like this:
1 AAA [timestamp]
2 AAA [timestamp]
3 BBB [timestamp]
4 CCC [timestamp]
5 CCC [timestamp]
6 CCC [timestamp]
7 DDD [timestamp]
8 DDD [timestamp]
9 EEE [timestamp]
....
And I want to convert it to:
1 AAA 1
2 AAA 2
4 CCC 1
5 CCC 2
6 CCC 3
7 DDD 1
8 DDD 2
...
The solution was:
CREATE PROCEDURE [dbo].[RankIt]
AS
BEGIN
SET NOCOUNT ON;
SELECT *, RANK() OVER(PARTITION BY col2 ORDER BY timestamp DESC) AS ranking
FROM MYTABLE;
END
You could try using ROW_NUMBER if you are using Sql Server 2005
DECLARE @Table TABLE(
ID INT,
Val VARCHAR(10)
)
INSERT INTO @Table SELECT 1,'AAA'
INSERT INTO @Table SELECT 2,'AAA'
INSERT INTO @Table SELECT 3,'BBB'
INSERT INTO @Table SELECT 4,'CCC'
INSERT INTO @Table SELECT 5,'CCC'
INSERT INTO @Table SELECT 6,'CCC'
INSERT INTO @Table SELECT 7,'DDD'
INSERT INTO @Table SELECT 8,'DDD'
INSERT INTO @Table SELECT 9,'EEE'
SELECT *,
ROW_NUMBER() OVER(PARTITION BY VAL ORDER BY Val)
FROM @Table