SQL: Deleting row which values already exist - sql

Question

I have a table that look like this:
ID | DATE | NAME | VALUE_1 | VALUE_2
1 | 27.11.2015 | Homer | A | B
2 | 27.11.2015 | Bart | C | B
3 | 28.11.2015 | Homer | A | C
4 | 28.11.2015 | Maggie | C | B
5 | 28.11.2015 | Bart | C | B
I currently delete duplicate rows (thank to this thread) using this code :
WITH cte AS
(SELECT ROW_NUMBER() OVER (PARTITION BY [VALUE_1], [VALUE_2]
ORDER BY [DATE] DESC) RN
FROM [MY_TABLE])
DELETE FROM cte
WHERE RN > 1
But this code don't delete exactly the lines I want. I would like to delete only rows which values already exist so in my example I would like to delete only line 5 because line 2 have the same values and is older.
Code to create my table and insert values:
CREATE TABLE [t_diff_values]
([id] INT IDENTITY NOT NULL PRIMARY KEY,
[date] DATETIME NOT NULL,
[name] VARCHAR(255) NOT NULL DEFAULT '',
[val1] CHAR(1) NOT NULL DEFAULT '',
[val2] CHAR(1) NOT NULL DEFAULT '');
INSERT INTO [t_diff_values] ([date], [name], [val1], [val2]) VALUES
('2015-11-27','Homer', 'A','B'),
('2015-11-27','Bart', 'C','B'),
('2015-11-28','Homer', 'A','C'),
('2015-11-28','Maggie', 'C','B'),
('2015-11-28','Bart', 'C','B');

Answer 1

You need to add one more CTE where you will index all islands and then apply your duplicate logic in second CTE:
DECLARE #t TABLE
(
ID INT ,
DATE DATE ,
VALUE_1 CHAR(1) ,
VALUE_2 CHAR(1)
)
INSERT INTO #t
VALUES ( 1, '20151127', 'A', 'B' ),
( 2, '20151128', 'C', 'B' ),
( 3, '20151129', 'A', 'B' ),
( 4, '20151130', 'A', 'B' );
WITH cte1
AS ( SELECT * ,
ROW_NUMBER() OVER ( ORDER BY date)
- ROW_NUMBER() OVER ( PARTITION BY VALUE_1, VALUE_2 ORDER BY DATE) AS gr
FROM #t
),
cte2
AS ( SELECT * ,
ROW_NUMBER() OVER ( PARTITION BY VALUE_1, VALUE_2, gr ORDER BY date) AS rn
FROM cte1
)
DELETE FROM cte2
WHERE rn > 1
SELECT *
FROM #t

Answer 2

Try this
CREATE TABLE [dbo].[Employee](
[ID] INT NOT NULL,
[Date] DateTime NOT NULL,
[VAL1] varchar(20) NOT NULL,
[VAL2] varchar(20) NOT NULL
)
INSERT INTO [dbo].[Employee] VALUES
(1,'2015-11-27 10:44:33.087','A','B')
INSERT INTO [dbo].[Employee] VALUES
(2,'2015-11-28 10:44:33.087','C','B')
INSERT INTO [dbo].[Employee] VALUES
(3,'2015-11-29 10:44:33.087','A','B')
INSERT INTO [dbo].[Employee] VALUES
(4,'2015-11-30 10:44:33.087','A','B')
with cte as(
select
*,
rn = row_number() over(partition by [VAL1], [VAL2]
ORDER BY [DATE] DESC),
cc = count(*) over(partition by [VAL1], [VAL2])
from [Employee]
)
delete
from cte
where
rn > 1 and rn < cc
select * from [Employee]

Answer 3

You could use this query:
WITH cte AS
(
SELECT RN = ROW_NUMBER() OVER (ORDER BY ID)
, *
FROM #data
)
DELETE FROM c1
--SELECT *
FROM CTE c1
INNER JOIN CTE c2 ON c1.RN +1 = c2.RN AND c1.VALUE_1 = c2.VALUE_1 AND c1.VALUE_2 = c2.VALUE_2
Here I order them by ID. If the next one (RN+1) has similar V1 and V2, it is deleted.
Output:
ID DATE VALUE_1 VALUE_2
1 2015-11-27 A B
2 2015-11-28 C B
4 2015-11-30 A B
Data:
declare #data table(ID int, [DATE] date, VALUE_1 char(1), VALUE_2 char(1));
insert into #data(ID, [DATE], VALUE_1, VALUE_2) values
(1, '20151127', 'A', 'B'),
(2, '20151128', 'C', 'B'),
(3, '20151129', 'A', 'B'),
(4, '20151130', 'A', 'B');