SQL statement for new vs duplicate data

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SQL statement for new vs duplicate data

Author

Topic

avitron2020
Starting Member

3 Posts

Posted - 2013-05-08 : 15:11:51

I'm looking for a SQL statement that will output 'new vs duplicate' data on a monthly bases, but I can't seem to come up SQL statement that can accomplish what I need. See example below to get a better idea of what I'm trying to do since this is hard to explain.

Example data:

id, num, create_date

// January
1, 34250, 2012-01-01
2, 25383, 2012-01-03
3, 93482, 2012-01-13
4, 52934, 2012-01-20
5, 34823, 2012-01-23

// Febuary
6, 09943, 2012-02-03
7, 32845, 2012-02-05
8, 34250, 2012-02-16 -- same num as id 1
9, 57325, 2012-02-28
10, 34823, 2012-02-28 -- same num as id 5

// March
11, 95234, 2012-03-05
12, 25383, 2012-03-03 -- same num as id 2
13, 32845, 2012-03-16 -- same num as id 7
14, 34250, 2012-03-28 -- same num as id 1, 8
15, 12412, 2012-03-28

// April
16, 53535, 2012-04-05
17, 23232, 2012-04-03
18, 57325, 2012-04-16 -- same num as id 9
19, 34250, 2012-04-28 -- same num as id 1, 8, 14
20, 12343, 2012-04-28

Example results from data above would be something like below (just looking for new vs duplicate numbers):

January: 5 new and 0 duplicates (no duplicates will ever be found the first month)
Febuary: 3 new and 2 duplicates (checks for duplicates in the month of Jan only)
March: 2 new and 3 duplicates (checks for duplicates in the month of Jan & Feb only)
April: 3 new and 2 duplicates (checks for duplicates in the month of Jan, Feb & March only)
...and so on for each month after that

Hopefully you can see what I'm trying to do. Also, the total amount (new + duplicate) will always be 5 based on the data above. Any real smart person out there that can help?

Note: I'm using SQL Server 2005

James K
Master Smack Fu Yak Hacker

3873 Posts

Posted - 2013-05-08 : 16:30:28

Give this a try

SELECT 
	YEAR(create_date),
	MONTH(create_date),
	COUNT(*) - COUNT(DISTINCT b.dups) AS New,
	COUNT(DISTINCT b.dups) AS Dups
FROM
	YourTable a
	OUTER APPLY
	(
		SELECT num FROM 
		YourTable b
		WHERE 
		b.create_date < DATEADD(mm,DATEDIFF(mm,0,a.create_date),0)
		--AND b.create_date >= DATEADD(yy,DATEDIFF(yy,0,a.create_date),0)
		AND a.num = b.num

	) b

visakh16
Very Important crosS Applying yaK Herder

52326 Posts

Posted - 2013-05-09 : 00:51:44

[code]
SELECT YEAR(create_date) AS [Year],
MONTH(create_date) AS [Month],
SUM(CASE WHEN Rn=1 THEN 1 ELSE 0 END) AS New,
SUM(CASE WHEN Rn > 1 THEN 1 ELSE 0 END) AS Duplicate
FROM
(
SELECT ROW_NUMBER() OVER (PARTITION BY num ORDER BY create_Date) AS Rn,*
FROM Table
)t
GROUP BY YEAR(create_date),
MONTH(create_date)
[/code]

------------------------------------------------------------------------------------------------------
SQL Server MVP
http://visakhm.blogspot.com/
https://www.facebook.com/VmBlogs

visakh16
Very Important crosS Applying yaK Herder

52326 Posts

Posted - 2013-05-09 : 01:03:30

quote:

Originally posted by James K

Give this a try

SELECT 
	YEAR(create_date),
	MONTH(create_date),
	COUNT(*) - COUNT(DISTINCTSUM(CASE WHEN b.dups>0 THEN 1 ELSE 0 END) AS New,
	COUNT(DISTINCT b.dups)SUM(CASE WHEN COALESCE(b.dups,0)=0 THEN 1 ELSE 0 END) AS Dups
FROM
	YourTable a
	OUTER APPLY
	(
		SELECT numCOUNT(*) AS dups FROM 
		YourTable b
		WHERE 
		b.create_date < DATEADD(mm,DATEDIFF(mm,0,a.create_date),0)
		--AND b.create_date >= DATEADD(yy,DATEDIFF(yy,0,a.create_date),0)
		AND a.num = b.num

	) b
GROUP BY YEAR(create_date),
	MONTH(create_date)

Some tweaks to get expected result

------------------------------------------------------------------------------------------------------
SQL Server MVP
http://visakhm.blogspot.com/
https://www.facebook.com/VmBlogs

avitron2020
Starting Member

3 Posts

Posted - 2013-05-09 : 10:39:33

Thanks. I'll try it out today and let you know how it works.

visakh16
Very Important crosS Applying yaK Herder

52326 Posts

Posted - 2013-05-10 : 00:17:37

welcome

------------------------------------------------------------------------------------------------------
SQL Server MVP
http://visakhm.blogspot.com/
https://www.facebook.com/VmBlogs

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