Solved

Ranking Based On Value

Posted on 2016-11-10
3
52 Views
Last Modified: 2016-11-13
I know how to do basic ranking, but is it possible to do it based on a sum of a value?

Using the small subset of data below, my goal is to sum up the "DescValue" column and every SUM of 5, I would increment the "Page" number. The "Page" must increment to never allow the sum of "DescValue" on a page to be greater than 5.

I am using MS SQL 2014.

Description            DescValue      Page
Line1                   1                  1
Line2                   2                  1
Line3                  1                  1
Line4                  2                  2
Line5                  3                  2
Line6                  2                  3
Line7                  4                  4
Line8                  1                  4
Line9                  1                  5
0
Comment
Question by:ScubeduFan
  • 2
3 Comments
 
LVL 28

Expert Comment

by:Pawan Kumar
ID: 41883326
ok got it , working. Good One Vitor :) and I totally take that back.

Regards,
Pawan
0
 
LVL 28

Accepted Solution

by:
Pawan Kumar earned 500 total points
ID: 41883678
Here is the solution, try

Table creation

--

Create table five
(
	 Description varchar(100)           
	,DescValue  int  	
)
GO

INSERT INTO five VALUES
('Line1'            ,       1   ),     
('Line2'            ,       2     ),             
('Line3'            ,      1     ),             
('Line4'            ,      2     ),             
('Line5'             ,     3     ),             
('Line6'             ,     2     ),             
('Line7'            ,      4      ),            
('Line8'            ,      1     ),             
('Line9'           ,       1     )



--

Open in new window


--

;WITH CTE As
(
	SELECT * , ROW_NUMBER() OVER (ORDER BY (SELECT 1)) rnk FROM Five 
)
,CTE1 AS
(
	SELECT Description , DescValue a1 , DescValue , rnk , 1 Lvl FROM CTE WHERE rnk = 1
	UNION ALL
	SELECT c1.Description , c1.DescValue a1, 
	CASE WHEN c1.DescValue + c.DescValue > 5 THEN c1.DescValue ELSE c1.DescValue + c.DescValue END DescValue , c1.rnk 	
	,CASE WHEN c1.DescValue + c.DescValue > 5 THEN Lvl + 1 ELSE Lvl END Lvl
	FROM CTE c1 INNER JOIN CTE1 c ON c.rnk + 1 = c1.rnk
)
SELECT Description , DescValue , lvl Page FROM CTE1
OPTION (MAXRECURSION 0)

GO



--

Open in new window


Output
------------------
Description      DescValue      Page
Line1               1                       1
Line2               3                       1
Line3               4                       1
Line4               2                       2
Line5               5                       2
Line6               2                       3
Line7               4                       4
Line8               5                       4
Line9               1                       5



Hope it helps !!
0
 

Author Closing Comment

by:ScubeduFan
ID: 41885385
Works great! Thank You!!!
0

Featured Post

PRTG Network Monitor: Intuitive Network Monitoring

Network Monitoring is essential to ensure that computer systems and network devices are running. Use PRTG to monitor LANs, servers, websites, applications and devices, bandwidth, virtual environments, remote systems, IoT, and many more. PRTG is easy to set up & use.

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

Suggested Solutions

This article explains how to reset the password of the sa account on a Microsoft SQL Server.  The steps in this article work in SQL 2005, 2008, 2008 R2, 2012, 2014 and 2016.
The Delta outage: 650 cancelled flights, more than 1200 delayed flights, thousands of frustrated customers, tens of millions of dollars in damages – plus untold reputational damage to one of the world’s most trusted airlines. All due to a catastroph…
Via a live example, show how to backup a database, simulate a failure backup the tail of the database transaction log and perform the restore.
Viewers will learn how to use the SELECT statement in SQL to return specific rows and columns, with various degrees of sorting and limits in place.

828 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question