Solved

Stored Procedure to return values from x to y with increment z

Posted on 2003-11-15
11
1,602 Views
Last Modified: 2008-03-06
This is simple (I think), just have not done anything like it yet

If a table has values x, y, and z

(where x is "low value", y is High value" and z is "increment")

in a row I am calling with variable "a", how do I return a set of rows that start with x, end with y, and move upwards with the increment z?

for example, x=2, y=10, z=2

the stored procedure returns

1- 2
2- 4
3- 6
4- 8
5- 10

thanks! It is urgent I learn this asap to forward my project.
0
Comment
Question by:JCMcNeil
  • 6
  • 4
11 Comments
 
LVL 6

Expert Comment

by:robertjbarker
ID: 9757708
Could you post the table structure, or at least the columns of interest, and indicate which column(s) contain the low, high, and increment?

I'm not getting the question.
0
 
LVL 22

Expert Comment

by:CJ_S
ID: 9758154
Not getting the question either. This isn't what you meant or is it?

declare @x int, @y int, @z int

set @x = 2
set @y = 10
set @z = 2

declare @current int
set @current = @x

declare @tmp table(id int identity(1,1) not null, [index] int)

while @current<=@y
begin
  insert into @tmp values(@current)
  set @current = @current + @z
end

select * from @tmp
0
 

Author Comment

by:JCMcNeil
ID: 9758185
CJ_S, it looks like you are getting the idea, but the sp you wrote didn't return any records, which I need.

Let me see if I can be more clear:

There is a table that contains "start" (x), "end" (y) and "increment" (z) values. A variable (a) calls the correct row.

upon selecting that row, new rows need to be returned that start with x and increase by increment z until they end up at y.

does that make more sense? Sorry I am not familiar enough with sql terminology to explain better.

0
 
LVL 22

Expert Comment

by:CJ_S
ID: 9758192
Works fine for me:

-- your parameter:
declare @a int


-- inner parameters
declare @x int, @y int, @z int

select @x = start, @y = end, @z = increment from YOURTABLE

declare @current int
set @current = @x

declare @tmp table(id int identity(1,1) not null, [index] int)

while @current<=@y
begin
 insert into @tmp values(@current)
 set @current = @current + @z
end

select * from @tmp
0
 
LVL 22

Expert Comment

by:CJ_S
ID: 9758196
Oh, one more change:
select @x = start, @y = end, @z = increment from YOURTABLE
should be:
select @x = start, @y = end, @z = increment from YOURTABLE where id=@a
0
Microsoft Certification Exam 74-409

Veeam® is happy to provide the Microsoft community with a study guide prepared by MVP and MCT, Orin Thomas. This guide will take you through each of the exam objectives, helping you to prepare for and pass the examination.

 

Author Comment

by:JCMcNeil
ID: 9758317
It all looks good but I keep getting "the stored procedure executed correctly but did not return any records"
0
 
LVL 22

Expert Comment

by:CJ_S
ID: 9758332
Here's a full sample which also creates a RES table with the values. This is a full example

drop table RES

create table RES
(
      id int identity(1,1) not null constraint RES_PK_id PRIMARY KEY,
      start int not null,
      [end] int not null,
      increment int not null
)

insert into RES values(4, 10, 2)


declare @a int
declare @x int, @y int, @z int

set @a = 1

select @x = start, @y = [end], @z = increment from RES

declare @current int
set @current = @x

declare @tmp table(id int identity(1,1) not null, [index] int)

while @current<=@y
begin
  insert into @tmp values(@current)
  set @current = @current + @z
end

select * from @tmp
0
 
LVL 22

Accepted Solution

by:
CJ_S earned 500 total points
ID: 9758336
oh, and it does work for me so if this does not return any results for you, you might want to check out the actual values in your table.
0
 

Author Comment

by:JCMcNeil
ID: 9758380
how strange, I copied the full example into a stored procedure and still got   "the stored procedure executed correctly but did not return any records"  

I am using sql server 2000 controlled through an ACCESS 2002 project and here is all the text from the SP:
 

ALTER PROCEDURE StoredProcedure19

AS
drop table RES

create table RES
(
id int identity(1,1) not null constraint RES_PK_id PRIMARY KEY,
start int not null,
[end] int not null,
increment int not null
)

insert into RES values(4, 10, 2)


declare @a int
declare @x int, @y int, @z int

set @a = 1

select @x = start, @y = [end], @z = increment from RES

declare @current int
set @current = @x

declare @tmp table(id int identity(1,1) not null, [index] int)

while @current<=@y
begin
insert into @tmp values(@current)
set @current = @current + @z
end

select * from @tmp





0
 

Author Comment

by:JCMcNeil
ID: 9758449
OK, it works when pulled into the datagrid, I just couldn't see it directly in the stored procedure- thanks!
0
 
LVL 22

Expert Comment

by:CJ_S
ID: 9758471
Glad that it's working!
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

Nowadays, some of developer are too much worried about data. Who is using data, who is updating it etc. etc. Because, data is more costlier in term of money and information. So security of data is focusing concern in days. Lets' understand the Au…
Let's review the features of new SQL Server 2012 (Denali CTP3). It listed as below: PERCENT_RANK(): PERCENT_RANK() function will returns the percentage value of rank of the values among its group. PERCENT_RANK() function value always in be…
Using examples as well as descriptions, and references to Books Online, show the documentation available for datatypes, explain the available data types and show how data can be passed into and out of variables.
Viewers will learn how to use the INSERT statement to insert data into their tables. It will also introduce the NULL statement, to show them what happens when no value is giving for any given column.

911 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

Need Help in Real-Time?

Connect with top rated Experts

16 Experts available now in Live!

Get 1:1 Help Now