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MS SQL Query to count records within 5 minute intervals

Posted on 2008-10-24
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Last Modified: 2011-10-19
I'm looking for a MS SQL query for SQL 2000 where I can count records within 5 minute intervals.  Once I have this data I'm going to use it to chart the numbers so I will want to write it to a new table and be able to go backward through my existing data.  Additionally I want to do this going forward, presumably with DTS.?

Thanks in advance.
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Question by:powercram
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19 Comments
 
LVL 39

Expert Comment

by:BrandonGalderisi
ID: 22800013
This will show you how you can round time down to 5 minute increments:




declare @DT datetime
set @dt=getdate()
select dateadd(n, (datediff(n, 0,@dt)/5)*5,0),@dt
set @dt=dateadd(n,1,getdate())
select dateadd(n, (datediff(n, 0,@dt)/5)*5,0),@dt
set @dt=dateadd(n,4,getdate())
select dateadd(n, (datediff(n, 0,@dt)/5)*5,0),@dt
set @dt=dateadd(n,7,getdate())
select dateadd(n, (datediff(n, 0,@dt)/5)*5,0),@dt

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LVL 32

Expert Comment

by:Daniel Wilson
ID: 22800014
Select  DateDiff(m, '1/1/2000', MyDatefield) / 5, count(DateDiff(m, '1/1/2000', MyDatefield) / 5)
from MyTable
Group by DateDiff(m, '1/1/2000', MyDatefield) / 5
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LVL 39

Expert Comment

by:BrandonGalderisi
ID: 22800029
DW: m is for month :)
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LVL 39

Expert Comment

by:BrandonGalderisi
ID: 22800032
mi or n is for minute
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LVL 32

Accepted Solution

by:
Daniel Wilson earned 250 total points
ID: 22800071
You're right ... corrected in the one I tested ... failed to correct in what I posted.  So ... if he prefers my solution ...

Select  DateDiff(n, '1/1/2000', MyDatefield) / 5, count(DateDiff(n, '1/1/2000', MyDatefield) / 5)
from MyTable
Group by DateDiff(n, '1/1/2000', MyDatefield) / 5

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LVL 6

Author Comment

by:powercram
ID: 22800105
DanielWilson,

Sorry for my ignorance but how would I name the columns so I can do an order by?
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LVL 6

Author Comment

by:powercram
ID: 22800121
I'm guessing the number in the first column is the time relative to 1900 or whatever?  How can I convert that back to the actual date/time?
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LVL 39

Expert Comment

by:BrandonGalderisi
ID: 22800122
The one that I have does that.  It gives you the time.
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LVL 39

Assisted Solution

by:BrandonGalderisi
BrandonGalderisi earned 250 total points
ID: 22800128
select dateadd(n, (datediff(n, 0,YourDateColumn)/5)*5,0), count(*)
from YourTable
group by dateadd(n, (datediff(n, 0,YourDateColumn)/5)*5,0)
order by dateadd(n, (datediff(n, 0,YourDateColumn)/5)*5,0)
0
 
LVL 6

Author Comment

by:powercram
ID: 22800151
Brandon,

This is all I get with yours.
Untitled.png
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LVL 39

Expert Comment

by:BrandonGalderisi
ID: 22800166
That was just an example of how it would round various times (the right column) down to the 5 minute increment.
0
 
LVL 6

Author Comment

by:powercram
ID: 22800198
This one is very close to what I'm after.

     select dateadd(n, (datediff(n, 0,YourDateColumn)/5)*5,0), count(*)
     from YourTable
     group by dateadd(n, (datediff(n, 0,YourDateColumn)/5)*5,0)
     order by dateadd(n, (datediff(n, 0,YourDateColumn)/5)*5,0)

It is only displaying the time periods where a record exists.  How can I get it to display the period even if it is zero (IE has no entry)?
0
 
LVL 39

Expert Comment

by:BrandonGalderisi
ID: 22800205
Are you on SQL 2000 or 2005?
0
 
LVL 6

Author Comment

by:powercram
ID: 22800217
SQL 2000.  I'm using SQL Server 2005 Management Studio.
0
 
LVL 39

Expert Comment

by:BrandonGalderisi
ID: 22800226
In order to do that, you need to have a time coordinate table.  We can fake that though if you have a numbers table.  If you have neither, we would have to generate numbers in the procedure and it would not be efficient.  Do you have a numbers table?

ie. a table that contains sequential numbers.
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LVL 6

Author Comment

by:powercram
ID: 22800257
I do have a numbers table.  Currently it only goes to 9000.
0
 
LVL 6

Author Comment

by:powercram
ID: 22800377
I think what we have will work for my purpose.  If you have time additional info on getting the 0 items would be nice but not necessary.

Thank you both for your (quick) help!
0
 
LVL 6

Author Closing Comment

by:powercram
ID: 31509821
Thank you both very much!
0
 
LVL 39

Expert Comment

by:BrandonGalderisi
ID: 22800589
You'll need more than 9000.  Basically you'll need n numbers where n is the number of 5 minute increments you will have from your earliest to your latest data point.

The basic principal is.
declare @minD datetime
@maxD datetime
select @mind = dateadd(n, (datediff(n, 0,min(YourDateColumn))/5)*5,0)
, @maxd = dateadd(n, (datediff(n, 0,max(YourDateColumn))/5)*5,0)
 
--To see the approximate numbers you need as of NOW, not counting the 288 increments you will need per day.
 
select datediff(d,@mind,@max)*288
 
 
 
--Your list of ALL dates would then be
 
select dateadd(n, (n-1)*5,@minD)
from YourNumbersTable
where dateadd(n, (n-1)*5,@minD) <= @MaxD
 
--and incorporated into your above query.  I'm using inc to refer to an increment (of 5).
 
select allinc.theInc,isnull(cnt,0) as Count
(select dateadd(n, (n-1)*5,@minD) theInc
from YourNumbersTable
where dateadd(n, (n-1)*5,@minD) <= @MaxD) as AllInc
left outer join 
(select dateadd(n, (datediff(n, 0,YourDateColumn)/5)*5,0) theinc, count(*) cnt
from YourTable
group by dateadd(n, (datediff(n, 0,YourDateColumn)/5)*5,0)
order by dateadd(n, (datediff(n, 0,YourDateColumn)/5)*5,0)) ThisInc
on allinc.theinc = thisinc.theinc

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