# Auto-increment without the need of a table.

I came across a situation where I need unique numbers for both a large and small set of data. The standard approach is to create a table with an identity column and do your inserts. For a table with tens of thousands of records, this might be the best approach, but there is another one: ROW_NUMBER().

The key to using the ROW_NUMBER() function is to provide a constant so you will get a contiguous sequence.  Let's look at sample data.

Format
AB
AL
CD
CE
GG
LI

Let's look at the function:

ROW_NUMBER() OVER (Partition by <group> order by <columns>)

I can first construct the function as follows

select Format,
ROW_NUMBER() OVER(PARTITION BY Format ORDER BY FORMAT) as id

However this will give us the following results:

id Format
1   AB
1   AL
1   CD
1   CE
1   GG
1   LI

The reason is your PARTITION BY creates a unique number for each format. To solve this problem you need to put a value that will never change over the life of the select. I use 1.

select Format,
ROW_NUMBER() OVER(PARTITION BY 1 ORDER BY FORMAT) as id

The new result set is:

id Format
1   AB
2   AL
3   CD
4   CE
5   GG
6   LI

# Grouping data by 15 minute intervals

After a rather long course change (I was doing little work with SQL Server and work with other RDBMS systems), I'm back working on Microsoft SQL Server and a few other Microsoft products. Here is an interesting problem presented to me a few days ago.

A colleague of mine need timed data grouped by 15 minute intervals. I found this to be an interesting problem and didn't see much online, so thought I'd post this quick tidbit for others to find. The technique can be used to group by any interval with a little tweaking.

The field in question is a datetime field. The code needed to group data in 15 min intervals is this:

datepart(mi,C001_TimeEnd)/15)*15

Let's break it down. I take the minutes of the datetime field and divide it by 15, which is my interval. That will give me a number between 0 and 3, because it is an integer/integer. I then multiple that by 15. The distribution is as follows:

• (00-14)/15 = 0*15=0
• (15-29)/15= 1*15 = 15
• (30-44)/15=2*15=30
• (45-59)=3*15=45

Datepart(mi,"2018-5-9 11:48:33")/15) = 48/15 = 3*15 = 45. Now add in the the date and hours.

left(convert(varchar,"2018-5-9 11:48:33",120),14)+RIGHT('00' + CONVERT(VARCHAR(2), (datepart(mi,"2018-5-9 11:48:33")/15)*15), 2) = "2018-5-9 11:45:00"

If you wanted to do a 10 min interval, this: datepart(mi,C001_TimeEnd)/15)*15 becomes this datepart(mi,C001_TimeEnd)/10)*10.

# SSWUG Expo..I'm up!

I am proud to say that one of the webcasts I taped last year for SSWUG is part of the Free Virtual Expo.  My SSIS Enhanced Threading Framework is being shown.  Click here to register.