It’s not often that we (as programmers) need random numbers, especially from a database. When you are testing your queries for performance, it’s best to use a large table to do it. Sometimes you just don’t have access to a large table to test with, so you may think about creating many rows in your existing table.

SQL Server has a rand() function that will return a random (fractional) number between 0 and 1.

For example:

Select Rand() -- 0.686350654426017

The problem with the rand function occurs when you use set based operations.

Select Rand() As RandomNumber From (Select 1 As NUM Union All Select 2 Union All Select 3) As Alias

RandomNumber ---------------------- 0.920057583532051 0.920057583532051 0.920057583532051

Not very random, is it? I mean, the number is random, but it’s also the same for each row. The purpose of this blog is to demonstrate how to get a random value in set based operations.

There is an interesting technique that you can use to accomplish this. There is a NewId() function in SQL Server that returns a GUID, for example: ’94344EE4-5D7A-45EB-9EBC-7A596B7F90F3′. NewId does work well for set based operations, for example:

Select Rand() As RandomNumber, NewId() As GUID From (Select 1 As NUM Union All Select 2 Union All Select 3) As Alias

RandomNumber GUID ---------------------- ------------------------------------ 0.130542187318667 D12D6274-CB93-4CDB-B848-9F0FA8B88614 0.130542187318667 C708217F-5204-4284-B76B-A55D87BEFBC0 0.130542187318667 B42D7973-B3E2-420F-8D4D-DC72442D0824

Notice that the ‘Random Number’ column is the same for each row, but the GUID column is different. We can use this interesting fact to generate random numbers by combining this with another function available in SQL Server. The CHECKSUM function will return an integer hash value based on its argument. In this case, we can pass in a GUID, and checksum will return an integer.

Select Rand() As RandomNumber, NewId() As GUID, Checksum(NewId()) As RandomInteger From (Select 1 As NUM Union All Select 2 Union All Select 3) As Alias

RandomNumber GUID RandomInteger ---------------------- ------------------------------------ -------------- 0.152440960483059 CDEB9F5D-E8A2-4FC7-A5A4-0E0A9CCF9786 84,364,212 0.152440960483059 9ABCEC8F-6C0C-4CEF-BA16-E8F0B50FB399 -1,317,220,961 0.152440960483059 6BD23217-F3FF-4F19-AA7F-C127CA7A4763 976,389,102

Before we continue, let’s take a look at the output, because there are some interesting observations that we need to consider before continuing. RandomInteger can be positive or negative because it’s limited to the range of an integer, so the values must fall between -2,147,483,648 and 2,147,483,647.

Most of the time, we want a random number within a certain range of numbers. In most languages, we simply multiply the result of the Rand() function to get this number. Since our RandomInteger is already a whole number, we really can’t do this. However, we could use the mod operator to guarantee a range of numbers. Mod is the remainder of a division operation, so if we mod a number by 10, we are guaranteed to get a number between -9 and +9. Unfortunately, this is a little misleading because there are 19 possible numbers we can get for this. So, to make sure we get a range to 10 numbers, we need to take the absolute value of the number, and then mod 10. Like this:

Select Rand() As RandomNumber, NewId() As GUID, Abs(Checksum(NewId())) % 10 As RandomInteger From (Select 1 As NUM Union All Select 2 Union All Select 3) As Alias

RandomNumber GUID RandomInteger ---------------------- ------------------------------------ ------------- 0.490769895131745 8237A3F3-98C6-4B31-8F4B-8A80ECC8FBFF 1 0.490769895131745 C231EB65-FA81-4536-B909-44FEC91953D9 9 0.490769895131745 F0AD28C5-8AE4-46A1-B8AC-D8C9F90A5747 4

Now, we are guaranteed to get a Random number between 0 and 9. Suppose you want to get a random number between 10 and 15. The range of numbers is 6 (10,11,12,13,14,15). The mod value for this needs to be 6, to get a number in the range of 0 to 5. Then, we add 10 to the result to get a number in the range of 10 to 15.

If you want to generate a random number between -5 and 5, donâ€™t be tempted to remove the absolute value function, because you will get numbers that appears to be random, but are not ‘as random’ as they should be.

In my database, I have a numbers table with 1,000,000 rows. When I run this code:

Select RandomNumber, Count(*) As NumberCount From ( Select Checksum(NewId()) % 6 As RandomNumber From Numbers ) As A Group By RandomNumber Order By RandomNumber

RandomNumber NumberCount ------------ ----------- -5 83539 -4 83443 -3 83372 -2 82931 -1 82950 0 166254 1 83207 2 83635 3 83741 4 83417 5 83511

Notice that 0 has double the number of occurrences. Instead, we should write it like this:

Select RandomNumber, Count(*) As NumberCount From ( Select Abs(Checksum(NewId())) % 11 - 5 As RandomNumber From Numbers ) As A Group By RandomNumber Order By RandomNumber

RandomNumber NumberCount ------------ ----------- -5 90889 -4 90794 -3 91365 -2 90476 -1 91104 0 90730 1 90895 2 90815 3 90762 4 91133 5 91037

Notice now that the values are ‘more’ random than the previous version (where 0 had double the number of occurrences). Since we are talking about random numbers, we wouldn’t expect there to be the same number of occurrences for each number, but with the previous version, there were double the number of zero’s, which isn’t truly random. This version has approximately the same number of zero’s as any other number, making it ‘more random’.

## 15 Comments

Interesting, why are you doing the % 11 – 5 part?

Because there are 11 possible numbers between -5 and 5. Since I am taking the absolute value, you get a range between 0 and 10, subtract 5 and your range becomes -5 to 5.

great tip.

very very helpful!

Great, this is exactly what i needed to server up questions in a random order. Thanks.

Great blog, thanks.

Using your technique I created a trigger on a table that I needed a random password for. The trigger is in SQL Server 2008. For this application I didn’t need a strong password so this solution worked very well.

Here is the code

CREATE TRIGGER [dbo].[trg_tblResponsesPW]

ON [dbo].[tblResponses]

AFTER INSERT

AS

BEGIN

DECLARE

@MyCursor CURSOR,

@intID int,

@PW varchar (8)

– SET NOCOUNT ON added to prevent extra result sets from

– interfering with SELECT statements.

SET NOCOUNT ON;

SET @MyCursor = CURSOR FAST_FORWARD

FOR

SELECT intID FROM inserted

– Loop through each record in the inserted table to get the unique ID field. In each loop update the table

– that the record was appended to with a new random password. I loop here so that when bulk inserts occur each new record get’s it’s own unique password

OPEN @MyCursor

FETCH NEXT FROM @MyCursor

– Populate the variables with the values from the SQL Statement

INTO @intID

WHILE @@FETCH_STATUS = 0

BEGIN

– The NewId function returns a GUID Like “D12D6274-CB93-4CDB-B848-9F0FA8B88614″. Here I pull the 6 characters

– from the right but I could pull from anywhere or pull any number of characters depending on how long I wanted

– my password to be.

SET @PW = Right(NewID(),6)

UPDATE tblResponses

SET txtPW = @PW

WHERE (intID = @intID)

– Now we are finished doing the work so move to the next inserted record

FETCH NEXT FROM @MyCursor

– and populate the next variable

INTO @intID

END

CLOSE @MyCursor

DEALLOCATE @MyCursor

END

The ABS(CHECKSUM(NewId())) is less random than is needed for some applications. There is an article that can be found online describing the differences in how “random” the different techniques are: Random Sampling in T-SQL

by Brian Connolly.

An alternative in terms of generating the numbers so that the same is not returned multiple times, is described well in an article “Returning Random Numbers in a SELECT statement” on Jeff’s SQL Server Blog, on the Sqlteam web site.

Thanks for posting this. In my case, I needed to fill a map table with random IDs from other tables. I was able to adapt your technique to achieve what I needed (identifiers munged):

DECLARE @count INT

SET @count = (SELECT COUNT(*) FROM [ProductType])

INSERT INTO [ProductTypeMap]

(

[ProductId],

[ProductTypeId],

)

SELECT

[p].[ProductId],

[t].[ProductTypeId]

FROM

(

SELECT

[ProductId],

(ABS(CHECKSUM(NEWID())) % @count + 1) AS [RowNumber]

FROM

[Product]

) AS [p]

LEFT JOIN

(

SELECT

[ProductTypeId],

(ROW_NUMBER() OVER (ORDER BY [ProductTypeId])) AS [RowNumber]

FROM

[ProductType]

) AS [p]

ON ([t].[RowNumber] = [p].[RowNumber])

This worked for me great. I just wanted to convert the integer returned by Checksum into a value from 0 to 1 so I could have a function similar to Rand() for which I came up with:

——–

create function my_rand(@in_id uniqueidentifier)

returns float as

begin

declare @x int = checksum(@in_id)

declare @v float = 2147483647

return (@x + @v + 1) / (2 * @v + 1)

end

——–

This returns 0 for the min value of an integer and 1 for the max, with everything else in between.

Then I use this as select my_rand(newid())…

This blog was awesome!!!

newid() cannot be called from a function. Please let me know if there is a way to get a random number generated in a function

How can i get without duplicate values in generating random numbers by using abs(checksum(newid()))%1000 + 5000

[…] a look at SQL Server – Set based random numbers which has a very detailed […]

NEWID() cannot be in a function. I wonder if there is a way to generate a random number using something like this from a function.

You can create a view that returns NewId, and then a function that returns from the view.

Create View vwGuid

AS

Select NewId() As NewId

GO

Create Function GetNewId()

Returns UniqueIdentifier

As

Begin

Return (Select NewId From vwGuid)

End

GO

Select dbo.GetNewId()