Comments on: SQL SERVER – Random Number Generator Script – SQL Query

By: amit shukla

amit shukla — Mon, 30 Jan 2012 11:15:23 +0000

Hi,
Pinal Dave

i had a problem i want to create 6 digit non repeating random number like 123456 if 122345 (wrong digits)
how can i code it

By: Jia

Jia — Wed, 07 Dec 2011 19:19:35 +0000

The chance of getting random integer number is not equal by using ROUND function. I guess FLOOR or CEILING is better choice -

SELECT @Random = FLOOR(@Lower + RAND() * (@Upper – @Lower + 1))

By: Vadivel Mohanakrishnan

Vadivel Mohanakrishnan — Sun, 06 Nov 2011 13:02:43 +0000

Check out the way to generate random data into a table as a whole –
http://vadivel.blogspot.com/2011/10/generate-test-random-data-for-table-in.html

By: Alex Munoz

Alex Munoz — Fri, 28 Oct 2011 00:53:45 +0000

An easy change to Method 1 to make it more accurate is to replace the ROUND line as follows:

SELECT @Random = FLOOR(((@Upper – @Lower + .9999999999) * RAND() + @Lower))

The entire method would then read:

DECLARE @Random INT
DECLARE @Upper INT
DECLARE @Lower INT

SET @Lower = 0 —- The lowest random number
SET @Upper = 14 —- The highest random number
SELECT @Random = FLOOR(((@Upper – @Lower + .9999999999) * RAND() + @Lower))
SELECT @Random

By: Rajesh

Rajesh — Fri, 14 Oct 2011 10:13:49 +0000

Hi,

I have a problem on random number and great some one can help.

We have a table which has 100 records and i want to generate random number between 40 and 70 for each record. So i need unique number generated for each record and it should be between 40 and 70. I dont mind if we get float values as well.

If anyone can answer it would be of immense help.

Regards
Rajesh

By: will

will — Thu, 13 Oct 2011 18:21:23 +0000

DECLARE @Random INT, @Upper INT, @Lower INT, @Count INT

set @Count = 0

DECLARE @1 int , @2 int, @3 int, @4 int, @5 int, @6 int

SELECT @1 = 0, @2 = 0, @3 = 0, @4 = 0, @5 = 0, @6 = 0

SET @Lower = 1 —- The lowest random number
SET @Upper = 6 —- The highest random number

WHILE @Count < 1000
BEGIN

SELECT @Random = ROUND(((@Upper – @Lower -1) * RAND() + @Lower), 0)

IF @RANDOM = 1 SET @1 = @1 + 1

IF @RANDOM = 2 SET @2 = @2 + 1

IF @RANDOM = 3 SET @3 = @3 + 1

IF @RANDOM = 4 SET @4 = @4 + 1

IF @RANDOM = 5 SET @5 = @5 + 1

IF @RANDOM = 6 SET @6 = @6 + 1

SET @Count = @Count + 1
END

SELECT @1 , @2 , @3 , @4 , @5 , @6

By: alwaysinmotion

alwaysinmotion — Wed, 05 Oct 2011 12:28:00 +0000

Hi Pinal,

Thank you so much for your blog. I am so often googling and finding the answers at your site. Please keep up the great work. Jean

By: madhivanan

madhivanan — Thu, 15 Sep 2011 07:56:09 +0000

It just generates the numbers between 0 and 1. It is similar to the one available in C language

By: madhivanan

madhivanan — Thu, 15 Sep 2011 07:46:17 +0000

insert into table(decimal_col)
select cast(rand() as decimal(8,6))

By: madhivanan

madhivanan — Thu, 15 Sep 2011 07:44:42 +0000

This code will do what you needed

select cast(rand() as decimal(8,6))

By: madhivanan

madhivanan — Thu, 15 Sep 2011 07:38:03 +0000

Yes it will work. Did you try?

By: Peacemaker

Peacemaker — Wed, 14 Sep 2011 15:50:47 +0000

Thanks Madhivanan:
are you sure this working?
Consider table name accounts
and columns name account_id

By: madhivanan

madhivanan — Wed, 14 Sep 2011 14:39:17 +0000

insert into table(col)
select * from
(
select cast(cast(newid() as varbinary(2)) as int) random_no
) as t where not exists (select * from table where col=random_no )

By: Peacemaker

Peacemaker — Wed, 14 Sep 2011 14:10:28 +0000

Hi,

I am trying to write a sql for DB2 table where I am using the following:

- Inserting into an existing table
- Using a random number which already does not exist in this table.
- Can’t use stored procedure.

Any assistance will be great.

By: DANI

DANI — Tue, 31 May 2011 15:33:09 +0000

what should i insert in A to get non repeated random number in Seat_No column

SqlCommand com = new SqlCommand(“UPDATE Table1

SET Seat_No =’” + ” A “+ “‘

Where Applicant_NIC=’” + ” ” + TextBox3.Text + ” ” + “‘”, conn);

By: Muhammad Adeel

Muhammad Adeel — Sat, 05 Feb 2011 20:10:44 +0000

Thanks Man, your post helped me so much

By: Jason Long

Jason Long — Tue, 25 Jan 2011 15:26:07 +0000

Actually you are correct. The 1st one is flawed as it’s written. It will in fact cut off the @Upper value. Also it fails to provide an even distribution of occurrences with the 1st and last numbers within the range.

The following fixes both issues.

DECLARE @Rand1 INT;
DECLARE @Random INT;
DECLARE @Upper INT;
DECLARE @Lower INT

SET @Lower = 5 —- The lowest random number
SET @Upper = 55 —- The highest random number
SET @Rand1 = ROUND((((@Upper + 1) – @Lower) * RAND() + @Lower), 0)
SET @Random = CASE WHEN @Rand1 = @Upper + 1 THEN @Lower ELSE @Rand1 END

Cheers,

Jason

By: Jason Long

Jason Long — Tue, 25 Jan 2011 15:19:12 +0000

If you want each of the 100 records to have a unique number, just assigned in a random order, don’t use the Rand function.

Use the ROW_NUMBER() and NEWID() functions… Like this.

UPDATE t1 SET Random = t2.RN
FROM TableName AS t1
INNER JOIN (
SELECT RowID, ROW_NUMBER() OVER(ORDER BY NEWID())AS RN FROM TableName) AS t2 ON t1.RowID = t2.RowID

By: madhivanan

madhivanan — Thu, 04 Nov 2010 12:31:26 +0000

Check the uniqueness

select ABS(CHECKSUM(NEWID())+999999)%1000000

By: madhivanan

madhivanan — Thu, 04 Nov 2010 12:24:59 +0000

select CHECKSUM(NEWID())%1000

By: madhivanan

madhivanan — Thu, 04 Nov 2010 12:23:14 +0000

Do you have any unique/primary key in the table?

By: Hazael

Hazael — Wed, 03 Nov 2010 19:45:27 +0000

UPDATE seeker_profile SET career_level_id= CEIL(RANDOM()*7);

Random from 1 to 7

By: ala'a bahar

ala'a bahar — Mon, 18 Oct 2010 06:34:32 +0000

hello i need a procedure to insert a random value and select the this randomly

By: kenji

kenji — Sat, 18 Sep 2010 12:24:42 +0000

Best site ever!

By: Sumit Kumar

Sumit Kumar — Fri, 20 Aug 2010 09:24:06 +0000

use it
select top 50 ques from questionnaire order by newid()