SQL SERVER – Random Number Generator Script – SQL Query

Random Number Generator

There are many methods to generate random number in SQL Server.

Method 1 : Generate Random Numbers (Int) between Rang
---- Create the variables for the random number generation

---- This will create a random number between 1 and 999
SET @Lower = 1 ---- The lowest random number
SET @Upper = 999 ---- The highest random number
SELECT @Random = ROUND(((@Upper - @Lower -1) * RAND() + @Lower), 0)
SELECT @Random

Method 2 : Generate Random Float Numbers
+ (
DATEPART(ss, GETDATE()) * 1000 )

Method 3 : Random Numbers Quick Scripts

---- random float from 0 up to 20 - [0, 20)
-- random float from 10 up to 30 - [10, 30)
SELECT 10 + (30-10)*RAND()
--random integer BETWEEN 0
AND 20 - [0, 20]
----random integer BETWEEN 10
AND 30 - [10, 30]
SELECT 10 + CONVERT(INT, (30-10+1)*RAND())

Method 4 : Random Numbers (Float, Int) Tables Based with Time

DECLARE @t TABLE( randnum float )
DECLARE @cnt INT; SET @cnt = 0
WHILE @cnt <=10000
@cnt = @cnt + 1
+ (
DATEPART(ss, GETDATE()) * 1000 )
randnum, COUNT(*)
GROUP BY randnum

Method 5 : Random number on a per row basis

---- The distribution is pretty good however there are the occasional peaks.
---- If you want to change the range of values just change the 1000 to the maximum value you want.
---- Use this as the source of a report server report and chart the results to see the distribution
SELECT randomNumber, COUNT(1) countOfRandomNumber
SELECT ABS(CAST(NEWID() AS binary(6)) %1000) + 1 randomNumber
FROM sysobjects) sample
GROUP BY randomNumber
ORDER BY randomNumber

Watch a 60 second video on this subject

Reference : Pinal Dave (http://blog.SQLAuthority.com), Simon Sabin (http://sqlblogcasts.com)

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107 thoughts on “SQL SERVER – Random Number Generator Script – SQL Query

  1. Good info, it helped me understand sql rand function better.
    However I don’t seem to find an answer to what I’m trying to do.
    I have a table that looks like this

    Var Number
    ff 0
    zap 0
    zf 0
    and so on…
    What I want to do is assign a random int positive number for each Number between a certain range.
    Is that possible?

  2. Hi!

    Do you know how to create a random


    XX are character rand()
    ### number [0 to 9] rand()

    I need to produce or generate at least the first 100 to 1000 count() and more later



  3. Hi,
    If I need to check the possibilities of numbers generated within given range. if I wanna generate alphanumeric 4 charater values like A2B8,A3A2,..etc. How I find out the number of possibilities



  4. Great article. I am a noob to sql and know it works but not quite sure. In regards to this bit.
    — This will create a random number between 1000 and 9999
    SET @Lower = 1000 — The lowest random number
    SET @Upper = 9999 — The highest random number
    SELECT @Random = Round(((@Upper – @Lower -1) * Rand() + @Lower), 0)
    SELECT @Random

    I am having a hard time understanding how this works. If I replace the upper with 9999 and the lower with 1000 I get (8998) * Rand() + 1000), 0
    This is the section I am not understanding how it would generate a 4 digit random number.

  5. Hi,
    I want to select top random generated 3 records from top 30 records order by some column name with out using temporary table or variable.Is it possible?
    Thanks in advance.

  6. In the random number generator between the range the higher end number, 999. Will never be generated. I tried this by lowering the scope to 1 as low 3 as hi and never got 3 i would get lots of 1’s and 2’s but no 3. So if you want to get the top number you need to add 1 to it to include it within the range.

    • 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

  7. Hi Pinal,

    I am working as a MSSQL as well as aOracle DBA. I have been looking for health check scripts on SQL Server for quite some time. Please advice where I can get good health check scripts

  8. I would like to find very least missed number from the table. I have a empID field in Employee table. The data would be like this..

    From above data, the least missed number is: 1

    So my query should return a value : 1

    How to do this.
    Hope you understand my question..

  9. Further to Michael’s comments, also be careful when using method 1 if you require equal probability, as for ranges with a difference of 3 or more the first and last numbers in the range are only c. half as likely to be generated as the ‘middle’ numbers, e.g. a sample of 10000 nos from 1 to 9 (ie @lower=1, @upper = 10):
    val: Occurrence (sample):
    1 610
    2 1296
    3 1299
    4 1275
    5 1244
    6 1225
    7 1212
    8 1235
    9 604

  10. It can be difficult to get a random integer within a specified range, with a good quality probability distribution.
    The following gives a pretty good distribution, plus is flexible when it comes to specifying the range.
    SET @UpperLimit = 1
    SET @LowerLimit = -1
    SET @Rand =
    SELECT @Rand
    Both the upper and lower limit values are returned. Negative ranges can be specified, but the @UpperLimit must always be greater than the @LowerLimit.
    It can easily be made into a UDF or SP.

  11. hi there…. thanks for the sharing, anyway I have this problem regarding to my objective in simulation (markov chain monte carlo). because it seems that every time I “run” the query, the random number seems to be differ than the latest one. Its giving me a hard time in analyzing the system that I simulate.
    Is there any possibility to randomize fixed number…..
    for example….

    1st run (range 1′ cut – 9′ cut)

    2nd run (range 1′ cut – 9′ cut)

    3rd run (range 2′ cut – 10′ cut)

    please reply as soon as possible.
    thanks for your information and sharing.

  12. To Sundar; a quick, dirty way of doing what you want would be something like…

    SELECT MIN(empID) + 1 AS lowestMissedValue
    FROM Employee
    WHERE ((SELECT TOP 1 (empID + 2)
    FROM Employee E
    WHERE E.empID = Employee.empID) Employee.empID))

    It’ll only find values that are missing above the minimum, though, so if your lowest value is 2000, then the first value it could return is 2001.

  13. Damn, it cut out part of the SQL. Basically, you’re retrieving the minimum value plus one, for values where the same minimum value plus TWO is less than or equal to the next highest value, meaning that there’s a gap of at least two numbers between that value and the one following it. You should be able to fill in the SQL gaps from that.

  14. Hi, I have a data table , in which i want only selected (4) records when i press next button the next set of selected (4) records should be displayed like paging option in grid view .. is there any query to perform this operation.. so that loading the entire table in to grid can be avoided… please give your comments… thanks in advance

  15. I have a stored procedure that selects a number of rows dependant on various input values –

    for example, say the query selects 12 different rows, how can I then rendomly select one of the twelve and drop the other 11?



  16. hi
    I want to select a four different random number from the four different column of the table and put in in the another column of the table. can any one help me to solve this.

  17. Pingback: SQLAuthority News - Best Articles on SQLAuthority.com Journey to SQL Authority with Pinal Dave

  18. The crucial trick is to add 0.5 (!)
    In this way we get no bias against returning the upper and the lower limits

    CREATE PROCEDURE GetRandomInteger
    @Min int,
    @Max int,
    @rv int OUTPUT
    DECLARE @Rand float

    SELECT @Rand = ((@Max – @Min + 1) * RAND(CONVERT(int, CONVERT(varbinary, NEWID())))) + .5 + (@Min – 1)

    SELECT @rv = ROUND(@Rand, 0)


  19. I have 400K records on my table. Now I wanted you pull out random records that falls every 5th records from top to bottom. How can I do that sql?

    I tried using select stmt (below) but I had message error on LIMIT function.

    Select stmt:
    select * from table_name order by RAND() limit 5

    error msg:
    Msg 102, Level 15, State 1, Line 1
    Incorrect syntax near ‘limit’.

  20. Hi – loved this article. Got an interesting twist, though. Regarding ‘Method 5 : Random number on a per row basis’ – if I do this:

    SELECT ABS(CAST(NEWID() AS binary(6)) %3)

    I will always get either a 0, 1, or 2 – perfect!

    However, what I am ultimately trying to do is to assign a mapped value to a field depending on the value of the random number generated above, like this:

    UPDATE MyTable
    SET MyField =
    CASE ABS(CAST(NEWID() AS binary(6)) % 3)
    WHEN 0 THEN ‘red’
    WHEN 1 THEN ‘blue’
    WHEN 2 THEN ‘green’
    WHERE KeyField = ‘somevalue’

    Apparently the CASE statement is getting confused at times, and although in most cases, the field will be be updated with ‘red,’ ‘blue,’ or ‘green,’ at least 30% of the time, the CASE statement returns NULL! I don’t know why this would possibly be happening, since using ABS(CAST(NEWID() AS binary(6)) % 3) without a CASE statement NEVER returns NULL!

    To test this yourself, first trying executing this several times:

    SELECT ABS(CAST(NEWID() AS binary(6)) % 3)

    You’ll see that you never get a NULL returned.

    Next, try executing this several times:

    SELECT CASE ABS(CAST(NEWID() AS binary(6)) % 3)
    WHEN 0 THEN ‘zero’
    WHEN 1 THEN ‘one’
    WHEN 2 THEN ‘two’

    You’ll many times that you get a NULL returned.


  21. Great lesson but how can I display all of these in one record?

    DECLARE @Random_Sales INT;
    DECLARE @Upper_Sales INT;
    DECLARE @Lower_Sales INT;
    DECLARE @Random_Profit INT;
    DECLARE @Upper_Profit INT;
    DECLARE @Lower_Profit INT;
    DECLARE @Random_Percent INT;
    DECLARE @Upper_Percent INT;
    DECLARE @Lower_Percent INT
    SET @Lower_Sales = 500000
    SET @Upper_Sales = 599999
    SET @Lower_Profit = 60000
    SET @Upper_Profit = 99999
    SET @Lower_Percent = 15
    SET @Upper_Percent = 20
    SELECT @Random_Sales = ROUND(((@Upper_Sales - @Lower_Sales -1) * RAND() + @Lower_Sales), 0)
    SELECT @Random_Sales
    SELECT @Random_Sales = ROUND(((@Upper_Sales - @Lower_Sales -1) * RAND() + @Lower_Sales), 0)
    SELECT @Random_Sales
    SELECT @Random_Profit = ROUND(((@Upper_Profit - @Lower_Profit -1) * RAND() + @Lower_Profit), 0)
    SELECT @Random_Profit
    SELECT @Random_Percent = ROUND(((@Upper_Percent - @Lower_Percent -1) * RAND() + @Lower_Percent), 0)
    SELECT @Random_Percent
  22. Hi

    I have table with 100 records, how can I update a field on this with random numbers from 1 thru 100, when I do this using RANd() I do get duplicate random numers in the file which does not help me. Any help?

    • 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 (

  23. Hello.

    I think the first script, is flawed.
    This one:
    —- Create the variables for the random number generation
    DECLARE @Random INT;
    DECLARE @Upper INT;
    DECLARE @Lower INT

    —- This will create a random number between 1 and 999
    SET @Lower = 1 —- The lowest random number
    SET @Upper = 999 —- The highest random number
    SELECT @Random = ROUND(((@Upper – @Lower -1) * RAND() + @Lower), 0)
    SELECT @Random

    If you set a lower limit of 1, and upper limit 2, it will always return one.

    If you remove the ‘-1′ from here: ROUND(((@Upper – @Lower -1) * RAND() + @Lower), 0) it returns 1 and 2.

    Maybe I am mistaken and this is a particular case.

    Thanks for the great site, I found many useful scripts here. :)

    Alex Luca

    • 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



  24. Great article. What I need is a way to product int char char int. Example: 0AA0 – Position 1 – a number between 1-9, Position 2 – a letter between A-Z, Position 3 – a letter between A-Z, Position 4 – a number between 1-9.

    Very new to SQL, T-SQL – so be kind. :-)

  25. You can solve random string generation using a mask and substitution. There are many ways to solve this issue, but if we break them down into 3 parts, it will make our solution more flexible.

    Part 1 – The randomizer view, this is nothing new, just generating a random float in a view so it can be used in a function.
    CREATE VIEW [dbo].[Randomizer] AS
    SELECT abs((convert(bigint,convert(binary(8),newid()))*0.000000000000001)%1.0) AS n

    Part 2 – A Random function that takes min and max parameters so that we can generate a random value within desired bounds.

    CREATE FUNCTION dbo.Rnd( @min bigint, @max bigint ) RETURNS bigint AS

    RETURN convert(bigint,
    (@max – @min + 1)
    (SELECT TOP 1 n FROM dbo.Randomizer)

    Part 3 – A generic mask substitution function so that we can ask specify any mask we want. I will only implement the very most basic set of substitutions. You can extend these in simple ways.
    CREATE FUNCTION dbo.RandomText( @mask varchar(64) ) RETURNS varchar(64) AS
    — Goal: Substatute mask tokens with random values
    — Inputs: @mask – a string with subtatution tokens and literals
    — Tokens: # – will be replaced with a random digit [0-9]
    — A – will be replaced with a random upper case english letter [A-Z]

    @mask = Stuff( @mask, start, 1, replacement )
    FROM (
    n AS start,
    CASE SubString( @mask, n, 1 )
    WHEN ‘#’ THEN char( dbo.Rnd(48,57) ) — [0-9]
    WHEN ‘A’ THEN char( dbo.Rnd(65,90) ) — [A-Z]
    END AS replacement
    FROM (
    SELECT (1+n1.n+n10.n) AS n
    (1+n1.n+n10.n) BETWEEN 1 AND Len(@mask)
    ) sequence
    SubString( @mask, n, 1 ) IN (‘#’,’A’)
    ) substatutions

    RETURN @mask


  26. Additionally you can replace the use of the Char() function with a SubString.

    SubString(‘0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ’, dbo.Rnd(1,36), 1)

    For a random digit or letter and that list could obviously be extended to include any characters you wish.

  27. Hi Dave,

    Is there a possibility to find a null present in table as there are many columns in my table is there any other possible way to find the null in my columns in my table.

    Eg. SLNO, Name, DOB,…….n’ column i have i can use isnull but i have to mention each and every column name in the isnull function i guess its pain taking process. so i need to know is there any possible way through which i can find that null values is present in my table. using a query.

    Awaiting for you reply.

    Cheers …

  28. hi Pinal,

    My qusetion is like sometimes there is an requirement to genearete the random integers numbers from 1 to 10 like.
    If i m creating the online exam application and want to generate the 5 random numbers between 1 to 10 so that only thoese question numbers can be displayed at a time.

    I have seen such cases where the random numbers my be also duplicate like 1 and again one and the processs is alos recurssive so it hangs the application and fails or stack over flow problem.

    i do not have the code snipet now but can u give som e solution for this problem with code.


  29. Hi i am having 1 column location ids ( int )

    My task is to get top 2 random rows from tat column

    I have query like this

    select top 2 locationid from profile order by newid()

    this one is good for 500K only

    But i am having 20 millions rows tables in that it takes 4 sec

    its creating a hell in tunning …
    Can any one help me out in this pls Thank u

  30. Hello Mohan,

    Using the ORDER BY clause is the reason of high execution time of your query. Instead of ORDER BY, use TABLESAMPLE option to get random rows.

    Pinal Dave

  31. Dave,

    I need you help in creating a random number between 0 and 0.999999
    but the methods you havce defined above are too complex for me to understand as i am new to sql.
    I want to create this random number fields also for new varibales with a tabel so that i can keep creating new random numbers when needed.
    please help me

    Thanks pete

    • Here’s one way to do this if you ever need multiple random numbers generated between 0 and 1.

      DECLARE @Counter int

      CREATE TABLE ##Random (Number decimal(5,2)) — or decimal(8,6))

      SET @Counter = 1

      WHILE @Counter < 11 –change the number 11 to how ever many random numbers you want generated

      INSERT INTO ##Random (Number)
      SELECT CAST(RAND() as decimal(5,2)) as Number — or decimal(8,6))

      SET @Counter = @Counter + 1

      SELECT *
      FROM ##Random

      –DROP TABLE ##Random

  32. Dave i forgot to mention that once the number is created it requires to be stored in a decimla 8,6 feild permanently

    thanks once again


  33. hi
    can someone help me i have 70 questionnaire records in my table and i want to show 50 records in random using stored procedure in ms sql server 2000..

  34. Hi.
    I want to generate 6 digits random number for employee ID’s in my company, so the ganerated numbers must be unique in table.

    any suggestion..


  35. 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);

  36. 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.

  37. 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

  38. 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

    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

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

  39. 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.


  40. 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))

  41. 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

  42. Hello, what would be the best way to pull a random value from a list of values. I just need to select one value from a set list I have?

  43. Please help me for below qery , I m checking series between two number
    This is my sql qry.

    where ( FromCOINumber@FromCOINumber))
    IF @count > 0

    SET @msg=@msg+’ From Limit belong to some another series ‘;

    print @msg

  44. as per my experience randomizing number result in duplicate such as number “23” repeating after creation . please elaborate
    E.g: on execurte 23 another execute 23 …

    Please Help!

  45. Pingback: SQL SERVER – Retrieving Random Rows from Table Using NEWID() « SQL Server Journey with SQL Authority

  46. Create table COrder
    AutoID bigint identity(1,1),
    RndID bigint default cast (Rand() * 3000 as Bigint) Unique,
    name varchar(10),
    CreatedOn Datetime default getdate()

    If I insert a record dynamically [using asp.net] in this table and previous random number is generated then what would happen…………how to solve this that after repeating rndid it regenerate script… help me………

  47. Pingback: SQL SERVER – Generate Random Values – SQL in Sixty Seconds #042 – Video « SQL Server Journey with SQL Authority

  48. i want to generate table
    create script
    alter script
    insert script
    update script
    by using command?
    can you please tell me how it is possible by using sql commands

  49. Hi All, Can I generate multiple Random numbers, because when i am writing [select rand()] it gives me one random number. and if i want 10 random numbers then what should i write.

  50. Pingback: SQL SERVER – Weekly Series – Memory Lane – #027 | SQL Server Journey with SQL Authority

  51. Hello, I have the same question:

    How to auto generate number between 10000 to 99999 without repeat

    I’m trying to modify your example:

    SELECT randomNumber , COUNT(1) countOfRandomNumber
    FROM (
    SELECT ABS(Checksum(NEWID()) %1000000) + 1 randomNumber
    FROM sysobjects) sample
    GROUP BY randomNumber
    ORDER BY randomNumber

    but I wonder how to set interval beetwen two numbers

  52. Ok,
    I have interval between 100000 and 999999

    SELECT randomNumber, COUNT(1) countOfRandomNumber
    FROM (
    SELECT ABS(CAST(NEWID() AS binary(6)) %900000) + 100000 randomNumber
    FROM sysobjects) sample
    GROUP BY randomNumber
    ORDER BY randomNumber


    SELECT randomNumber, COUNT(1) countOfRandomNumber, COUNT(randomNumber)
    FROM (
    select ABS(CHECKSUM(NEWID()) %900000) + 100000 randomNumber
    FROM sysobjects) sample
    GROUP BY randomNumber
    ORDER BY randomNumber

    but now I think about second problem. I want 500 random number but query gives me only about 90.

  53. Hello,

    I have a car table. I want to set the carQuestionId for EACH car record to be either 1, 2, or 3. I cannot figure out how to apply your @Random code to get this to work for carQuestionId.

    Pseudo Code:
    update car set carQuestionId = (1, 2 or 3)
    where carStatus = ‘g’ and carType = ‘Honda’

    Please help.

  54. Hello,

    I want to solve this problem can any one help ?

    -How to generate 1000 records randomly in Database?

    Details Question:

    User will insert 2000 as input and in database one field(Input) will be there.
    In this field 1000 record should be generated randomly between 0.50 to 3.00 i.e. records can be any number between 0.50 to 3.00 and importantly the SUM of these 1000 records must be 2000.

  55. Pinal et al,

    I have a slightly more complicated Rand question. I have two tables: one that stores promotion records and a second that is used to calculate a random daily winner from the first table. Their structures are:

    table1 (
    rec_id (PK, int, not null),
    promo_id (int, not null),
    date (date, not null),
    user_id (nvarchar(10), not null)

    table2 (
    id (PK, int, not null),
    PromoID (int, not null),
    rec_id (int, not null)

    So obviously looking at this, table one has a PK for the record id that is incremented by 1 on each insert. I might have any number of promotions running simultaneously and need to create a random winner (table1.rec_id) on a daily basis for each promotion (table1.promo_id) and insert the results into table2.

    I tried:
    INSERT INTO prT_Daily_Winner (PromoID, rec_id)
    (abs(checksum(NewId())) % (SELECT COUNT(*) FROM prT_Promo_Records b WHERE b.promo_id=a.promo_id AND CONVERT(varchar,[date],101)=CONVERT(varchar,GETDATE(), 101)) +1) as RandNum
    dbo.prT_Promo_Records a
    CONVERT(varchar,[date],101)=CONVERT(varchar,GETDATE(), 101)
    GROUP BY promo_id

    The problem I run into is that setting the upper and lower based on my rec_id doesn’t work since I may get results where the Random number doesn’t correlate to the specific promotion going on.

    Any assistance would be appreciated. Thank you!

  56. Dave, Thanks so much for providing this. Using your Method 1, I specified upper and lower bounds to be 1 and 16. I ran your formula 1000 times in a loop and did not get 16 returned once! I really need to use your formula in a game I am writing, but need 16 returned sometimes! How can I accomplish this with your Method 1? Thanks, Bruce Wilson

    DECLARE @Random INT;
    DECLARE @Upper INT;
    DECLARE @Lower INT;
    DECLARE @intLoop INT;

    SET @Lower = 1 —- The lowest random number
    SET @Upper = 16 —- The highest random number
    SET @intLoop = 1
    WHILE @intLoop <= 1000
    —- This will create a random number between 1 and 999
    SET @Random = ROUND(((@Upper – @Lower -1) * RAND() + @Lower), 0)

    IF @Random = 16
    PRINT 'Returned 16'

    SET @intLoop = @intLoop + 1

  57. Yes, -1 in first method is redundant, it should be removed, because now it will never return the upper value. It is enough just to remove it:

    Method 1 : Generate Random Numbers (Int) between Rang
    —- Create the variables for the random number generation
    DECLARE @Random INT;
    DECLARE @Upper INT;
    DECLARE @Lower INT

    —- This will create a random number between 1 and 999
    SET @Lower = 1 —- The lowest random number
    SET @Upper = 999 —- The highest random number
    SELECT @Random = ROUND(((@Upper – @Lower) * RAND() + @Lower), 0)
    SELECT @Random

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