How can I fill a column with random numbers in SQL? I get the same value in every row

Solution 1:

Instead of rand(), use newid(), which is recalculated for each row in the result. The usual way is to use the modulo of the checksum. Note that checksum(newid()) can produce -2,147,483,648 and cause integer overflow on abs(), so we need to use modulo on the checksum return value before converting it to absolute value.

UPDATE CattleProds
SET    SheepTherapy = abs(checksum(NewId()) % 10000)
WHERE  SheepTherapy IS NULL

This generates a random number between 0 and 9999.

Solution 2:

If you are on SQL Server 2008 you can also use

 CRYPT_GEN_RANDOM(2) % 10000

Which seems somewhat simpler (it is also evaluated once per row as newid is - shown below)

DECLARE @foo TABLE (col1 FLOAT)

INSERT INTO @foo SELECT 1 UNION SELECT 2

UPDATE @foo
SET col1 =  CRYPT_GEN_RANDOM(2) % 10000

SELECT *  FROM @foo

Returns (2 random probably different numbers)

col1
----------------------
9693
8573

Mulling the unexplained downvote the only legitimate reason I can think of is that because the random number generated is between 0-65535 which is not evenly divisible by 10,000 some numbers will be slightly over represented. A way around this would be to wrap it in a scalar UDF that throws away any number over 60,000 and calls itself recursively to get a replacement number.

CREATE FUNCTION dbo.RandomNumber()
RETURNS INT
AS
  BEGIN
      DECLARE @Result INT

      SET @Result = CRYPT_GEN_RANDOM(2)

      RETURN CASE
               WHEN @Result < 60000
                     OR @@NESTLEVEL = 32 THEN @Result % 10000
               ELSE dbo.RandomNumber()
             END
  END