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

Sql Server

Sql Server Problem Overview


UPDATE CattleProds
SET SheepTherapy=(ROUND((RAND()* 10000),0))
WHERE SheepTherapy IS NULL

If I then do a SELECT I see that my random number is identical in every row. Any ideas how to generate unique random numbers?

Sql Server Solutions


Solution 1 - Sql Server

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 - Sql Server

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  

Solution 3 - Sql Server

While I do love using CHECKSUM, I feel that a better way to go is using NEWID(), just because you don't have to go through a complicated math to generate simple numbers .

ROUND( 1000 *RAND(convert(varbinary, newid())), 0)

You can replace the 1000 with whichever number you want to set as the limit, and you can always use a plus sign to create a range, let's say you want a random number between 100 and 200, you can do something like :

100 + ROUND( 100 *RAND(convert(varbinary, newid())), 0)

Putting it together in your query :

UPDATE CattleProds 
SET SheepTherapy= ROUND( 1000 *RAND(convert(varbinary, newid())), 0)
WHERE SheepTherapy IS NULL

Solution 4 - Sql Server

I tested 2 set based randomization methods against RAND() by generating 100,000,000 rows with each. To level the field the output is a float between 0-1 to mimic RAND(). Most of the code is testing infrastructure so I summarize the algorithms here:

-- Try #1 used
(CAST(CRYPT_GEN_RANDOM(8) AS BIGINT)%500000000000000000+500000000000000000.0)/1000000000000000000 AS Val
-- Try #2 used
RAND(Checksum(NewId()))
-- and to have a baseline to compare output with I used
RAND() -- this required executing 100000000 separate insert statements

Using CRYPT_GEN_RANDOM was clearly the most random since there is only a .000000001% chance of seeing even 1 duplicate when plucking 10^8 numbers FROM a set of 10^18 numbers. IOW we should not have seen any duplicates and this had none! This set took 44 seconds to generate on my laptop.

Cnt     Pct
-----   ----
 1      100.000000  --No duplicates

SQL Server Execution Times: CPU time = 134795 ms, elapsed time = 39274 ms.

IF OBJECT_ID('tempdb..#T0') IS NOT NULL DROP TABLE #T0;
GO
WITH L0   AS (SELECT c FROM (VALUES (1),(1),(1),(1),(1),(1),(1),(1),(1),(1),(1),(1),(1),(1),(1),(1)) AS D(c))  -- 2^4  
    ,L1   AS (SELECT 1 AS c FROM L0 AS A CROSS JOIN L0 AS B)    -- 2^8  
    ,L2   AS (SELECT 1 AS c FROM L1 AS A CROSS JOIN L1 AS B)    -- 2^16  
    ,L3   AS (SELECT 1 AS c FROM L2 AS A CROSS JOIN L2 AS B)    -- 2^32  
SELECT TOP 100000000 (CAST(CRYPT_GEN_RANDOM(8) AS BIGINT)%500000000000000000+500000000000000000.0)/1000000000000000000 AS Val
  INTO #T0
  FROM L3;

 WITH x AS (
     SELECT Val,COUNT(*) Cnt
      FROM #T0
     GROUP BY Val
)
SELECT x.Cnt,COUNT(*)/(SELECT COUNT(*)/100 FROM #T0) Pct
  FROM X
 GROUP BY x.Cnt;

At almost 15 orders of magnitude less random this method was not quite twice as fast, taking only 23 seconds to generate 100M numbers.

Cnt  Pct
---- ----
1    95.450254    -- only 95% unique is absolutely horrible
2    02.222167    -- If this line were the only problem I'd say DON'T USE THIS!
3    00.034582
4    00.000409    -- 409 numbers appeared 4 times
5    00.000006    -- 6 numbers actually appeared 5 times 

SQL Server Execution Times: CPU time = 77156 ms, elapsed time = 24613 ms.

IF OBJECT_ID('tempdb..#T1') IS NOT NULL DROP TABLE #T1;
GO
WITH L0   AS (SELECT c FROM (VALUES (1),(1),(1),(1),(1),(1),(1),(1),(1),(1),(1),(1),(1),(1),(1),(1)) AS D(c))  -- 2^4  
    ,L1   AS (SELECT 1 AS c FROM L0 AS A CROSS JOIN L0 AS B)    -- 2^8  
    ,L2   AS (SELECT 1 AS c FROM L1 AS A CROSS JOIN L1 AS B)    -- 2^16  
    ,L3   AS (SELECT 1 AS c FROM L2 AS A CROSS JOIN L2 AS B)    -- 2^32  
SELECT TOP 100000000 RAND(Checksum(NewId())) AS Val
  INTO #T1
  FROM L3;

WITH x AS (
    SELECT Val,COUNT(*) Cnt
     FROM #T1
    GROUP BY Val
)
SELECT x.Cnt,COUNT(*)*1.0/(SELECT COUNT(*)/100 FROM #T1) Pct
  FROM X
 GROUP BY x.Cnt;

RAND() alone is useless for set-based generation so generating the baseline for comparing randomness took over 6 hours and had to be restarted several times to finally get the right number of output rows. It also seems that the randomness leaves a lot to be desired although it's better than using checksum(newid()) to reseed each row.

Cnt  Pct
---- ----
1    99.768020
2    00.115840
3    00.000100  -- at least there were comparitively few values returned 3 times

Because of the restarts, execution time could not be captured.

IF OBJECT_ID('tempdb..#T2') IS NOT NULL DROP TABLE #T2;
GO
CREATE TABLE #T2 (Val FLOAT);
GO
SET NOCOUNT ON;
GO
INSERT INTO #T2(Val) VALUES(RAND());
GO 100000000

WITH x AS (
    SELECT Val,COUNT(*) Cnt
     FROM #T2
    GROUP BY Val
)
SELECT x.Cnt,COUNT(*)*1.0/(SELECT COUNT(*)/100 FROM #T2) Pct
  FROM X
 GROUP BY x.Cnt;

Solution 5 - Sql Server

require_once('db/connect.php');

//rand(1000000 , 9999999);

$products_query = "SELECT id FROM products";
$products_result = mysqli_query($conn, $products_query);
$products_row = mysqli_fetch_array($products_result);
$ids_array = [];

do
{
	array_push($ids_array, $products_row['id']);
}
while($products_row = mysqli_fetch_array($products_result));

/*
echo '<pre>';
print_r($ids_array);
echo '</pre>';
*/
$row_counter = count($ids_array);

for ($i=0; $i < $row_counter; $i++)
{ 
	$current_row = $ids_array[$i];
	$rand = rand(1000000 , 9999999);
	mysqli_query($conn , "UPDATE products SET code='$rand' WHERE id='$current_row'");
}

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Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionNibblyPigView Question on Stackoverflow
Solution 1 - Sql ServerAndomarView Answer on Stackoverflow
Solution 2 - Sql ServerMartin SmithView Answer on Stackoverflow
Solution 3 - Sql ServerSegev -CJ- ShmueliView Answer on Stackoverflow
Solution 4 - Sql ServerbielawskiView Answer on Stackoverflow
Solution 5 - Sql ServerVaso NadiradzeView Answer on Stackoverflow