Quickly Generate A Random Number Between 1 and 9By Strictly-Software
As you should know by now the RAND() function in SQL has the downside that it will return the same value if the same seed is provided to it.
E.G if I call this code multiple times I get the same number for each result:
SELECT RAND(1) AS [RandomNumber]
This always returns 0.713591993212924 for me.
Therefore when people use RAND() they often end up coming up with various ways to generate a unique seed value as well.
One of the quickest ways I have found to return a random number is by using the NEWID() function which generates a random GUID.
If I wanted to return a random record from a recordset I would use the NEWID() function in my ORDER BY statement.
I use this a lot in my Jobboard code when I want to return a random banner from a banner group like so:
SELECT TOP 1 BannerPK, BannerURL, RedirectURL FROM BANNERS ORDER BY NEWID()
However if I wanted to return a single number between 1 and 9 I could use the following code.
This basically converts the GUID to a VARBINARY value before converting it to a VARCHAR and then taking off the the first character with a LEFT statement.
I have found that this gives a larger proportion of 1's than any other number but it is useful if you want a quick way in SQL to generate a single number (or more if you extend the numbers returned with a string function) e.g LEFT(val,2) or use a MID or RIGHT. Then this code could do the trick.
DECLARE @RANDNO INT SELECT @RANDOMNO = CAST(LEFT(CAST(ABS(CAST(CAST(NEWID() AS VARBINARY) AS INT)) AS VARCHAR),1) AS INT)