Generating random strings with T-SQL

Question

If you wanted to generate a pseudorandom alphanumeric string using T-SQL, how would you do it? How would you exclude characters like dollar signs, dashes, and slashes from

Answer 1

So I liked a lot of the answers above, but I was looking for something that was a little more random in nature. I also wanted a way to explicitly call out excluded characters. Below is my solution using a view that calls the CRYPT_GEN_RANDOM to get a cryptographic random number. In my example, I only chose a random number that was 8 bytes. Please note, you can increase this size and also utilize the seed parameter of the function if you want. Here is the link to the documentation: https://docs.microsoft.com/en-us/sql/t-sql/functions/crypt-gen-random-transact-sql

CREATE VIEW [dbo].[VW_CRYPT_GEN_RANDOM_8]
AS
SELECT CRYPT_GEN_RANDOM(8) as [value];

The reason for creating the view is because CRYPT_GEN_RANDOM cannot be called directly from a function.

From there, I created a scalar function that accepts a length and a string parameter that can contain a comma delimited string of excluded characters.

CREATE FUNCTION [dbo].[fn_GenerateRandomString]
( 
    @length INT,
    @excludedCharacters VARCHAR(200) --Comma delimited string of excluded characters
)
RETURNS VARCHAR(Max)
BEGIN
    DECLARE @returnValue VARCHAR(Max) = ''
        , @asciiValue INT
        , @currentCharacter CHAR;

    --Optional concept, you can add default excluded characters
    SET @excludedCharacters = CONCAT(@excludedCharacters,',^,*,(,),-,_,=,+,[,{,],},\,|,;,:,'',",<,.,>,/,`,~');

    --Table of excluded characters
    DECLARE @excludedCharactersTable table([asciiValue] INT);

    --Insert comma
    INSERT INTO @excludedCharactersTable SELECT 44;

    --Stores the ascii value of the excluded characters in the table
    INSERT INTO @excludedCharactersTable
    SELECT ASCII(TRIM(value))
    FROM STRING_SPLIT(@excludedCharacters, ',')
    WHERE LEN(TRIM(value)) = 1;

    --Keep looping until the return string is filled
    WHILE(LEN(@returnValue) < @length)
    BEGIN
        --Get a truly random integer values from 33-126
        SET @asciiValue = (SELECT TOP 1 (ABS(CONVERT(INT, [value])) % 94) + 33 FROM [dbo].[VW_CRYPT_GEN_RANDOM_8]);

        --If the random integer value is not in the excluded characters table then append to the return string
        IF(NOT EXISTS(SELECT * 
                        FROM @excludedCharactersTable 
                        WHERE [asciiValue] = @asciiValue))
        BEGIN
            SET @returnValue = @returnValue + CHAR(@asciiValue);
        END
    END

    RETURN(@returnValue);
END

Below is an example of the how to call the function.

SELECT [dbo].[fn_GenerateRandomString](8,'!,@,#,$,%,&,?');

~Cheers

Answer 2

For one random letter, you can use:

select substring('ABCDEFGHIJKLMNOPQRSTUVWXYZ',
                 (abs(checksum(newid())) % 26)+1, 1)

An important difference between using newid() versus rand() is that if you return multiple rows, newid() is calculated separately for each row, while rand() is calculated once for the whole query.

Answer 3

I'm not expert in T-SQL, but the simpliest way I've already used it's like that:

select char((rand()*25 + 65))+char((rand()*25 + 65))

This generates two char (A-Z, in ascii 65-90).