Here's a modified form of the script. It takes an argument @uniquechars. If set to 1, gradually shrinks the @string variable by removing the chosen character. The exception is if your @string is less than your @passlen, it won't work (so for large passwords, you have no choice but to use repeating characters).

-- This generates a random password, defaulting to 10 characters

create procedure msdb.dbo.GetPass

@passlen int = 10, @uniquechars int = 0, @charset int = 0 -- 2 is alphanumeric + special characters,

-- 1 is alphanumeric, 0 is alphabetical only

as

set nocount on

if (@passlen > 8000 or @passlen < 1) -- Let's not go crazy here

select @passlen = 10

declare @password varchar(8000), @string varchar(256), @numbers varchar(10), @extra varchar(50),

@stringlen int, @index int, @passval varchar(1)

-- no 1, l, I, 0, O which can cause confusion

select @string = 'ABCDEFGHJKLMNPQRSTUVWXYZabcdefghijkmnopqrstuvwxyz' -- same as @charset = 0

select @numbers = '23456789'

select @extra = '>_!@#$%&=?<' -- add more special characters if you want

if @charset = 2

select @string = @string + @numbers + @extra

else if @charset = 1

select @string = @string + @numbers

-- else assume @extra is 0 and @string is just letters. Feel free to modify these criteria as you see fit

select @stringlen = len(@string)

select @password = ''

-- This check is if your password length exceeds the number of unique characters. If so, the @uniquechars setting

-- you turned on (1) is turned back off.

if ((@stringlen < @passlen) and (@uniquechars = 1))

select @uniquechars = 0

while (@passlen > 0)

begin

-- For the random part here, use rand() or, preferably, newid()

select @index = (abs(checksum(newid())) % @stringlen) + 1

select @passval = substring(@string, @index, 1)

if @uniquechars = 1 -- modified section that removes repeating letters

select @string = replace(@string, @passval, '')

select @password = @password + @passval

select @passlen = @passlen - 1

end

select @password

go