Generating Random numbers, but not repeating any, is there an easy way to do this?

[eprunner06]

How can I generate random numbers but not repeat any that have already been generated. Is there an easy way to do this at all?
Thanks!
Eric

[DiGiTaIErRoR]

Add the numbers you have generated randomly into a string, then use instr to check if that number has been generated, if so, randomly generate a new one.

10
x= int(rnd*255)
if instr(usednums,x) > 0 then
goto 10
else
usednums = x & "-" & usednums
end if

[YoungBuck]

If you know the range of numbers you want the random numbers to be generated between you can just add all those values to an array and shuffle them around like a deck of cards and then select a new random number by moving to next item in the array. For example...

Private Sub ShuffleArray(ByRef arr(), ByVal iShuffleTimes As Integer)
Dim iCt As Integer
Dim iRandElem As Integer
Dim vTemp
 
'assure we don't get the same set of random numbers
Randomize
 
For iCt = 1 To iShuffleTimes
    
    iRandElem = (Rnd * (UBound(arr) - LBound(arr))) + LBound(arr) ' get an element within the bounds of the array randomly
    vTemp = arr(LBound(arr)) ' assign the value of the first element in the array to the temporary variable
    arr(LBound(arr)) = arr(iRandElem) ' shuffle the values
    arr(iRandElem) = vTemp
    
Next iCt
 
End Sub
 
 
Private Sub Form_Load()
 Dim arr()
 Dim i As Integer
 Dim str As String
 
 ReDim arr(1 To 1000)
 
 ' load the values 1-1000 into an array
 For i = 1 To 1000
    arr(i) = i
 Next i
 
 ' shuffle there positions in the array around a bit
 ShuffleArray arr, 3000
 
 ' get the first 10 numbers, guaranteed not to be repeating
 For i = 1 To 10
    str = str & arr(i) & vbCrLf
 Next i
 
 ' display the 10 random numbers
 MsgBox "Here are 10 random numbers " & vbCrLf & str
 
End Sub

[Resilience]

Add the numbers you have generated randomly into a string, then use instr to check if that number has been generated, if so, randomly generate a new one.

10
x= int(rnd*255)
if instr(usednums,x) > 0 then
goto 10
else
usednums = x & "-" & usednums
end if

i know this is old but could someone elaborate on this post. It seems to me if the variable "usednums" holds the value of the generated number and the instr() checks to see if it matches the number that was generated, wouldnt it always go back to "10" on the goto loop? thanks

[mebhas]

digital error created another error here.
say we want to randomize 20 numbers, 1-20.
wouldn't instr(usednums,x) point to the digit 1 that is in 10 thru 19 if x=1?

[dglienna]

In VB, a random number is between 0 and .9999, so you have to multiply it to get a range of numbers. If you check for the closest Integer, you can get a number between 0 and the highest number -1.

x=rnd*2

.9999 * 2 = 1.9998

if you say this, you get this

Int(.9999 * 2) = 1

and if you don't want 0, you have to add 1, so you use this:

Int(.9999 * 2) +1 = 2

[randem]

eprunner06,

Each time you generate a number add it to a collection. A collection will keep them unique.