Comments on: Random number generation http://itknowledgeexchange.techtarget.com/itanswers/random-number-generation/ Sat, 28 Nov 2009 10:51:15 +0000 http://wordpress.org/?v=2.6.2 By: Mrdenny http://itknowledgeexchange.techtarget.com/itanswers/random-number-generation/#comment-53288 Mrdenny Tue, 20 May 2008 00:30:33 +0000 #comment-53288 I'm curious, what's the purpose behind this code? Check out my SQL Server blog "[A href="http://itknowledgeexchange.techtarget.com/sql-server/"]SQL Server with Mr Denny[/A]" for more SQL Server information. I’m curious, what’s the purpose behind this code?

Check out my SQL Server blog “SQL Server with Mr Denny” for more SQL Server information.

]]> By: Willmage http://itknowledgeexchange.techtarget.com/itanswers/random-number-generation/#comment-53282 Willmage Mon, 19 May 2008 17:29:58 +0000 #comment-53282 The above answer will give you a random letter/number output, but it doesn't take into account the original number the user started with. Here is an answer I have for this question: Dim Letters, OrigNum Dim TextNum, ID, TempHold Dim RandNum, FinalAns Letters = "abcdefghijklmnopqrstuvwxyz" OrigNum = 123456 Randomize Rem ** the following randomizes the original number RandNum = "" TextNum = Trim(Str(OrigNum)) For i = 1 To 5 ID = Int((Len(TextNum) * Rnd) + 1) RandNum = RandNum & Mid(TextNum, ID, 1) TempHold = Left(TextNum, ID - 1) TempHold = TempHold & Mid(TextNum, ID + 1, Len(TextNum) - ID) TextNum = TempHold Next RandNum = RandNum & TextNum Rem ** the following creates your new random character/number string FinalAns = "" For i = 1 To 10 If Len(FinalAns) + Len(RandNum) = 10 Then FinalAns = FinalAns & Left(RandNum, 1) RandNum = Right(RandNum, Len(RandNum) - 1) Else If Len(RandNum) = 0 Then FinalAns = FinalAns & Mid(Letters, Int(26 * Rnd) + 1, 1) Else If ((Int(Rnd * 2) + 1) Mod 2) <> 0 Then FinalAns = FinalAns & Left(RandNum, 1) RandNum = Right(RandNum, Len(RandNum) - 1) Else FinalAns = FinalAns & Mid(Letters, Int(26 * Rnd) + 1, 1) End If End If End If Next Hope that helps! The above answer will give you a random letter/number output, but it doesn’t take into account the original number the user started with. Here is an answer I have for this question:

Dim Letters, OrigNum
Dim TextNum, ID, TempHold
Dim RandNum, FinalAns

Letters = “abcdefghijklmnopqrstuvwxyz”
OrigNum = 123456

Randomize
Rem ** the following randomizes the original number
RandNum = “”
TextNum = Trim(Str(OrigNum))
For i = 1 To 5
ID = Int((Len(TextNum) * Rnd) + 1)
RandNum = RandNum & Mid(TextNum, ID, 1)
TempHold = Left(TextNum, ID - 1)
TempHold = TempHold & Mid(TextNum, ID + 1, Len(TextNum) - ID)
TextNum = TempHold
Next
RandNum = RandNum & TextNum
Rem ** the following creates your new random character/number string
FinalAns = “”
For i = 1 To 10
If Len(FinalAns) + Len(RandNum) = 10 Then
FinalAns = FinalAns & Left(RandNum, 1)
RandNum = Right(RandNum, Len(RandNum) - 1)
Else
If Len(RandNum) = 0 Then
FinalAns = FinalAns & Mid(Letters, Int(26 * Rnd) + 1, 1)
Else
If ((Int(Rnd * 2) + 1) Mod 2) <> 0 Then
FinalAns = FinalAns & Left(RandNum, 1)
RandNum = Right(RandNum, Len(RandNum) - 1)
Else
FinalAns = FinalAns & Mid(Letters, Int(26 * Rnd) + 1, 1)
End If
End If
End If
Next

Hope that helps!

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