serial number generator program

hello,
I need some help pls: i want to create a program that will generate a serial number composed of 8 carattercts (3 letters followed by 5 numbers ---> i.e. FRT78136). Unfortunately i don't know how to use visualbasic very well; the last program i learnt was gwbasic...using gwbasic i would have used the randomize function. The program should change the rnd function in accordance to a sspecific yesr (i.e. when i am prompted for the 'randomize seed number' this should be a specific year such as 2010).

thank you for your kind help
Debora

Search the Code Bank section on the forum for "serial generator", and check out Planet Source Code.

Real quick sample

Code:

---

Option Explicit

Private Sub Command1_Click()
    MsgBox GenerateSerialNumber
End Sub

Private Function GenerateSerialNumber() As String
Dim i As Integer
Dim strChar(7) As String

    Randomize
   
    For i = 0 To 7
        If i < 3 Then
            strChar(i) = Chr(Int(Rnd() * 26) + 65)
        Else
            strChar(i) = Int(Rnd() * 10)
        End If
    Next i
   
    GenerateSerialNumber = Join(strChar, "")
End Function

---

Not sure if this improves the speed of MarkT's code, but it does seem to knock out 4 looping iterations:

Code:

---

Option Explicit

Private Sub Command1_Click()
    MsgBox GenerateSerialNumber
End Sub

Private Function GenerateSerialNumber() As String
Dim strChar(3) As String, i As Integer
Randomize
For i = 0 To 3
    If i < 3 Then
        strChar(i) = Chr(Int(Rnd() * 26) + 65)
    Else
        strChar(i) = Format$(Int(Rnd() * 100000), "00000")
    End If
Next i
GenerateSerialNumber = Join(strChar, "")
End Function

---

Quote:

---

Originally Posted by Code Doc

Not sure if this improves the speed of MarkT's code, but it does seem to knock out 4 looping iterations:

---

I thought about doing it that way but decided to do a random digit as each spot. If you just do one random number for the last five digit you will end up with a lot of low numbers that will have a number of zeros padding them.

Quote:

---

Originally Posted by MarkT

I thought about doing it that way but decided to do a random digit as each spot. If you just do one random number for the last five digits you will end up with a lot of low numbers that will have a number of zeros padding them.

---

Hmmm... I generated 100,000 random 5-digit numbers this way several times and the average is always very near 50,000 and sometimes over 50,000. To me, that indicates no small number bias whatsoever. Give it a try:

Code:

---

Private Sub Command1_Click()
Dim Sum As Single, Num As Long
Num = 100000
For i = 1 To Num
    Call GenerateSerialNumber
    Sum = Sum + CSng(Right$(GenerateSerialNumber, 5))
Next
MsgBox Format$(Sum / Num, "##,###.##")
End Sub

Private Function GenerateSerialNumber() As String
Dim strChar(3) As String
For i = 0 To 3
    If i < 3 Then
        strChar(i) = Chr(Int(Rnd() * 26) + 65)
    Else
        strChar(i) = Format$(Int(Rnd() * 100000), "00000")
    End If
Next i
GenerateSerialNumber = Join(strChar, "")
End Function

Private Sub Form_Load()
Randomize
End Sub

---

If a distribution of random 5-digit numbers were biased toward small numbers (and thus not random), would not the mean of the distribution have to be significantly less than the theoretical mean of 50,000?

Well I'd probably use something more like:

Code:

---

Option Explicit

Private Type GUID
    Data1_Lo As Integer
    Data1_Hi As Integer
    Data2 As Integer
    Data3 As Integer
    Data4_0To3 As Long
    Data4_4To7 As Long
End Type

Private Declare Function CoCreateGuid Lib "OLE32.DLL" (pGuid As GUID) As Long

Private Function GenSerial() As String
    Dim udtGUID As GUID
    Dim intA As Integer
   
    If CoCreateGuid(udtGUID) Then
        GenSerial = "*error*"
    Else
        intA = Asc("A")
        With udtGUID
            GenSerial = _
                    Chr$(intA + (.Data1_Lo And &H7FFF) Mod 26) _
                  & Chr$(intA + (.Data2 And &H7FFF) Mod 26) _
                  & Chr$(intA + (.Data3 And &H7FFF) Mod 26) _
                  & Format$((.Data4_4To7 And &H7FFFFFFF) Mod 100000, "00000")
        End With
    End If
End Function

Private Sub cmdGenerate_Click()
    txtSerial.Text = GenSerial()
End Sub

---

So, what is the difference between :

Format$((.Data4_4To7 And &H7FFFFFFF) Mod 100000, "00000")

and

Format$(Int(Rnd() * 100000), "00000")

I've tried both and the mean of the results of thousands of repetitions looks to be the same --> 50,000.

No extra variable or loop.

Code:

---

Option Explicit

Private Sub Form_Load()
    Randomize
End Sub

Private Sub cmdGetSN_Click()
    txtSN.Text = RndSN()
End Sub

Function RndSN() As String
    '-- no guaranty SN is unique
    RndSN = Chr$(Int(Rnd * 26) + 65) & Chr$(Int(Rnd * 26) + 65) & _
            Chr$(Int(Rnd * 26) + 65) & Format(Int(Rnd * 100000), "00000")
End Function

---

Quote:

---

Originally Posted by anhn

No extra variable or loop.

Code:

---

Option Explicit

Private Sub Form_Load()
    Randomize
End Sub

Private Sub cmdGetSN_Click()
    txtSN.Text = RndSN()
End Sub

Function RndSN() As String
    '-- no guaranty SN is unique
    RndSN = Chr$(Int(Rnd * 26) + 65) & Chr$(Int(Rnd * 26) + 65) & _
            Chr$(Int(Rnd * 26) + 65) & Format(Int(Rnd * 100000), "00000")
End Function

---

---

I considered that, but I figured everybody would yell at me for "slow" string concatenation. I thought MarkT's Join() of the simple array might be faster. Maybe Merri can test them.

Good one, AnHn. :thumb:

Why would speed even be an issue here? Nobody is going to need thousands of these at once.


I completely missed the "year" requirement. I'm not sure where that comes in.

Speed is not an issue here but for CodeDoc:
My function RndSN() is much simpler and about 30% faster than your function GenerateSerialNumber() in Post#6.

Edit: If you want a faster function (but longer code) then this one is 3 times faster:

Code:

---

Function RndSN2() As String
    '-- no guaranty SN is unique
    Dim n As Long
    RndSN2 = "00000000"
    n = Int(Rnd * 100000)
    Mid$(RndSN2, 8 - Len(n)) = n
    Mid$(RndSN2, 1, 1) = Chr$(65 + Int(Rnd * 26))
    Mid$(RndSN2, 2, 1) = Chr$(65 + Int(Rnd * 26))
    Mid$(RndSN2, 3, 1) = Chr$(65 + Int(Rnd * 26))
End Function

---

Quote:

---

Originally Posted by anhn

Speed is not an issue here but for CodeDoc:
My function RndSN() is much simpler and about 30% faster than your function GenerateSerialNumber() in Post#6.

Edit: If you want a faster function (but longer code) then this one is 3 times faster:

Code:

---

Function RndSN2() As String
    '-- no guaranty SN is unique
    Dim n As Long
    RndSN2 = "00000000"
    n = Int(Rnd * 100000)
    Mid$(RndSN2, 8 - Len(n)) = n
    Mid$(RndSN2, 1, 1) = Chr$(65 + Int(Rnd * 26))
    Mid$(RndSN2, 2, 1) = Chr$(65 + Int(Rnd * 26))
    Mid$(RndSN2, 3, 1) = Chr$(65 + Int(Rnd * 26))
End Function

---

---

Truly exceptional. Mid$() replace blows both simple string concatenation and Join() away. Not sure exactly why but it does. :thumb:

Dilettante said, "I completely missed the "year" requirement. I'm not sure where that comes in."
--------------
Neither do I. I'm not sure what that means to the problem. Perhaps it has something to do wth seeding Randomize, but the whole concept seemed nebulous. So, I did not address it either.

Maybe a new repeating series each year?

To me it just looks like the OP was using the year as the seed for the Rnd or Randomize seed.