by GAUTAM (1 Submission)
Category: String Manipulation
Compatability: VB 6.0
Difficulty: Unknown Difficulty
Originally Published: Fri 11th January 2002
Date Added: Mon 8th February 2021
Rating: 
(1 Votes)
THIS PROGRAM DETERMINES THE TYPE OF CREDIT CARD ie: MASTER,VISA,AMERICAN,DISCOVER,etc ON THE BASIS OF THE CREDIT CARD NUMBER WHICH HAS TO BE
API Declarations
'DRAW A COMMAND BUTTON ON Form1
VBcoders
'*CARD TYPES *PREFIX *WIDTH
'American Express 34, 37 15
'Diners Club 300 to 305, 36 14
'Carte Blanche 38 14
'Discover 6011 16
'EnRoute 2014, 2149 15
'JCB 3 16
'JCB 2131, 1800 15
'Master Card 51 to 55 16
'Visa 4 13, 16
'Just in case nothing is found
CreditCardType = "Unknown Credit Card"
'Remove all spaces and dashes from the passed string
CardNo = Replace(Replace(CardNo, " ", ""), "-", "")
'Check that the minimum length of the string isn't less
'than fourteen characters and -is- numeric
If Len(CardNo) < 14 Or Not IsNumeric(CardNo) Then Exit Function
'Check the first two digits first
Select Case CInt(Left(CardNo, 2))
Case 34, 37
CreditCardType = "American Express"
Case 36
CreditCardType = "Diners Club"
Case 38
CreditCardType = "Carte Blanche"
Case 51 To 55
CreditCardType = "This is a Master Credit Card Number."
Case Else
'None of the above - so check the
'first four digits collectively
Select Case CInt(Left(CardNo, 4))
Case 2014, 2149
CreditCardType = "EnRoute"
Case 2131, 1800
CreditCardType = "JCB"
Case 6011
CreditCardType = "Discover"
Case Else
'None of the above - so check the
'first three digits collectively
Select Case CInt(Left(CardNo, 3))
Case 300 To 305
CreditCardType = "American Diners Club"
Case Else
'None of the above -
'so simply check the first digit
Select Case CInt(Left(CardNo, 1))
Case 3
CreditCardType = "JCB"
Case 4
CreditCardType = "This is a Visa Credit Card Number."
End Select
End Select
End Select
End Select
End Function
Private Sub Command1_Click()
MsgBox CreditCardType("5404607010803842") 'THE CREDIT CARD NUMBER HAS TO BE
PASSED AS AN ARGUMENT AS SHOWN ABOVE.
End Sub
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