by Chris Morton (5 Submissions)
Category: String Manipulation
Compatability: VB.NET
Difficulty: Unknown Difficulty
Originally Published: Sun 16th October 2005
Date Added: Mon 8th February 2021
Rating: 
(1 Votes)
Validate phone number, email address and south african ID number
VBcoders
Public Function ValidateEmail(ByVal Email As String) As Boolean
Dim i As Integer
Dim atplace As Integer
Dim dotplace As Integer
Dim atcount As Integer
Dim atcountboo As Boolean
Dim dotcount As Integer
For i = 1 To Len(Email)
If Mid(Email, i, 1) = "@" Then 'Checks where the location of the @ symbol is and if there is more than 1
atcount = atcount + 1
atplace = i
If atcount > 1 Then
Return False
End If
End If
If Mid(Email, i, 1) = "." Then
dotcount = dotcount + 1
dotplace = i
End If
If InStr("[email protected]_", LCase(Mid(Email, i, 1))) > 0 Then 'checks for illegal charaters
Else
Return False
End If
Next
If dotplace < atplace Or Len(Email) = dotplace Then 'checks to see whether the is a . after the @ and if something follows after the .
Return False
End If
If atcount <> 1 Then 'checks if there are invalid or no @ symbols
atcountboo = False
Else
atcountboo = True
End If
If Mid(Email, 1, 1) = "@" Or Mid(Email, Len(Email), 1) = "@" Then 'checks if the first or last charater is @
Return False
End If
If Mid(Email, atplace + 1, 1) = "." Then 'checks if the character after @ is a .
Return False
End If
If dotcount >= 1 And atcountboo = True Then 'checks if there is at least 1 . and an @ exists in the right place
Return True
Else
Return False
End If
End Function
Public Function ValidateIDNumber(ByVal IDNumber As String, ByVal DateofBirth As Date) As Boolean
Dim First6Numbers() As String
Dim i As Integer
Dim numbers(2) As Integer
Dim DOB As Integer
If IsNumeric(IDNumber) = True And Len(IDNumber) = 13 Then
Else
Return False
End If
First6Numbers = Split(DateofBirth.ToShortDateString, "/")
For i = 0 To First6Numbers.Length - 1
numbers(i) = First6Numbers(i)
Next
DOB = Mid(Format(numbers(0), "0000") & Format(numbers(1), "00") & Format(numbers(2), "00"), 3, 6)
If Mid(IDNumber, 1, 6) = DOB Then
Return True
Else
Return False
End If
End Function
Public Function ValidatePhoneNumber(ByVal PhoneNumber As String) As Boolean
Dim i As Integer
Dim pluscount As Integer
For i = 1 To Len(PhoneNumber)
If InStr("1234567890- +()", LCase(Mid(PhoneNumber, i, 1))) > 0 Then 'checks for illegal charaters
Else
Return False
End If
Next
For i = 1 To Len(PhoneNumber)
If Mid(PhoneNumber, i, 1) = "+" Then 'Checks if there is more than 1 "+"
pluscount += 1
If pluscount > 1 Then
Return False
End If
End If
Next
Return True
End Function
End Module
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