How do I check for a valid email address before sending?

Posted on 2010-09-01
Last Modified: 2012-06-21
Hello experts;

I have an Access database with code that sends out emails, the pointy end of the code looks like;

                        Set objOutlookRecip = objOutlookMsg.Recipients.Add(clientemail)
                        objOutlookRecip.Type = olTo

The problem is sometimes people have entered invalid email addresses, and when it gets to the send line, the code breaks.

How do I check BEFORE sending that the email address is valid according to Outlook?
Question by:OzoneFriendly
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LVL 53

Accepted Solution

Dhaest earned 125 total points
ID: 33575536
Validate email addresses using VB.NET

Validate email address with Regular Expressions

Check an Email Address for Validity (VB.NET and VB6)

Effective Email Address Validation
LVL 92

Assisted Solution

by:Patrick Matthews
Patrick Matthews earned 125 total points
ID: 33575561

One way would be to use Regular Expressions to see if the email address "looks" valid. You can add the function below to a regular VBA module; the function is fully described in my article

You can then use the RegExpFind function to validate the email address:

If RegExpFind(clientemail, "[\w-]+(\.[\w-]+)*@[\w-]+(\.[\w-]+)*\.[A-Za-z]{2,4}", 1) = "" Then
'code for invalid email address
End If

Function RegExpFind(LookIn As String, PatternStr As String, Optional Pos, _

    Optional MatchCase As Boolean = True, Optional ReturnType As Long = 0, _

    Optional MultiLine As Boolean = False)


    ' Function written by Patrick G. Matthews.  You may use and distribute this code freely,

    ' as long as you properly credit and attribute authorship and the URL of where you

    ' found the code


    ' For more info, please see:



    ' This function relies on the VBScript version of Regular Expressions, and thus some of

    ' the functionality available in Perl and/or .Net may not be available.  The full extent

    ' of what functionality will be available on any given computer is based on which version

    ' of the VBScript runtime is installed on that computer


    ' This function uses Regular Expressions to parse a string (LookIn), and return matches to a

    ' pattern (PatternStr).  Use Pos to indicate which match you want:

    ' Pos omitted               : function returns a zero-based array of all matches

    ' Pos = 1                   : the first match

    ' Pos = 2                   : the second match

    ' Pos = <positive integer>  : the Nth match

    ' Pos = 0                   : the last match

    ' Pos = -1                  : the last match

    ' Pos = -2                  : the 2nd to last match

    ' Pos = <negative integer>  : the Nth to last match

    ' If Pos is non-numeric, or if the absolute value of Pos is greater than the number of

    ' matches, the function returns an empty string.  If no match is found, the function returns

    ' an empty string.  (Earlier versions of this code used zero for the last match; this is

    ' retained for backward compatibility)


    ' If MatchCase is omitted or True (default for RegExp) then the Pattern must match case (and

    ' thus you may have to use [a-zA-Z] instead of just [a-z] or [A-Z]).


    ' ReturnType indicates what information you want to return:

    ' ReturnType = 0            : the matched values

    ' ReturnType = 1            : the starting character positions for the matched values

    ' ReturnType = 2            : the lengths of the matched values


    ' If you use this function in Excel, you can use range references for any of the arguments.

    ' If you use this in Excel and return the full array, make sure to set up the formula as an

    ' array formula.  If you need the array formula to go down a column, use TRANSPOSE()


    ' Note: RegExp counts the character positions for the Match.FirstIndex property as starting

    ' at zero.  Since VB6 and VBA has strings starting at position 1, I have added one to make

    ' the character positions conform to VBA/VB6 expectations


    ' Normally as an object variable I would set the RegX variable to Nothing; however, in cases

    ' where a large number of calls to this function are made, making RegX a static variable that

    ' preserves its state in between calls significantly improves performance


    Static RegX As Object

    Dim TheMatches As Object

    Dim Answer()

    Dim Counter As Long


    ' Evaluate Pos.  If it is there, it must be numeric and converted to Long


    If Not IsMissing(Pos) Then

        If Not IsNumeric(Pos) Then

            RegExpFind = ""

            Exit Function


            Pos = CLng(Pos)

        End If

    End If


    ' Evaluate ReturnType


    If ReturnType < 0 Or ReturnType > 2 Then

        RegExpFind = ""

        Exit Function

    End If


    ' Create instance of RegExp object if needed, and set properties


    If RegX Is Nothing Then Set RegX = CreateObject("VBScript.RegExp")

    With RegX

        .Pattern = PatternStr

        .Global = True

        .IgnoreCase = Not MatchCase

        .MultiLine = MultiLine

    End With


    ' Test to see if there are any matches


    If RegX.Test(LookIn) Then


        ' Run RegExp to get the matches, which are returned as a zero-based collection


        Set TheMatches = RegX.Execute(LookIn)


        ' Test to see if Pos is negative, which indicates the user wants the Nth to last

        ' match.  If it is, then based on the number of matches convert Pos to a positive

        ' number, or zero for the last match


        If Not IsMissing(Pos) Then

            If Pos < 0 Then

                If Pos = -1 Then

                    Pos = 0



                    ' If Abs(Pos) > number of matches, then the Nth to last match does not

                    ' exist.  Return a zero-length string


                    If Abs(Pos) <= TheMatches.Count Then

                        Pos = TheMatches.Count + Pos + 1


                        RegExpFind = ""

                        GoTo Cleanup

                    End If

                End If

            End If

        End If


        ' If Pos is missing, user wants array of all matches.  Build it and assign it as the

        ' function's return value


        If IsMissing(Pos) Then

            ReDim Answer(0 To TheMatches.Count - 1)

            For Counter = 0 To UBound(Answer)

                Select Case ReturnType

                    Case 0: Answer(Counter) = TheMatches(Counter)

                    Case 1: Answer(Counter) = TheMatches(Counter).FirstIndex + 1

                    Case 2: Answer(Counter) = TheMatches(Counter).Length

                End Select


            RegExpFind = Answer


        ' User wanted the Nth match (or last match, if Pos = 0).  Get the Nth value, if possible



            Select Case Pos

                Case 0                          ' Last match

                    Select Case ReturnType

                        Case 0: RegExpFind = TheMatches(TheMatches.Count - 1)

                        Case 1: RegExpFind = TheMatches(TheMatches.Count - 1).FirstIndex + 1

                        Case 2: RegExpFind = TheMatches(TheMatches.Count - 1).Length

                    End Select

                Case 1 To TheMatches.Count      ' Nth match

                    Select Case ReturnType

                        Case 0: RegExpFind = TheMatches(Pos - 1)

                        Case 1: RegExpFind = TheMatches(Pos - 1).FirstIndex + 1

                        Case 2: RegExpFind = TheMatches(Pos - 1).Length

                    End Select

                Case Else                       ' Invalid item number

                    RegExpFind = ""

            End Select

        End If


    ' If there are no matches, return empty string



        RegExpFind = ""

    End If



    ' Release object variables


    Set TheMatches = Nothing


End Function

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Assisted Solution

YohanF earned 125 total points
ID: 33575566
best thing is to, or a way you can use is regular expressions. I have attached a piece of code I am using, and I should warn you that its not Perfect in terms of checking every bit of validity. However it checks most of it.

Public Function validateEmailAddress(EmailAddress As String) As String
    Dim reEx As RegExp
    Dim reEx1 As RegExp
    Set reEx = New RegExp
    Set reEx1 = New RegExp
    reEx.Pattern = "\b[A-Z0-9._%+-]+@[A-Z0-9.-]+\.[A-Z]{2,6}\b"
    reEx.IgnoreCase = True
    reEx1.Pattern = "\b[\s]\b"
    reEx1.IgnoreCase = True
    If reEx.Test(EmailAddress) = True And reEx1.Test(EmailAddress) = False Then
            validateEmailAddress = "EMAILVERIFIED: Email address is valid"
        validateEmailAddress = "EMAILVALIDATE: Invalid Email Syntax, please correct the email format and try again" 
    End If
End Function

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Author Comment

ID: 33575583
Cool, I'll try and implement one of those solutions and get back to you. I was hoping there was a simple "outlook" command to do it, but this is cool. :-)

Out of interest, would any of these validate as okay the following string;;

I know some instances might be formatted like that, which Outlook WILL accept, but will that code?

Expert Comment

ID: 33575592
Mine wont, It will only process an email at a time.
LVL 92

Expert Comment

by:Patrick Matthews
ID: 33575600
If you add recipients one at a time, no, passing >1 address will not work.  However, this will:    objOutlookMsg.To = ";;"
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LVL 11

Assisted Solution

kbirecki earned 125 total points
ID: 33585190
You can use the Split() function to separate out each address btw the semicolons, process each address individually with one of the methods the others provided, and then concatenate them back together.

Expert Comment

ID: 33592398
Basically, you cannot check if an email address is valid, but you can check if it is proper in "form" e.g someone@domain.tld

1 either you connect to the email domain server and check
2. try sending the email, some servers will reject the email if the email is not balid (ie, does not exist)

To check for valid form, use regular expression
check here for example

Author Comment

ID: 33872191
I will check out these options and see which one will work in my situation soon. Thanks for the advice.

Author Comment

ID: 34708135
I really must get around to trying these solutions. I've been soooo busy. :-(
LVL 53

Expert Comment

ID: 34709276
>> I really must get around to trying these solutions

As far as I see, all possible solutions are provided.
LVL 142

Expert Comment

by:Guy Hengel [angelIII / a3]
ID: 34959143
This question has been classified as abandoned and is being closed as part of the Cleanup Program. See my comment at the end of the question for more details.

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