Error attaching file to outlook

[rnm89]

I am having a problem when I try to attach a file to my outlook. If I direct a path, no problem. Below is where the error is. Thanks

Option Explicit

Dim FileToAttach As String
Dim ActualNumberOfOccurrences As Integer
Dim AnsParseStr As String
Dim position As Integer
Dim NumCharInString As Integer
Dim checkit As Boolean
Dim RetVal As Variant

Private Sub cmdBrowse_Click()

'***********************************************************************
'Give the file selection window a title.
    cdlOpen.DialogTitle = "Pick A File To Attach To Outlook Express"
    'The file selection window will start in the
    'applications directory.
    cdlOpen.InitDir = App.Path
        
    'Show all files
    cdlOpen.Filter = "All Files (*.*)|*.*"
                       
    'Open the file selection window.
    cdlOpen.ShowOpen
    
    FileToAttach = cdlOpen.FileName

    position = 1

Do Until position = 0

   position = InStr(position + 1, FileToAttach, "\")
   ActualNumberOfOccurrences = ActualNumberOfOccurrences + 1
  
Loop
  
'Parse the string to get the part after the last backslash \
 AnsParseStr = ParseStr(FileToAttach, "\", ActualNumberOfOccurrences, ActualNumberOfOccurrences)
  

 '****************************************************************
    'Check to see if a file has been selected.
    Select Case FileToAttach
    
       Case vbNullString
            'Do not allow empty strings.
            checkit = False
            Exit Sub
            
        Case Else
            checkit = True
            lblShortPath.Visible = True
            lblShortPath = AnsParseStr
            
    End Select

End Sub

Private Sub cmdEnd_Click()
    Unload Me
End Sub

Private Sub Command1_Click()

Dim o
Dim m

Set o = CreateObject("Outlook.Application")
Set m = o.CreateItem(0)

m.To = "[email protected]"
m.Subject = "This is the Subject"
m.Body = "File Attached"

	
	
m.Attachments.Add = lblShortPath.Caption

m.Display

End Sub

Function ParseStr(ByVal Text, ByVal separator, ByVal start As Integer, _
ByVal toEnd As Integer) As String
    
    Dim i As Integer, Temp As String, result As String
    Dim ParseStrBegin As Integer, t As Integer, Count As Integer
    Dim ParseStrEnd As Integer, Found As Integer
    
    ParseStr = ""
    If Text = "" Then Exit Function
    If separator = "" Then Exit Function
    If Not (start > 0) Then start = 1
    If toEnd < start Then toEnd = start
    'Find first instance of the separator
     t = InStr(1, Text, separator)
    
    'If no occurence return original string and exit
    If t = 0 Then
    ParseStr = Text
    Exit Function
    End If

    'If first ParseStr, return left most data and exit

    If (start = 1) And (start = toEnd) Then
    If t = 1 Then
        ParseStr = ""
        Exit Function
    Else
        ParseStr = Left(Text, t - 1)
        Exit Function
    End If
    End If
    
    ParseStrBegin = 1
    For i = 1 To start - 1
       t = InStr(ParseStrBegin, Text, separator)
       If t = 0 Then Exit For
       ParseStrBegin = t + 1
       Next i
    
    ' If there is no separator exit function with "" result
    If t = 0 Then Exit Function
    
    'If only one ParseStr to return, find it and exit
    If start = toEnd Then
    t = InStr(ParseStrBegin, Text, separator)
    If t = 0 Then t = Len(Text) + 1
    result = Left(Text, t - 1)
    ParseStr = Right(result, t - ParseStrBegin)
    Exit Function
    End If
    
    'Find last ParseStr then exit

    ParseStrEnd = t + 1
    If start = 1 Then start = 2
    For i = start To toEnd

    t = InStr(ParseStrEnd, Text, separator)
    If t = 0 Then
        t = Len(Text) + 1
        Exit For
    End If
    ParseStrEnd = t + 1
    Next i
    If t = 0 Then t = Len(Text) + 1
    result = Left(Text, t - 1)
    ParseStr = Right(result, t - ParseStrBegin)

End Function

[emartinho]

You need to tell us what error you get or what exactly is happening at the point your program has a problem. That will give us all a better idea of what's going on.

Thanks.
-EM