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WSH Open Outlook with .attachment (Zip-File)

Posted on 2014-03-24
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Last Modified: 2014-03-25
Hello Experts,

I've a simple wsh/vbs Script which doesn't work. In "Code-Example 1" Outlook doesn't start so I think there must be a problem with the DFile.

My Script create a Zip Archiv and I want to send this Archiv via Outlook respectively open Outlook with the right attachment.

DFile is the 'guilty' item (i.g. DFile = C:\Verschlüsselung\10000\Archiv-10000.Zip"

' Code-Example 1
...
ScriptPath = Left(WScript.ScriptFullName, InStrRev(WScript.ScriptFullName, "\"))
sStr=Split(ScriptPath,"\")
FolderN=sStr(UBound(sStr)-1)
DFile = ScriptPath & "Archiv-" & FolderN & "-" & ".Zip"
...
Function Send2Outlook (DFile) 
Set objOutl = CreateObject("Outlook.Application")
Set objMailItem = objOutl.CreateItem(olMailItem)
Set attachment = objMailItem.Attachments
attachment.Add DFile, olByValue, 4, "Great"
objMailItem.Display
...
End Function

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If I use a simple Filename in attachment.Add it works.

 
' Code-Example 1
...
Function Send2Outlook () 
Set objOutl = CreateObject("Outlook.Application")
Set objMailItem = objOutl.CreateItem(olMailItem)
Set attachment = objMailItem.Attachments
attachment.Add "C:\123456\123456.txt", olByValue, 4, "Great"
objMailItem.Display
End Function
...

Open in new window


Why doesn't DFile work in Attachment.Add Statement? It's an $ which the absolute Filename and Filepath.
I use ByRef and ByVal, delimiter chr(34) but in "Code-Example 1" Outlook doesn't start?

Best Regards.
reredok
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Question by:reredok
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9 Comments
 
LVL 72

Expert Comment

by:Qlemo
ID: 39950048
Maybe it is because you append another dash ("-") to the filename? Your created zip file is searched as C:\Verschlüsselung\10000\Archiv-10000-.Zip, and probably does not exist.

It might be generated the same, but that is something we cannot see here.
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Author Comment

by:reredok
ID: 39950118
The file exists.
The same file can be send via Outlook manually.
I added the pathname/filename hard-coded in script and it works fine.
I presume that Outlook doesn't start because there must be a diffence between "C:\123456\123456.txt" and "C:\Verschlüsselung\10000\Archiv-10000.Zip" regarding the Outlook.Attachments Statement.
Further I used the Filesystemobject in Function call ... with no success.
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LVL 72

Expert Comment

by:Qlemo
ID: 39950174
To clarify: You used "C:\Verschlüsselung\10000\Archiv-10000-.Zip" (with the trailing dash) hardcoded, and it worked?
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LVL 4

Author Comment

by:reredok
ID: 39950263
Yes.
C:\Verschlüsselung\10000\Archiv-10000.Zip
not
C:\Verschlüsselung\10000\Archiv-10000-.Zip
sry

the vbs code create a Zip Archiv with winrar:
i.e.
Zip-Archiv: Archiv-10001-2014-03-24_115211.Zip
Syntax: a custom Nummer 10001
Date: 2014-03-24
TimeStamp: 115211
Extension: Zip
Result: C:\Verschlüsselung\Archiv-10001-2014-03-24_115211.Zip

works:
attachment.Add "C:\Verschlüsselung\Archiv-10001-2014-03-24_115211.Zip", olByValue, 4, "Great"

works not:
attachment.Add DFile, olByValue, 4, "Great"
where DFile = "C:\Verschlüsselung\Archiv-10001-2014-03-24_115211.Zip"

works not:
attachment.Add chr(34) & DFile & chr(34), olByValue, 4, "Great"

What's wrong with DFile???
0
 
LVL 72

Expert Comment

by:Qlemo
ID: 39950504
It's best to debug that in Outlook VBA, and that is what I will try to do now.

Edit: Works in VBA, as expected. I get the "Der Pfad ist nicht vorhanden" error for non-existing files, and the file attached in a new mail if it exists, with this simple VBA code:
Public Sub atttest()
Dim DFile As String, objMailItem As MailItem
  ' DFile = "C:\Verschlüsselung\Archiv-10001-2014-03-24_115211.Zip"
  DFile = "C:\temp\ee\1.txt"
  Set objMailItem = CreateItem(olMailItem)
  objMailItem.Attachments.Add DFile, olByValue, 4, "Great"
  objMailItem.Display
End Sub

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What's your default eMail type? HTML, Text or RTF? (The latter has some special features, and should NOT be used).
0
 
LVL 72

Expert Comment

by:Qlemo
ID: 39950568
Same with VBS:
Option Explicit

const olMailItem = 0
const olByValue = 1

Dim DFile, objOutlook, objMailItem

' DFile = "C:\Verschlüsselung\Archiv-10001-2014-03-24_115211.Zip"
DFile = "C:\temp\ee\1.txt"
set objOutlook = CreateObject("Outlook.Application")
Set objMailItem = objOutlook.CreateItem(olMailItem)
objMailItem.Attachments.Add DFile, olByValue, 4, "Great"
objMailItem.Display

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So it must be something else, or RTF.
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LVL 4

Author Comment

by:reredok
ID: 39950595
it's wsh vbs code which is controlled by a "file or more file drag 'n' drop" Event. Debug is not so easy and it's impossible for my opinion in Outlook VBA.
I don't understand why
CreateObject("Outlook.Application").CreateItem(olMailItem).Attachments deals so different between a $ "C:\Verschlüsselung\Archiv-10001-2014-03-24_115211.Zip" and a variant DFile
Which kind of Input is needed? In ms documentation it must be a variant
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Accepted Solution

by:
Qlemo earned 2000 total points
ID: 39950687
"It must be a variant" cannot be true. In almost all cases the real type of the var is what you need. The file parameter can be a string or an Outlook object (hence the var type of variant is allowed here). For other types the Attachment.Add should result in an error message.

First step to disect the culprit is to do exactly what I did for test - use a very limited test case and code. If that works, you can try to integrate that into some more code of you, and so on.

BTW, if you write
DFile = "C:\Verschlüsselung\Archiv-10001-2014-03-24_115211.Zip"

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that isn't a variant anymore, but a string. Keep in mind you can't use (force) types for vars in VBS.

On another note, why are you providing the "4" as position? I would expect to have a 1 here, as you did not fill in anything in the mail body yet.
0
 
LVL 4

Author Closing Comment

by:reredok
ID: 39952836
I rebuild my Code and found a Problem with DFile which was only declared in a vbs function and not in called code. So the scope of this var maybe not well defined... maybe outlook had same problems with this fact. So I <Dim DFile> in "Main-Code" and transmit it up to every called function so to say as dummy
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