Re: help with AddRecipToContacts procedure by Sue Mosher
- From: "Sue Mosher [MVP-Outlook]" <suemvp@xxxxxxxxxxxxxxx>
- Date: Thu, 31 May 2007 21:28:12 -0400
I'm glad you like the code. You can check the value of Recipient.AddressEntry.Type to determine whether the recipient has an SMTP address.
The sample code at http://www.outlookcode.com/d/code/autobcc.htm shows how to add a Bcc recipient using the same Application.ItemSend event in Outlook VBA. You'd just need to combine the two into one procedure.
--
Sue Mosher, Outlook MVP
Author of Configuring Microsoft Outlook 2003
http://www.turtleflock.com/olconfig/index.htm
and Microsoft Outlook Programming - Jumpstart for
Administrators, Power Users, and Developers
http://www.outlookcode.com/jumpstart.aspx
"Richard T" <RichardT@xxxxxxxxxxxxxxxxxxxxxxxxx> wrote in message news:6F7AAF5F-67FA-4779-9CE3-237FC2B06CE2@xxxxxxxxxxxxxxxx
One last thing. I am also using a rule on my machine to add my own email.
address in the CC field to each message that I send. When I put Sue's code
into place on my machine, this rule has stopped functioning.
Is there a way to make the rule work again? Or better yet, have this code
insert my email address as a BCC instead so I can delete the rule?
Thanks again!
-- Richard
"Richard T" wrote:
I am a newbie to Outlook VBA, so pardon me if this is a stupid question.
I am using the code by Sue Mosher that's found here
(http://www.outlookcode.com/d/code/autoaddrecip.htm) to automatically add
recipients of my outgoing mail to my contacts folder. This is working and I
am very excited about it. Many thanks to Sue for making this code available.
However, I am in an Exchange environment and when I send mail to another
Exchange user, I end up with a contact that has an email address like this:
"/o=Exchange1/ou=First Administrative Group/cn=Recipients/cn=jondoe"
Is there a way to modify the AddRecipToContacts procedure so that it only
adds those recipients who are not exchange users? Better yet, is there a way
to get the Exchange user's actual email address?
Thanks in advance,
Richard
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