OK, I think I found it. Looks like you've got an extraneous parameter is your WinZip command line, between the zip file name and the target path. The line

oVar.WinZipCmndLn = ALLT(oVar.WinZipLoc) + ;
                    " -e -o " + ;
                    ALLT(oVar.ZipFname) + " " + ALLT(oVar.ZipTargetPath)

should read

oVar.WinZipCmndLn = ALLT(oVar.WinZipLoc) + ;
                    " -e -o " + ;
                    ALLT(oVar.ZipFname) + ALLT(oVar.ZipTargetPath)

Try it and see.

>nope. nothing under the path column...
>
>using the msmapi class, the files that are attached are temporily stored in the c:\temp\ directory. once i .fetch() the attachment, I am attempting to extract the zip file....
>
>oVar.ZipFname = allt(m.att_file)
>
>***********oVar.Zipfname equals "c:\temp\z442001.zip"
>
>oVar.WinZipCmndLn = ALLT(oVar.WinZipLoc) + ;
> " -e -o " + ;
> ALLT(oVar.ZipFname) + " " + ALLT(oVar.ZipTargetPath)
>
>oProcess = CREATEOBJECT("API_APPRUN", allt(oVar.WinZipCmndLn),"","NOR")
>oProcess.LaunchAppAndWait()
>WAIT CLEAR
>oVar.ZipStatus = oProcess.CheckProcessExitCode()
>oProcess = .NULL.
>
>RELEASE CLASSLIB PROCESS