Shell function problem (MS Access)

[Guest]

Hi

Trying to do this within Excel VBA but having a problem with the following
shell command. The command line switch can't open the named file.

Dim retVal
retVal = Shell("C:\Program Files\Microsoft Office\Office\MSACCESS.EXE S:\My
Folder\MyFile.mdb", 1)

The problem stems, I assume, from the space in the pathname (something over
which I have no control)

Is there a known solution to this issue or is there a more correct way to
type the pathname and filename to overcome the problem. i've tried several
combinations of parentheses with no luck

Any help appreciated

Michael Bond

 

[Rick Rothstein \(MVP - VB\)]

Trying to do this within Excel VBA but having a problem with the following

shell command. The command line switch can't open the named file.

Dim retVal
retVal = Shell("C:\Program Files\Microsoft Office\Office\MSACCESS.EXE
S:\My
Folder\MyFile.mdb", 1)

The problem stems, I assume, from the space in the pathname (something
over
which I have no control)

Give this a try...

retVal = Shell("""C:\Program Files\Microsoft Office\Office\MSACCESS.EXE""
""S:\My Folder\MyFile.mdb""", 1)

Rick

 

[Guest]

Michael,

Try enclosing the path and filename with double quotes. like this:

retVal = Shell("C:\Program Files\Microsoft Office\Office\MSACCESS.EXE
""S:\My Folder\MyFile.mdb""", 1)

 

[Helmut Weber]

Hi Michael,

strTmp = "C:\Program Files\Microsoft Office\Office\"
strtmp = strtmp & "MSACCESS.EXE S:\My Folder\MyFile.mdb"
strtmp = chr(34) & strtmp & chr(34)

retVal = Shell(strtmp, 1)

--

Helmut Weber

Win XP, Office 2003
"red.sys" & Chr$(64) & "t-online.de"

 

[Guest]

Rick

thanks, worked a treat...that was one combination I didn't try! Appreciate
the help and spead of response.

Regards

Michael Bond

 

[Guest]

Thanks Vergel

Appreciate the help. Got it working now.

Regards

Michael Bond

 

[Guest]

Helmut

Thanks

Michael