Open file in VBA
 

Hello,

When I walk through the following statement I get a run time error 1004
that says that the particular file couls not be found.

I used the following statement:

sFilename = Dir("H:\Risk_Management\Special_Review\Upload_file s\*.xls")
Workbooks.Open Filename:=sFilename

Do I forgot something?

Thanks for your assistance!!

Rgds,
Robert

Open file in VBA
 

Hi Robert,

dim sFilename as string
sFilename = Dir("H:\Risk_Management\Special_Review\Upload_file s\*.xls")
if len(sfilename)<0 then
Workbooks.Open Filename:=sFilename
else
'no file path returned from
dir("H:\Risk_Management\Special_Review\Upload_file s\*.xls")
end if

Regads,
Ivan

Open file in VBA
 

Another possibility:
Dir doesn't return the path in sFileName, so

Dim sPath as String, sFileName as String
sPath = "H:\Risk_Management\Special_Review\Upload_file s\"
sFileName = Dir(sPath & "*.xls")
if sFileName < "" then
workbooks.Open Filename:=sPath & sFileName
Else
msgbox "No excel workbooks"
End if

--
Regards,
Tom Ogilvy


*.xls")
Workbooks.Open Filename:=sFilename

"Robert" wrote:

Hello,

When I walk through the following statement I get a run time error 1004
that says that the particular file couls not be found.

I used the following statement:

sFilename = Dir("H:\Risk_Management\Special_Review\Upload_file s\*.xls")
Workbooks.Open Filename:=sFilename

Do I forgot something?

Thanks for your assistance!!

Rgds,
Robert