GetOpenFilename - Error

L

Les Stout

Hi all,
I have tried the code below, but it does not like the second line on the
Filt. If i only use one it runs, but with the other i get a method
failed message? Can somebody tell me what the problem could be please


Filt = "Comma Seperated Files (*.csv), *.csv" & _
"Excel Files (*.xls),*.xls"
Title = "Please select the supplier file to process"
ChDrive "C"
ChDir "C:\Recon\A_BackUpFiles\From_Supplier"

FileName = Application.GetOpenFilename(FileFilter:=Filt,
Title:=Title)



Les Stout
 
L

Les Stout

Have found the problem; it was a comma missing after the last CSV !!


Filt = "Comma Seperated Files (*.csv), *.csv," & _
"Excel Files (*.xls),*.xls"


Les Stout
 

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