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#1
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VBA project - OpenExcelFile
I would like this VBA - OpenExcelFile
to Open the Directory c:\files where would I input a command to start open this folder ? ============== Sub OpenExcelFile() Dim vFile As Variant 'Showing Excel Open Dialog Form vFile = Application.GetOpenFilename("Limit Scan File (*.xlx)," & _ "*.xlx", 1, "Select Excel File", "Open", False) 'If Cancel then exit If TypeName(vFile) = "Boolean" Then Exit Sub End If 'Open selected file Workbooks.Open vFile End Sub ============== Thanks... |
#2
Posted to microsoft.public.excel.misc
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VBA project - OpenExcelFile
Try this:
Sub OpenExcelFile() Dim vFile As Variant ChDir "C:\FILES" .. .. .. .. Micky "It's Me Again" wrote: I would like this VBA - OpenExcelFile to Open the Directory c:\files where would I input a command to start open this folder ? ============== Sub OpenExcelFile() Dim vFile As Variant 'Showing Excel Open Dialog Form vFile = Application.GetOpenFilename("Limit Scan File (*.xlx)," & _ "*.xlx", 1, "Select Excel File", "Open", False) 'If Cancel then exit If TypeName(vFile) = "Boolean" Then Exit Sub End If 'Open selected file Workbooks.Open vFile End Sub ============== Thanks... . |
#3
Posted to microsoft.public.excel.misc
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VBA project - OpenExcelFile
That worked...
Thanks for the Help... In article , ?B?157Xmdeb15DXnCAo157Xmden15kpINeQ15HXmdeT158=?= <micky-a*at*tapuz.co.il wrote: Try this: Sub OpenExcelFile() Dim vFile As Variant ChDir "C:\FILES" .. Micky I would like this VBA - OpenExcelFile to Open the Directory c:\files where would I input a command to start open this folder ? ============== Sub OpenExcelFile() Dim vFile As Variant 'Showing Excel Open Dialog Form vFile = Application.GetOpenFilename("Limit Scan File (*.xlx)," & _ "*.xlx", 1, "Select Excel File", "Open", False) 'If Cancel then exit If TypeName(vFile) = "Boolean" Then Exit Sub End If 'Open selected file Workbooks.Open vFile End Sub ============== |
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