opening a csv with Excel VBA

Question

I am attempting open a csv file with this code and it keeps giving me an "error 52 bad file name or number"

 Sub ShowFileDialog()
   Dim x As String
     Dim FF1 As Integer
    Dim dlgOpen As FileDialog
    Set dlgOpen = Application.FileDialog( _
        msoFileDialogFilePicker)
    With dlgOpen

        .Show
    End With

x = CStr(dlgOpen.SelectedItems(1))
MsgBox x

Open x For Input As #FF1

Do While Not EOF(FF1)

Line Input #FF1, inputdata

Dim lineData() As String
lineData() = Split(inputdata, ",")



Loop
Close #FF1
End Sub

The debugger is highlighting the Open for X line but I am feeding it the path name as a string

You never set FF1 to a file no. FF1 = FreeFile before opening. — Jacob
– Jacob, Commented Oct 12, 2012 at 14:45

Daniel · Accepted Answer · 2012-10-12 14:49:47Z

1

Add this line directly before the line causing the error:

FF1 = FreeFile

Because Open For Input Requires a number between 1 and 512, which should be obtained via a call to FreeFile.

answered Oct 12, 2012 at 14:49

Daniel

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