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我正在嘗試編寫一個代碼以將工作表複製到打開的工作簿。但最後我得到了路徑錯誤。將工作表複製到另一個工作簿 - 路徑錯誤
代碼看起來像這樣;
Sub Storyboard_Ekle()
Dim DosyaSec As Office.FileDialog
Set DosyaSec = Application.FileDialog(msoFileDialogFilePicker)
With DosyaSec
.AllowMultiSelect = False
.Title = "Lütfen yeni eklenecek Storyboard dosyasini seçiniz."
.Filters.Clear
.Filters.Add "Excel Macro-Enabled Workbook", "*.xlsm"
.Filters.Add "Excel Workbook", "*.xlsx"
.Filters.Add "All Files", "*.*"
If .Show = True Then
YeniSB = .SelectedItems(1)
End If
Dim YeniStoryBoard As Workbook
Dim AnaDosya As Workbook
Dim YeniStoryBoard_Sheet As Worksheet
Dim AnaDosya_Sheet As Worksheet
Application.ScreenUpdating = False
Set AnaDosya = ThisWorkbook
YeniStoryBoard.Sheets("Storyboard").Copy After:=ThisWorkbook.Sheets("Kunye") '-> This gives error
YeniStoryBoard.Close
Set YeniStoryBoard_isim = Sheets("Storyboard")
YeniStoryBoard_isim.Name = "StoryboardXXYYZZ"
End With
End Sub
我要做出起的代碼進行一些修改,但是這並不正常工作。 ?:(
任何建議
是在同一個目錄中的工作簿? –
樣子YeniStoryBoard_isim&YeniStoryBoard是兩個不同的變量witht他從未後者爲S-等 –
@DougCoats - 完全一樣的想法:) – Vityata