Question

我是VBA的新手。我需要从其他位置打开现有文件，并将每个工作表保存到工作表名称（选项卡名称）名称中的不同csv文件中。我尝试过这样的事情。

Sub SplitFile()
Dim vPath As String
vPath = Application.ActiveWorkbook.Path
Application.ScreenUpdating = False
Application.DisplayAlerts = False
For Each xWs In ThisWorkbook.Sheets
    xWs.Copy
    Application.ActiveWorkbook.SaveAs Filename:=vPath & "\" & xWs.Name & ".csv", FileFormat:=xlCSV, CreateBackup:=False
    Application.ActiveWorkbook.Close False
Next
Application.DisplayAlerts = True
Application.ScreenUpdating = True
End Sub

这适用于当前活动的工作表，但我想在其他路径中处理另一个工作表。

我试图分配一个保存文件路径但却抛出错误的变量。

请帮帮我。在此先感谢。

Answer 1

您需要创建一个工作簿变量并为其分配一个工作簿对象：

Sub SplitFile()

Dim vPath As String
Dim wb As Workbook
Dim wbPath As Variant

wbPath = Application.GetOpenFileName("Excel Files (*.xls*), *.xls*")

If wbPath = False Then Exit Sub

Set wb = Workbooks.Open(wbPath)

vPath = wb.Path
Application.ScreenUpdating = False
Application.DisplayAlerts = False
For Each xWs In wb.Sheets
    xWs.Copy
    Application.ActiveWorkbook.SaveAs Filename:=vPath & "\" & xWs.Name & ".csv", FileFormat:=xlCSV, CreateBackup:=False
    Application.ActiveWorkbook.Close False
Next
Application.DisplayAlerts = True
Application.ScreenUpdating = True
End Sub

正如您所看到的，一旦您分配了工作簿，您只需在代码中引用该代码即可。

VBA宏打开Excel文件并将工作表保存到单独的CSV文件中

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