Question

我很抱歉，因为我是vba代码的新手。我遇到以下代码的问题。我正在尝试打开一个文件，然后将该文件中的所有工作表复制到当前工作簿。它当前返回运行时错误424

Private Sub testGetFile()

Dim wb1, wb2 As Workbook
Dim sh As Worksheet
Set wb1 = ThisWorkbook
Set wb2 = Application.GetOpenFilename

For Each sh In wb2
sh.Copy After:=wb1.Sheets(wb1.Sheets.Count)
Next

End Sub

Answer 1

您没有正确使用GetOpenFilename。由于它不是Workbook对象，因此无法将其直接设置为工作簿。

我清理过的代码还有其他几个地方。

Option Explicit

Private Sub testGetFile()

    Dim wb1 As Workbook
    Set wb1 = ThisWorkbook

    Dim fileToOpen As Variant
    fileToOpen = Application.GetOpenFilename

    If fileToOpen = False Then
        Exit Sub
    Else
        Dim wb2 As Workbook
        Set wb2 = Workbooks.Open(fileToOpen)
    End If

    Dim sh As Worksheet
    For Each sh In wb2.Worksheets
        sh.Copy After:=wb1.Sheets(wb1.Sheets.Count)
    Next

End Sub

Answer 2

wb2是一本工作簿，GetOpenFilename返回String。例如，您必须获取返回的String并使用它将其设置为wb2：

Dim fileName    As String
fileName = Application.GetOpenFilename()
Workbooks.Open fileName

GetOpenFilename MSDN

复制工作表时的运行时错误424

2 个答案: