Question

运行下面的代码时，根据打开的工作簿，我得到不同的结果。子程序位于与Master Sheet.xlsm

相关的模块中

如果仅打开Master Sheet.xlsm，则代码将正确运行，即消息框显示（逗号分隔第一和第二个消息框）：母版纸，已转移案件03-09-18
如果Master Sheet.xlsm和transferred cases 03-09-18.xlsx都打开了，但是transferred cases 03-09-18.xlsx处于第二打开状态，则消息框显示：已转移案例03-09-18，已转移案例03-09- 18
如果Master Sheet.xlsm和transferred cases 03-09-18.xlsx都打开了，但Master Sheet.xlsm处于第二打开状态，则消息框显示：“母版纸，母版纸”

Sub foo()
    Dim x As Workbook
    Dim y As Workbook

   '## Open both workbooks first:
   Set x = Workbooks.Open("C:\Users\owner\Documents\ExelatecOutput\Master Sheet.xlsm")
   Set y = Workbooks.Open("C:\Users\owner\Documents\ExelatecOutput\transferred cases 03-09-18.xlsx")

   'Now, copy what you want from x:
   MsgBox x.Name
   MsgBox y.Name

End Sub

为什么变量x和y分配不正确。

Answer 1

仅需注意，您可以通过传递工作簿名称的功能来检查工作簿是否已经打开。

Public Function BookOpen(strBookName As String) As Boolean

    Dim oBk As Workbook
    On Error Resume Next
    Set oBk = Workbooks(strBookName)
    On Error GoTo 0
    If oBk Is Nothing Then
        BookOpen = False
    Else
        BookOpen = True
    End If

End Function

如果返回true，则可以set x = Workbooks("your workbook name")

Answer 2

Workbooks.Open始终返回上一个打开的文件（即使它不是传入的参数）。这可能是错误的文档，也可能是Excel IMO中的错误。

您不需要检查文件是否已打开，因为打开已经打开的文件不会引发错误，但是您稍后需要设置变量：

Workbooks.Open "C:\Users\owner\Documents\ExelatecOutput\Master Sheet.xlsm": Set x = Workbooks("Master Sheet.xlsm") ' or Set x = ActiveWorkbook since Open will activate it
Workbooks.Open "C:\Users\owner\Documents\ExelatecOutput\transferred cases 03-09-18.xlsx": Set y = Workbooks("transferred cases 03-09-18.xlsx") ' or Set y = ActiveWorkbook since Open will activate it

Answer 3

使用Workbook分配给Workbooks.Open()变量时，意味着要打开的工作簿已关闭。否则，如果所有工作簿都已打开，则它将使用最后打开的Workbooks.Open()工作簿或代码所在的工作簿。

因此，请确保在尝试打开工作簿之前先将它们关闭。打开之前，您需要执行以下两个操作：

Detect whether Excel workbook is already open

如果有打开的工作簿，则将其关闭：

Closing Excel Application using VBA

另一个更快的选项是显式分配变量，如Vincent G所述，更快，因为您不会关闭已打开的变量Excel文件：

Sub TestMe()

    Dim x As Workbook
    Dim y As Workbook
    Dim xPath As String: xPath = "C:\Book1.xlsx"
    Dim yPath As String: yPath = "C:\Book2.xlsx"

    Workbooks.Open xPath
    Set x = Workbooks(Split(xPath, "\")(UBound(Split(xPath, "\"))))
    Workbooks.Open yPath
    Set y = Workbooks(Split(yPath, "\")(UBound(Split(yPath, "\"))))

    Debug.Print x.Name
    Debug.Print y.Name

End Sub

代码Split(xPath, "\")(UBound(Split(xPath, "\")))的一部分采用\分割数组的最后一个元素。

在Visual Basic中打开两个工作簿时，工作簿变量未正确分配

3 个答案: