打开文件和刷新数据连接的VBA语法错误

不知道这里发生了什么…在excel vba中创build一个子例程，打开并通过传递给它的值来激活文件。 显然我做错了什么…不知道是什么。

Sub openBook(ByVal fName As String, ByVal activate As Boolean) Application.Workbooks.Open(fName, 0, False) '= Required here? End Sub 

编辑有工作，只是想检查，以确保这是正确的语法下面看到更新的代码：

 Sub openBook(ByVal fileName As String, ByVal refresh As Boolean) Dim wb As Workbook Set wb = Workbooks.Open(fileName, 0, False) If refresh = True Then wb.RefreshAll End If End Sub 

Answer 1

Workbooks.Open的语法是

expression.Open(FileName, UpdateLinks, ReadOnly, Format, Password, WriteResPassword, IgnoreReadOnlyRecommended, Origin, Delimiter, Editable, Notify, Converter, AddToMru, Local, CorruptLoad)

这是你正在尝试？

 Sub Sample() openBook "C:\MyFile.xlsx", False, True End Sub Sub openBook(fileName As String, UpdtLink As Boolean, RdOnly As Boolean) Application.Workbooks.Open fileName, UpdtLink, RdOnly End Sub 

编辑

如果你想传递0/False和1/True那么你将不得不改变

 Sub openBook(fileName As String, UpdtLink As Boolean, RdOnly As Boolean) 

至

 Sub openBook(fileName As String, UpdtLink As Variant, RdOnly As Variant) 

跟随评论

反正也有激活该工作簿在同一行或将另一行代码是必需的？ – 1分钟前的metsales

你为什么要激活它？ 尽可能避免.Activate 。 你可能想看到这个

话虽如此，如果你想激活它，那么你必须使用这样的代码

 Sub Sample() openBook "C:\MyFile.xlsx", False, True End Sub Sub openBook(fileName As String, UpdtLink As Boolean, RdOnly As Boolean) Dim wb As Workbook Set wb = Application.Workbooks.Open(fileName, UpdtLink, RdOnly) wb.Activate End Sub 

但是，下面是我build议根据我以前的build议，不要使用.Activate

 Dim wb As Workbook Sub Sample() openBook "C:\MyFile.xlsx", False, True DoEvents With wb ' '~~> Do something with the workbook here ' End With End Sub Sub openBook(fileName As String, UpdtLink As Boolean, RdOnly As Boolean) Set wb = Application.Workbooks.Open(fileName, UpdtLink, RdOnly) End Sub 

Answer 2

尝试像这样（未经testing）

 Sub openBook(ByVal fileName As String, ByVal activate As Boolean) Application.Workbooks.Open fileName:=filename End Sub