Question

在VBA for Excel中将参数分配给.open methon时遇到问题。下面的代码是我所拥有的，我所需要的只是将指定的工作簿打开为只读且不可编辑。

请问如何设置参数才能正常工作！

Dim source As Workbook
Dim projekt, BOM As String
Dim count As Integer
Dim LastBOM As Long

LastBOM = Ark1.Range("I" & Rows.count).End(xlUp).Row

For count = 16 To LastBOM
BOM = Range("I" & count) 'List of Excel workbooks in active workbook
projekt = "[Path to file on drive]\" & BOM & ".xlsx"

Workbooks.Open (projekt) '<= This works and open above specified workbook 
                         '   but should be read only and not editable

Workbooks.Open(projekt,ReadOnly:=True,Editable:=False) '<= None of these work
Workbooks.Open(projekt,,True,,,,,,,False,,,,,)
Workbooks.Open(projekt,True,False)
Workbooks.Open("projekt",True,False)
Workbooks.Open(Filename:="projekt",True,False)
Workbooks.Open(Filename:=projekt,ReadOnly:=True,Editable:=False)
Workbooks.Open(Filename:="projekt",ReadOnly:=True,Editable:=False)
Next count

Answer 1

阅读文档始终是一个好主意：Workbooks.Open Method (Excel)并使用ReadOnly参数：

Workbooks.Open FileName:=projekt, ReadOnly:=True, Editable:=False

如果不需要返回值，也可以删除括号。或使用括号将工作簿返回到变量中：

Dim OpenedWorkbook As Workbook
Set OpenedWorkbook = Workbooks.Open(FileName:=projekt, ReadOnly:=True, Editable:=False)

我建议使用最后一个选项，因为这样您便可以轻松访问打开的工作簿，例如，使用：

OpenedWorkbook.Worksheets("Sheet1").Range("A1")

VBA中的excel .open方法参数

1 个答案: