Question

重写我的问题：

Sub Path()
Dim path As Range
Dim shPivot As Worksheet
Set shPivot = ActiveWorkbook.Sheets("Pivot")
Set path = shPivot.Range("E12").Value
Set wkbFrom = Workbooks.Open("S:\_Supply Chain\Weekly Rpts Supplier and Buyer\" & path & "\SUPPLIER_01_00028257_KIK CUSTOM PRODUCTS GAINSVILLE_21-OCT-12.xls")

路径是单元格中的日期。当此单元格更改时，我希望目录根据路径更改。

Answer 1

两件事

将路径声明为字符串而不是范围。
在路径

这是你在尝试的吗？

Dim path As String
path = shPivot.Range("E12").Value
Set wkbFrom = Workbooks.Open("S:\_Supply Chain\Weekly Rpts Supplier and Buyer\" & _
              format(path,"DD-MM-YYYY") & _
              "\SUPPLIER_01_00028257_KIK CUSTOM PRODUCTS GAINSVILLE_21-OCT-12.xls")

<强>后续

在这种情况下使用此

Dim path As String
path = shPivot.Range("E12").Value
Set wkbFrom = Workbooks.Open("S:\_Supply Chain\Weekly Rpts Supplier and Buyer\" & _
              path & _
              "\SUPPLIER_01_00028257_KIK CUSTOM PRODUCTS GAINSVILLE_21-OCT-12.xls")

如果有任何不需要的空格，则必须使用TRIM

Dim path As String
path = shPivot.Range("E12").Value
Set wkbFrom = Workbooks.Open("S:\_Supply Chain\Weekly Rpts Supplier and Buyer\" & _
              Trim(path) & _
              "\SUPPLIER_01_00028257_KIK CUSTOM PRODUCTS GAINSVILLE_21-OCT-12.xls")

使用VBA代码中的单元格数据打开目录

1 个答案: