Question

我试图将excel文件1的单元格（1,1）复制到excel文件2的单元格（1,1）。但假设我已经在单元格（2,20）中放置了要打开的文件的名称，并且我想分配变量j = cells（2,20）并在复制文件的代码中使用它。我似乎遇到了问题。

这是我的代码：

Sub Copy_Workbook()

j = Cells(2, 20)

Workbooks.Open ("C:\Users\GNPOWER\Desktop\TRADERS\Jonel\practice\data fetching\" & j & ".xlsx")

Workbooks("Practice_Copy_From.xlsm").Worksheets("Sheet1").Cells(1, 1) = _
Workbooks(" & j & "&.xlsx").Worksheets("Sheet1").Cells(1, 1).Value

End Sub

我错过了宣言或什么？

我在运行程序时得到范围9的下标。

Answer 1

您不必在工作簿名称声明中添加明确的引号。在扩展名前面还有一个＆符号＆amp; .xlsx＆＃34;我删除了。代码是：

Sub Copy_Workbook()

Dim j As String, wb As Workbook
j = Cells(2, 20).Value2
Set wb = Workbooks.Open("C:\Users\GNPOWER\Desktop\TRADERS\Jonel\practice\data fetching\" & j & ".xlsx")
ThisWorkbook.Worksheets("Sheet1").Cells(1, 1).Value2 = wb.Worksheets("Sheet1").Cells(1, 1).Value2

End Sub

Answer 2

您甚至不需要打开“源”文件

Option Explicit

 Sub Copy_Workbook()        
    Dim pathName As String, fileName As String

    fileName = Cells(2, 20).Value '<--| retrieve the file name from the currently active worksheet cell "A1"
    pathName = "C:\Users\GNPOWER\Desktop\TRADERS\Jonel\practice\data fetching\" '<--| set the folder path where "source" workbooks resides

    With Workbooks("Practice_Copy_From.xlsm").Worksheets("Sheet1").Cells(1, 1) '<--| reference your "target" cell
        .Value = "='" & pathName & "[" & fileName & "]Sheet1'!$A$1" '<--| write a formula that references the proper cell in the proper worksheet of the proper workbook
        .Value = .Value '<--| get rid of formula and leave value only
    End With
End Sub

在复制excel中将变量分配给excel文件

2 个答案: