再次求助,文件复制工作簿问题

VictorHuzw

感谢赵老师的答复!

那执行宏按钮复制文件后,如何直接自动打开这个文件呢?  如何修改这段代码

即打开这个文件: arr(m, 1) & ".xlsx"

1. Private Sub CommandButton1_Click()
   
2.     Dim cnn, arr, m&, f$, SQL$
   
3.     If ActiveCell.Value = "" Then Exit Sub
   
4.     If ActiveCell.Column = 1 Or ActiveCell.Row < 3 Then Exit Sub
   
5.     arr = [a1].CurrentRegion
   
6.     m = ActiveCell.Row
   
7.     f = ThisWorkbook.Path & "" & arr(m, 1) & ".xlsx"
   
8.     If Dir(f) <> "" Then Kill f
   
9.     FileCopy ThisWorkbook.Path & "\分类模板.xlsx", f
   
10.     Set cnn = CreateObject("ADODB.Connection")
    
11.     cnn.Open "Provider=Microsoft.Ace.OLEDB.12.0;Extended Properties='Excel 12.0;HDR=No';Data Source=" & f
    
12.     SQL = "update [Sheet1$b1:b1] set f1='" & arr(m, 1) & "'"
    
13.     cnn.Execute SQL
    
14.     SQL = "update [Sheet1$c2:c2] set f1=" & arr(m, 2)
    
15.     cnn.Execute SQL
    
16.     SQL = "update [Sheet1$c3:c3] set f1=" & arr(m, 4)
    
17.     cnn.Execute SQL
    
18.     SQL = "update [Sheet1$c4:c4] set f1=" & arr(m, 6)
    
19.     cnn.Execute SQL
    
20.     MsgBox "已生成" & arr(m, 1) & ".xlsx"
    
21.     cnn.Close
    
22.     Set cnn = Nothing
    
23. End Sub

复制代码

补充内容 (2016-3-12 20:10):
求助

补充内容 (2016-3-12 20:10):
求助

LMY123

SQL = "update [Sheet1$c2:c2] set f1=" & arr(m, 2)