主题：选择模板

老师，

根据条件选择打印模板，是在菜单的Click中写入代码还是？

'Dim dr As DataRow
'If dr("客户简称") = "A客户" Then
Dim Book As New XLS.Book(ProjectPath & "Attachments\A客户合同明细.xls")
Dim fl As String = ProjectPath & "Reports\QD" & Format(Date.Today,"yyyyMMdd") & ".xls"
Dim Sheet As XLS.Sheet = Book.Sheets(0)
Sheet(4,14).Value = "A" & Format(Date.Today, "yyMMdd") '写入合同编号
Book.Build() '生成细节区
Book.Save(fl) '保存工作簿
Dim Proc As New Process '打开工作簿
Proc.File = fl
Proc.Start()

Else

Dim Book As New XLS.Book(ProjectPath & "Attachments\B客户合同明细.xls")
Dim fl As String = ProjectPath & "Reports\QD" & Format(Date.Today,"yyyyMMdd") & ".xls"
Dim Sheet As XLS.Sheet = Book.Sheets(0)
Sheet(4,14).Value = "B" & Format(Date.Today, "yyMMdd") '写入合同编号
Book.Build() '生成细节区
Book.Save(fl) '保存工作簿
Dim Proc As New Process '打开工作簿
Proc.File = fl
Proc.Start()

End If

 呃，都可以

Dim dr As Row = Tables("表A").Current
If dr("客户简称") = "A客户" Then
    Dim Book As New XLS.Book(ProjectPath & "Attachments\A客户合同明细.xls")
    Dim fl As String = ProjectPath & "Reports\QD" & Format(Date.Today,"yyyyMMdd") & ".xls"
    Dim Sheet As XLS.Sheet = Book.Sheets(0)
    Sheet(4,14).Value = "A" & Format(Date.Today, "yyMMdd") '写入合同编号
    Book.Build() '生成细节区
    Book.Save(fl) '保存工作簿
    Dim Proc As New Process '打开工作簿
    Proc.File = fl
    Proc.Start()
   
Else
   
    Dim Book As New XLS.Book(ProjectPath & "Attachments\B客户合同明细.xls")
    Dim fl As String = ProjectPath & "Reports\QD" & Format(Date.Today,"yyyyMMdd") & ".xls"
    Dim Sheet As XLS.Sheet = Book.Sheets(0)
    Sheet(4,14).Value = "B" & Format(Date.Today, "yyMMdd") '写入合同编号
    Book.Build() '生成细节区
    Book.Save(fl) '保存工作簿
    Dim Proc As New Process '打开工作簿
    Proc.File = fl
    Proc.Start()
   
End If