命名表&建立超级链接

0

的问题

所以我创造一个新片的正是如此:

Dim Naming As String
Dim ProjectNumber As String
Dim ws As Worksheet

ThisWorkbook.Sheets("Project Template").Copy After:=Sheets(Sheets.Count)
Set ws = ActiveSheet

Naming = CallLaunched.Value & " - " & ProjectLead.Value

ws.Name = Naming

这给了我一张名为:"月21-萨姆*史密斯"

以后在结束码我想要的超链接,新的片从另一片所以我用这个:

ActiveSheet.Hyperlinks.Add Anchor:=ws.Range("A" & LastRow), Address:="", SubAddress:=Naming & "!A1", TextToDisplay:=ProjectNumber

这一切工作,但当我点击超级链接我得到的"参考是不是有效的".

我不能为了我的生活弄清楚为什么不是超链接引用我的新片。

任何帮助,将不胜感激,

谢谢,

苏西

excel hyperlink vba
2021-11-18 21:56:21
1

最好的答案

0

片名与空间需要的报价:

SubAddress:= "'" & Naming & "'!A1"
2021-11-18 21:59:57

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