Question

我试图将工作表的代码名称分配给可用的有时它会正确获取代码名称，有时它不会，变量保持为空。

 Dim sCodeName As String
 sCodeName = Worksheets(atar).CodeName

atar是包含工作表名称的变量。当我停止代码运行，并在调试模式下，它工作正常。可能是什么原因？

Answer 1

没有限定条件的

Worksheets与ActiveWorkbook相同，也许您在某些情况下假设哪些工作簿处于活动状态。

建议完全限定工作表。因此，请引用Workbook对象，例如wbFoo，然后使用sCodeName = wbFoo.Worksheets(atar).CodeName

Answer 2

您可以尝试使用下面的代码来验证atar中某个工作表名称中是否存在ThisWorkbook。

Option Explicit

Sub GetWorksheetCodeName()

Dim sCodeName As String
Dim atar As String
Dim Sht As Worksheet

'atar = "Sheet3" '<-- for tests only

' loop through all worksheets in ThisWorkbook
For Each Sht In ThisWorkbook.Worksheets
    If Sht.Name Like atar Then
        sCodeName = Worksheets(atar).CodeName
        Exit For
    End If
Next Sht

End Sub

Answer 3

创建多个工作表但有时不使用工作表中的宏仅使用VBA代码运行

 Sub WSCreate()
    Dim WSA As String
    Dim WSB As String
    Dim WS As Worksheet

        WSA = "SheetA"
        WSB = "SheetB"

        Application.DisplayAlerts = False
        On Error Resume Next

'DELETE MULTIPLE WORKSHEETS  (IF Already Exist)
    Set WS = Nothing
    Set WS = Sheets(WSA)
        WS.Delete
    Set WS = Nothing
    Set WS = Sheets(WSB)
        WS.Delete
    Set WS = Nothing


'CREATE MULTIPLE WORKSHEETS    
   With Application.ThisWorkbook
        .Worksheets.Add.Name = WSA  'WSA = SheetA
        .VBProject.VBComponents(Worksheets(WSA).CodeName).Name = WSA
        .Worksheets.Add.Name = WSB  'WSB = SheetB
        .VBProject.VBComponents(Worksheets(WSB).CodeName).Name = WSB
    End With
     If Err <> 0 Then Exit Sub
    End Sub

未分配工作表CodeName

3 个答案: