Question

我有函数返回当前工作表的列表：

Function getListOfSheetsW() As Variant
  Dim i As Integer
  Dim sheetNames() As Variant

  ReDim sheetNames(1 To Sheets.Count)
  For i = 1 To Sheets.Count
    sheetNames(i) = Sheets(i).name
  Next i

  getListOfSheetsW = sheetNames
End Function

然后我的函数返回TRUE或FALSE，具体取决于needle是否在haystack。

Function IsInArray2(ByVal needle As String, haystack() As String) As Boolean
  Dim element As Variant
  For Each element In haystack
    If element = needle Then
      IsInArray = True
      Exit Function
    End If
  Next element
  IsInArray = False
End Function

我的目标是创建新的子程序，首先检查具有给定名称的工作表是否已经存在，如果没有，则创建新工作表。我试过以下：

Sub CreateNewSheet(ByVal dstWSheetName As String)
  Dim srcWSheetName As String

  ' Dim sheetNames() As String
  Dim sheetNames() As Variant

  sheetNames = getListOfSheetsW()
  Dim sheetCount As Integer

  If IsInArray2(dstWSheetName, sheetNames) Then
    MsgBox "Sheet with following name: " & dstWSheetName & " already exists"
  Else
    srcWSheetName = ActiveSheet.name
    sheetCount = Sheets.Count

    ' CREATE NEW SHEET
    ' Worksheets(dstWsheetName).Delete
    Sheets.Add.name = dstWSheetName
    ' Q: why 6 instead of 5
    ' Worksheets("Test").Move after:=Worksheets("Sheet5")
    Worksheets(dstWSheetName).Move After:=Worksheets(sheetCount + 1)

    ' SWITCH TO SRC SHEET
    Worksheets(srcWSheetName).Activate
  End If
End Sub

我这样称呼它：

Sub CallCreateNewSheet()
  Call CreateNewSheet("test")
End Sub

我想问题出在Dim sheetNames() As String或Dim sheetNames() As Variant。

当我使用Dim sheetNames() As String时，我得到了

运行时错误'13'：类型不匹配

当我使用Dim sheetNames() As Variant时，我得到：

编译错误：类型不匹配：数组或用户定义的预期类型

我有类似问题before，但定义sheetNames因为数组在这里没有帮助。问题是什么？两个不同的错误是什么意思？

Answer 1

如果从类型化数组切换到变体数组，您将避免所有这些问题。

在第一个功能中，删除以下行：

 Dim sheetNames() As Variant

更改第二个功能的定义行：

Function IsInArray2(ByVal needle As String, haystack() As String) As Boolean

......对此：

Function IsInArray2(ByVal needle As String, haystack) As Boolean

在您的子网站中，更改此行：

Dim sheetNames() As Variant

......对此：

Dim sheetNames

Answer 2

新脚本如何：

Sub NewSheetByName(SName as String)
  Dim oldSheet as Object

  For Each oldSheed in ThisWorkbook.Sheets
    if oldSheet.Name = Sname Then
      MsgBox "Sheet with following name: " & SName & " already exists"
      Exit Sub
    End If 
  Next

  oldSheet = ActiveSheet
  Sheets.Add.Name = SName
  ActiveSheet.Move , Worksheets(Sheets.Count)

  oldSheet.Activate

End Sub

Answer 3

变量必须是sinc。

在两个过程中以相同的方式声明变量sheetNames：

Sub CreateNewSheet(ByVal dstWSheetName As String)和

Function getListOfSheetsW() As Variant

将其声明为：Dim sheetNames() As String

另请注意，Function IsInArray2始终返回False。要更正此问题，请在函数正文中将IsInArray替换为IsInArray2。

总是拥有

是一个好习惯

Option Explicit

在模块的开头。

然而，它会省去验证工作表是否存在的所有麻烦，只是为了将目标工作表分配给变量，它会产生错误，如果工作表不存在，变量将不返回任何内容。试试这个：

Dim Wsh As Worksheet
On Error Resume Next
Set Wsh = Workbook(x).Worksheets("Test")
On Error GoTo 0
If Wsh Is Nothing Then Add Worksheet

将数组传递给函数时，Variant与String类型不匹配

3 个答案: