无效的过程调用或参数在拆分mac excel vba

我不知道为什么，但这个cet = Split(strCSV, " - ")导致运行时错误5：无效的过程调用或参数。

  strCSV = Trim(cWk.Range("P" & i)): dt = CDate(CLng(cWk.Range("H" & i))) 

第一复仇者（+ S $ 1.50）/盒：无 – 只购买3个及以上的帽子 – 免费盒子“

下面的代码完美的工作在Windows上，但它会抛出上述线错误

 Option Explicit Option Compare Text Sub Get_Data() Application.ScreenUpdating = False Dim fName, wkB2 As Workbook, cWk As Worksheet, xWk As Worksheet, frowC As Long, i As Long, j As Long, ch As String, num As String Dim strCSV As String, dt As Date, shtName As String, cet, temp As String, rng As Range, cel As Range, cl As String, rw As Long, toF As String On Error GoTo Err fName = Application.GetOpenFilename If fName <> False Then Set wkB2 = Workbooks.Open(fName): Set cWk = wkB2.Worksheets(1): frowC = cWk.Range("P" & Rows.Count).End(xlUp).Row 'Cap Style:Baseball - CC / Number:04 / Color:Grey(+S$2) / Box:none - Only Purchase 3 caps and above - Free Box 'Cap Style:SnapBack - CC / Number:04 / Color:Grey(+S$1.50) / Box:none - Only Purchase 3 caps and above - Free Box For i = 2 To frowC strCSV = Trim(cWk.Range("P" & i)): dt = CDate(CLng(cWk.Range("H" & i))) If strCSV <> "" And IsDate(dt) Then 'ERROR cet = Split(strCSV, " - "): temp = cet(LBound(cet)): cet = Split(temp, ":"): shtName = Trim(cet(UBound(cet))) For Each xWk In ThisWorkbook.Worksheets If shtName = Trim(xWk.Name) Then Set rng = xWk.Range("E3:BD3") For Each cel In rng If cel.Value = dt Then cet = Split(cel.Address, "$"): cl = cet(UBound(cet) - 1): Exit For End If Next cel cet = Split(strCSV, "Number:"): temp = cet(UBound(cet)): cet = Split(temp, "/"): num = Trim(cet(LBound(cet))) cet = Split(strCSV, " / "): temp = cet(LBound(cet)): cet = Split(temp, " - "): ch = Trim(cet(UBound(cet))): ch = ch & "-" & num Debug.Print "Ch is " & ch Set rng = xWk.Range("A1:A" & xWk.Range("A" & Rows.Count).End(xlUp).Row) For Each cel In rng If cel.Value = ch Then rw = cel.Row: Exit For End If Next cel cet = Split(strCSV, "Color:"): temp = cet(UBound(cet)): cet = Split(temp, "("): toF = Trim(cet(LBound(cet))) For j = rw To rw - 10 Step -1 If Trim(xWk.Range("B" & j)) = toF Then rw = j: Exit For End If Next j Debug.Print "Address is: " & cl & rw & " for row " & i xWk.Range(cl & rw) = cWk.Range("O" & i) Exit For End If Next xWk End If Next i wkB2.Close False Else Exit Sub End If Application.ScreenUpdating = True MsgBox "Done" Exit Sub Err: MsgBox Err.Description End Sub 

Answer 1

更新：SplitString现在可以处理多字符分隔符。

我们得出结论，旧版本的Mac Office使用相当于VB5。 由于Split函数是在VB6中引入的。 正在抛出一个Invalid procedure call or argument ，因为Split函数在VB5中不可用。

解决方法是创build一个像Spli一样工作的自定义函数。

拆分replacefunction

 Function SplitString(Text As String, Delimiter As String) Dim arr() As String, s As String Dim i As Long, iEnd As Long, iStart As Long, length As Long length = Len(Delimiter) ReDim Preserve arr(0) iStart = 1 Do iEnd = InStr(Mid(Text, iStart), Delimiter) - 1 If iEnd = -1 Then ReDim Preserve arr(i) arr(i) = Mid(Text, iStart) Exit Do Else ReDim Preserve arr(i) arr(i) = Mid(Text, iStart, iEnd) iStart = iStart + iEnd + length i = i + 1 End If Loop Until iStart = 0 SplitString = arr End Function 

这是我跑的testing

 Sub BatchTest() Dim strCSV As String, Temp As String, Delimiter As String Dim a strCSV = "Cap Style Snapback - CD / Number 07 / Color First Avenger(+S$1.50) / Box none - Only Purchase 3 caps and above - Free Box" a = SplitString(strCSV, "/") TestSplit strCSV, " / " TestSplit strCSV, " /" TestSplit strCSV, "/" TestSplit strCSV, " Color First" End Sub Sub TestSplit(Text As String, Delimiter As String) Dim arr As Variant, sReplcement As String arr = SplitString(Text, Delimiter) sReplcement = Replace(Text, Delimiter, "|") Debug.Print sReplcement Debug.Print Join(arr, "|") Debug.Print sReplcement = Join(arr, "|") End Sub 

testing的结果

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 Sub TestRegEx() MsgBox RegexExtract("sdi 99090 dfddf sdi 5666", "(sdi \d+)", ", ") = "sdi 99090, sdi 5666" End Sub Function RegexExtract(ByVal text As String, _ ByVal extract_what As String, _ Optional seperator As String = "") As String Dim i As Long, j As Long Dim result As String Dim allMatches As Object Dim RE As Object Set RE = CreateObject("vbscript.regexp") RE.Pattern = extract_what RE.Global = True Set allMatches = RE.Execute(text) For i = 0 To allMatches.count - 1 For j = 0 To allMatches.Item(i).submatches.count - 1 result = result & seperator & allMatches.Item(i).submatches.Item(j) Next Next If Len(result) <> 0 Then result = Right(result, Len(result) - Len(seperator)) End If RegexExtract = result End Function