减lessstring并保持零不变

我正在使用格式为“XXXX-000-000-000”的单元格值。

• 第一个连字符之前的所有内容都需要删除，我可以这样做。
• string的其余部分需要减less到整数，用连字符和任何额外的零删除。

我很难将零位保持在正确的位置。

• AD12-002-020-34需要如下所示：2-20-34
• CA1-002-101-001需要如下所示：2-101-1
• AD12-002-020-10需要如下所示：2-20-10

例如：

dim ir as range ir = "AD12-002-020-100" ir1 = InStr(ir, "-") ir2 = InStrRev(ir, "-") ir.Offset(0, 1) = Mid(ir, ir1 + 1, ir2 - ir1 + 3) 

这给了我：002-020-100

build议？ 先谢谢你！

Answer 1

此外，这将作为一个UDF（用户定义的function）

 Function STRIP(r As String) If InStr(1, r, "-00", vbTextCompare) Then r = Replace(r, "-00", "-") End If If InStr(1, r, "-0", vbTextCompare) Then r = Replace(r, "-0", "-") End If Dim v As Variant, s As String, i As Long v = Split(r, "-") For i = 1 To UBound(v) s = s & "-" & v(i) Next i STRIP = Right(s, Len(s) - 1) End Function 

你可以简单地使用=STRIP(A1)单元格来调用，其中A1是你想要分割的单元格的引用

例：

Answer 2

考虑：

 Sub dural() Dim s As String s = "AD12-002-020-34" s = Replace(s, "-0", "-") s = Replace(s, "-0", "-") ary = Split(s, "-") ary(0) = "" s = Mid(Join(ary, "-"), 2) MsgBox s End Sub 

Answer 3

我也可以玩吗？ ：p

一个class轮

Debug.Print Mid(Replace(Replace(Replace(Mid(sString, InStr(1, sString, "-")), "-000", "-"), "-00", "-"), "-0", "-"), 2)

其中sString是有效的"XXXX-000-000-000"string

Answer 4

一个正则expression式的方法

 Sub TestCleanNames() Debug.Print CleanString("AD12-002-020-34") Debug.Print CleanString("CA1-002-101-001") Debug.Print CleanString("AD12-002-020-10") End Sub Function CleanString(strIn As String) As String Dim objRegex As Object Set objRegex = CreateObject("vbscript.regexp") With objRegex 'remove first portion .Pattern = "(.+?)-" CleanString = .Replace(strIn, vbNullString) .Global = True 'trim leasfing zeroes .Pattern = "(\-|^)(0)+([1-9])" CleanString = .Replace(CleanString, "$1$3") End With End Function 

Answer 5

UDF是没有必要的（但显然是一个更好的主意！）：

=LEFT(MID(A1,FIND("-",A1)+1,LEN(A1)),FIND("-",MID(A1,FIND("-",A1)+1,LEN(A1)))-1)*1&"-"&VALUE(MID(MID(A1,FIND("-",A1)+1,LEN(A1)),FIND("-",MID(A1,FIND("-",A1)+1,LEN(A1)))+1,FIND("-",MID(A1,FIND("-",A1)+1,LEN(A1)),FIND("-",MID(A1,FIND("-",A1)+1,LEN(A1)))+1)-FIND("-",MID(A1,FIND("-",A1)+1,LEN(A1)))-1))&"-"&VALUE(MID(MID(A1,FIND("-",A1)+1,LEN(A1)),FIND("-",MID(A1,FIND("-",A1)+1,LEN(A1)),FIND("-",MID(A1,FIND("-",A1)+1,LEN(A1)))+1)+1,LEN(A1)))

Answer 6

 Sub x() 'Technically, you don't have to declare variables, but it prevents typos: Dim ir As String Dim ir1 As String Dim ir2 As String Dim ir3 As String ir = "AD12-002-020-100" 'First, get rid of the first 5 characters: ir = Right(ir, Len(ir) - 5) 'Isolate each section. Convert to Int to get rid of leading zeros: ir1 = CInt(Left(ir, 3)) ir2 = CInt(Right(Left(ir, 7), 3)) ir3 = CInt(Right(ir, 3)) 'Return the result: MsgBox ir1 & "-" & ir2 & "-" & ir3 End Sub 

Answer 7

我以为我会把这个帽子戴在戒指上。 请注意，所有版本都假定要转换的单元格位于A单元格的A列中。

作为一个macros观：

 Sub tgr() Dim arrResults() As String Dim varText As Variant Dim varPart As Variant Dim ResultIndex As Long With Range("A1", Cells(Rows.Count, "A").End(xlUp)) ReDim arrResults(1 To .Rows.Count, 1 To 1) For Each varText In .Value ResultIndex = ResultIndex + 1 For Each varPart In Split(Mid(varText, InStr(varText, "-") + 1), "-") arrResults(ResultIndex, 1) = arrResults(ResultIndex, 1) & "-" & Val(varPart) Next varPart arrResults(ResultIndex, 1) = Mid(arrResults(ResultIndex, 1), 2) Next varText .Value = arrResults End With End Sub 

作为UDF：

 Function tgrUDF(sText As String) As String Dim varPart As Variant For Each varPart In Split(Mid(sText, InStr(sText, "-") + 1), "-") tgrUDF = tgrUDF & "-" & Val(varPart) Next varPart tgrUDF = Mid(tgrUDF, 2) End Function 

作为一个公式：

= – MID（SUBSTITUTE（A1，“ – ”，REPT（“”，99）），99,99） ），99 * 2,99）“ – ”＆ – MID（SUBSTITUTE（A1，“ – ”，REPT（“”，99）），99 * 3,99）