如何從 Excel 中的文本字符串中調用正確的數字？

Question

考慮以下文本字符串：

(*4,14)(7,15)(10,13)(9,12)-(1,8)(2,6)-5,3-11

我的目標是計算該字符串中每個單獨數字之前有多少個左括號（“（”）、括號外的逗號和連字符（例如，數字 10 前面的 3 個左括號、6 個左括號和前面的 3 個連字符） 11）。

我當前的解決方案是首先調用每個單獨數字前面的剩余文本字符串，只需=LEFT(A1,(FIND("1",A1,1)-1)) ，但碰巧 Excel 會調用出現的字符串在第一個“1”之前（即(*4, )，而不是從字符串中的實際數字“1”中調用剩余的字符串（即(*4,14)(7,15)(10,13)(9,12)-( )。

旁注，關於如何計算括號外逗號數量的任何想法？

幫助將不勝感激！

Answer 1

如果您有帶有FILTERXML function 的 Excel 版本（Windows Excel 2013+），您可以使用：

=SUM(LEN(FILTERXML("<t>" & SUBSTITUTE(SUBSTITUTE(A1,"(","<s>"),")","</s>") & "</t>","//t")))- LEN(SUBSTITUTE(FILTERXML("<t>" & SUBSTITUTE(SUBSTITUTE(A1,"(","<s>"),")","</s>") & "</t>","//t"),",",""))

該公式創建了一個 xml ，其中s節點是括號內的內容，而t節點是其他所有內容。

如果您沒有FILTERXML function，最好使用 VBA 解決方案。 這取決於您的 Excel 版本，以及是 Windows 還是 MAC。

Answer 2

計數字符

Option Explicit

Function countChars(SourceString As String, SourceNumber As Variant, _
  CountChar As String, Optional countRight As Boolean = False) As Long

    Dim NumberDouble As Double
    Dim NumberString As String
    Dim NumberLength As Long
    Dim StringLength As Long
    Dim CurrentStart As Long
    Dim CurrentFound As Long
    Dim i As Long
    Dim isFound As Boolean

    StringLength = Len(SourceString)

    If VarType(SourceNumber) = 8 Then
        If Not IsNumeric(SourceNumber) Then _
          Exit Function   ' SourceNumber is not numeric.
    End If
    NumberDouble = Val(SourceNumber)
    If NumberDouble <> Int(NumberDouble) Then _
      Exit Function       ' SourceNumber is not an integer.
    NumberString = CStr(NumberDouble)
    NumberLength = Len(NumberString)

    CurrentStart = 1
    Do
        CurrentFound = InStr(CurrentStart, SourceString, NumberString)
        GoSub checkNumber
        If isFound Then
            GoSub countTheChars
            Exit Do
        End If
        CurrentStart = CurrentFound + 1
    Loop Until CurrentFound = 0

Exit Function

countTheChars:  ' Can be written better.
    If Not countRight Then
        For i = 1 To CurrentFound - 1
            If Mid(SourceString, i, 1) = CountChar Then
                countChars = countChars + 1
            End If
        Next i
    Else
        For i = CurrentFound + 1 To StringLength
            If Mid(SourceString, i, 1) = CountChar Then
                countChars = countChars + 1
            End If
        Next i
    End If

checkNumber:  ' Check for adjacent numbers.
   Select Case CurrentFound
       Case 0: Exit Function  ' NumberString (initially) not found.
       Case 1                 ' NumberString found at the beginning.
           isFound = Not _
             IsNumeric(Mid(SourceString, CurrentFound + NumberLength, 1))
       Case StringLength - NumberLength + 1   ' NumberString found at the end.
           isFound = Not _
             IsNumeric(Mid(SourceString, CurrentFound - 1, 1))
       Case Else               ' NumberString found in the middle.
           isFound = Not _
             IsNumeric(Mid(SourceString, CurrentFound + NumberLength, 1)) _
             And Not IsNumeric(Mid(SourceString, CurrentFound - 1, 1))
   End Select
Return

End Function