如何从 Excel 中的文本字符串中调用正确的数字？

Question

Consider the following text string:

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

My goal is to count how many left brackets ("("), commas outside brackets, and hyphens before each individual number in this string (eg, 3 left brackets in front of the number 10, 6 left brackets and 3 hyphens in front of 11).

My current solution is to first recall the remaining text string in front of each individual number, simply =LEFT(A1,(FIND("1",A1,1)-1)) , but it happens that Excel will recall the string appeared before the first "1" (ie, (*4, ), instead of recalling the remaining string from the actual number "1" in the string (ie, (*4,14)(7,15)(10,13)(9,12)-( ).

Side note, any idea on how to count the number of commas that are outside of brackets?

Help would be much appreciate!

Answer 1

If you have a version of Excel with the FILTERXML function (Windows Excel 2013+), you can use:

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

The formula creates an xml where the s nodes are what's included inside the parentheses, and the t node is everything else.

If you don't have the FILTERXML function, a VBA solution would be best. Which depends on your version of Excel, and whether it is Windows or MAC.

Answer 2

Count Chars

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