Extracting Text Between Brackets with Regex

Question

In sentences like:

\"[x] Alpha

[33] Beta\"

I extract an array of bracketed data as ([x], [33])

using VBA regex Pattern:

\"(\\         


        

Answer 1

Try this:

\[(x)\]|\[(\d*)\]

What you don't want to be captured, don't put them inside (). this is used for grouping

Explanation

You will get x and 33 in $1 and $2

Dot Net Sample

Alright, I prepared it for you , although far away from vb for long. Lots of it might be not needed, yet it might help you to understand it better

Imports System.Text.RegularExpressions

Module Example
   Public Sub Main()
      Dim text As String = "[x] Alpha      [33] Beta]"
      Dim pattern As String = "\[(x)\]|\[(\d*)\]"

      ' Instantiate the regular expression object.
      Dim r As Regex = new Regex(pattern, RegexOptions.IgnoreCase)

      ' Match the regular expression pattern against a text string.
      Dim m As Match = r.Match(text)
      Dim matchcount as Integer = 0
      Do While m.Success
         matchCount += 1
         Console.WriteLine("Match" & (matchCount))
         Dim i As Integer
         For i = 1 to 2
            Dim g as Group = m.Groups(i)
            Console.WriteLine("Group" & i & "='" & g.ToString() & "'")
            Dim cc As CaptureCollection = g.Captures
            Dim j As Integer 
            For j = 0 to cc.Count - 1
              Dim c As Capture = cc(j)
               Console.WriteLine("Capture" & j & "='" & c.ToString() _
                  & "', Position=" & c.Index)
            Next 
         Next 
         m = m.NextMatch()
      Loop
   End Sub
End Module

Answer 2

Use capturing around the subpatterns that will fetch you your required value.

Use

"\[(x)\]|\[(\d*)\]"

(or \d+ if you need to match at least 1 digit, as * means zero or more occurrences, and + means one or more occurrences).

Or, use the generic pattern to extract anything inside the square brackets without the brackets:

"\[([^\][]+)]"

Then, access the right Submatches index by checking the submatch length (since you have an alternation, either of the submatch will be empty), and there you go. Just change your for loop with

For Each oMatch In .Execute(SourceString)
    ReDim Preserve arrMatches(lngCount)
    If Len(oMatch.SubMatches(0)) > 0 Then
        arrMatches(lngCount) = oMatch.SubMatches(0)
    Else
        arrMatches(lngCount) = oMatch.SubMatches(1)
    End If
    ' Debug.Print arrMatches(lngCount) ' - This outputs x and 33 with your data
    lngCount = lngCount + 1
Next

Answer 3

Array Without Regex:

For Each Value In Split(SourceString, Chr(13))
  ReDim Preserve arrMatches(lngCount)
  arrMatches(lngCount) = Split(Split(Value, "]")(0), "[")(1)
  lngCount = lngCount + 1
Next

Answer 4

With Excel and VBA you can strip the brackets after the regex extraction:

Sub qwerty()

    Dim inpt As String, outpt As String
    Dim MColl As MatchCollection, temp2 As String
    Dim regex As RegExp, L As Long

    inpt = "38c6v5hrk[x]537fhvvb"

    Set regex = New RegExp
    regex.Pattern = "(\[x\])|(\[\d*\])"
    Set MColl = regex.Execute(inpt)
    temp2 = MColl(0).Value

    L = Len(temp2) - 2
    outpt = Mid(temp2, 2, L)

    MsgBox inpt & vbCrLf & outpt
End Sub