Extracting Text Between Brackets with Regex - regex

Question

In sentences like:
"[x] Alpha
[33] Beta"
I extract an array of bracketed data as ([x], [33])
using VBA regex Pattern:
"(\[x\])|(\[\d*\])"
I cannot extract directly the array of un-bracketed data as (x, 33)
using web resources advice for pattern
"(?<=\[)(.*?)(?=\])"
Is this a VBA specific problem (i.e. limits on its implementation of Regex)
or did I misunderstand 'looking forward and backward' patterns?
Public Function Regx( _
ByVal SourceString As String, _
ByVal Pattern As String, _
Optional ByVal IgnoreCase As Boolean = True, _
Optional ByVal MultiLine As Boolean = True, _
Optional ByVal MatchGlobal As Boolean = True) _
As Variant
Dim oMatch As Match
Dim arrMatches
Dim lngCount As Long
' Initialize to an empty array
arrMatches = Array()
With New RegExp
.MultiLine = MultiLine
.IgnoreCase = IgnoreCase
.Global = MatchGlobal
.Pattern = Pattern
For Each oMatch In .Execute(SourceString)
ReDim Preserve arrMatches(lngCount)
arrMatches(lngCount) = oMatch.Value
lngCount = lngCount + 1
Next
End With
Sub testabove()
Call Regx("[x] Alpha" & Chr(13) & _
"[33] Beta", "(\[x\])|(\[\d*\])")
End Sub

Answer 1

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 2

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

Answer 3

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 4

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