Regex with Non-capturing Group - regex

I am trying to understand Non-capturing groups in Regex.
If I have the following input:
He hit the ball. Then he ran. The crowd was cheering! How did he feel? I felt so energized!
If I want to extract the first word in each sentence, I was trying to use the match pattern:
^(\w+\b.*?)|[\.!\?]\s+(\w+)
That puts the desired output in the submatch.
Match $1
He He
. Then Then
. The The
! How How
? I I
But I was thinking that using non-capturing groups, I should be able to get them back in the match.
I tried:
^(?:\w+\b.*?)|(?:[\.!\?]\s+)(\w+)
and that yielded:
Match $1
He
. Then Then
. The The
! How How
? I I
and
^(?:\w+\b.*?)|(?:[.!\?]\s+)\w+
yielded:
Match
He
. Then
. The
! How
? I
What am I missing?
(I am testing my regex using RegExLib.com, but will then transfer it to VBA).

A simple example against string "foo":
(f)(o+)
Will yield $1 = 'f' and $2 = 'oo';
(?:f)(o+)
Here, $1 = 'oo' because you've explicitly said not to capture the first matching group. And there is no second matching group.
For your scenario, this feels about right:
(?:(\w+).*?[\.\?!] {2}?)
Note that the outermost group is a non-capturing group, while the inner group (the first word of the sentence) is capturing.

The following constructs a non-capturing group for the boundary condition, and captures the word after it with a capturing group.
(?:^|[.?!]\s*)(\w+)
It's not clear from youf question how you are applying the regex to the text, but your regular "pull out another until there are no more matches" loop should work.

This works and is simple:
([A-Z])\w*
VBA requires these flag settings:
Global = True 'Match all occurrences not just first
IgnoreCase = False 'First word of each sentence starts with a capital letter
Here's some additional hard-earned info: since your regex has at least one parenthesis set, you can use Submatches to pull out only the values in the parenthesis and ignore the rest - very useful. Here is the debug output of a function I use to get Submatches, run on your string:
theMatches.Count=5
Match='He'
Submatch Count=1
Submatch='H'
Match='Then'
Submatch Count=1
Submatch='T'
Match='The'
Submatch Count=1
Submatch='T'
Match='How'
Submatch Count=1
Submatch='H'
Match='I'
Submatch Count=1
Submatch='I'
T
Here's the call to my function that returned the above:
sText = "He hit the ball. Then he ran. The crowd was cheering! How did he feel? I felt so energized!"
sRegEx = "([A-Z])\w*"
Debug.Print ExecuteRegexCapture(sText, sRegEx, 2, 0) '3rd match, 1st Submatch
And here's the function:
'Returns Submatch specified by the passed zero-based indices:
'iMatch is which match you want,
'iSubmatch is the index within the match of the parenthesis
'containing the desired results.
Function ExecuteRegexCapture(sStringToSearch, sRegEx, iMatch, iSubmatch)
Dim oRegex As Object
Set oRegex = New RegExp
oRegex.Pattern = sRegEx
oRegex.Global = True 'True = find all matches, not just first
oRegex.IgnoreCase = False
oRegex.Multiline = True 'True = [\r\n] matches across line breaks, e.g. "([\r\n].*)" will match next line + anything on it
bDebug = True
ExecuteRegexCapture = ""
Set theMatches = oRegex.Execute(sStringToSearch)
If bDebug Then Debug.Print "theMatches.Count=" & theMatches.Count
For i = 0 To theMatches.Count - 1
If bDebug Then Debug.Print "Match='" & theMatches(i) & "'"
If bDebug Then Debug.Print " Submatch Count=" & theMatches(i).SubMatches.Count
For j = 0 To theMatches(i).SubMatches.Count - 1
If bDebug Then Debug.Print " Submatch='" & theMatches(i).SubMatches(j) & "'"
Next j
Next i
If bDebug Then Debug.Print ""
If iMatch < theMatches.Count Then
If iSubmatch < theMatches(iMatch).SubMatches.Count Then
ExecuteRegexCapture = theMatches(iMatch).SubMatches(iSubmatch)
End If
End If
End Function

Related

How to match escaped group signs {&date:dd.\{mm\}.yyyy} but not {&date:dd.{mm}.yyyy} with vba and regex

I'm trying to create a pattern for finding placeholders within a string to be able to replace them with variables later. I'm stuck on a problem to find all these placeholders within a string according to my requirement.
I already found this post, but it only helped a little:
Regex match ; but not \;
Placeholders will look like this
{&var} --> Variable stored in a dictionary --> dict("var")
{$prop} --> Property of a class cls.prop read by CallByName and PropGet
{#const} --> Some constant values by name from a function
Generally I have this pattern and it works well
Dim RegEx As Object
Set RegEx = CreateObject("VBScript.RegExp")
RegEx.pattern = "\{([#\$&])([\w\.]+)\}"
For example I have this string:
"Value of foo is '{&var}' and bar is '{$prop}'"
I get 2 matches as expected
(&)(var)
($)(prop)
I also want to add a formating part like in .Net to this expression.
String.Format("This is a date: {0:dd.mm.yyyy}", DateTime.Now());
// This is a date: 05.07.2019
String.Format("This is a date, too: {0:dd.(mm).yyyy}", DateTime.Now());
// This is a date, too: 05.(07).2019
I extended the RegEx to get that optional formatting string
Dim RegEx As Object
Set RegEx = CreateObject("VBScript.RegExp")
RegEx.pattern = "\{([#\$&])([\w\.]+):{0,1}([^\}]*)\}"
RegEx.Execute("Value of foo is '{&var:DD.MM.YYYY}' and bar is '{$prop}'")
I get 2 matches as expected
(&)(var)(DD.MM.YYYY)
($)(prop)()
At this point I noticed I have to take care for escapet "{" and "}", because maybe I want to have some brackets within the formattet result.
This does not work properly, because my pattern stops after "...{MM"
RegEx.Execute("Value of foo is '{&var:DD.{MM}.YYYY}' and bar is '{$prop}'")
It would be okay to add escape signs to the text before checking the regex:
RegEx.Execute("Value of foo is '{&var:DD.\{MM\}.YYYY}' and bar is '{$prop}'")
But how can I correctly add the negative lookbehind?
And second: How does this also works for variables, that should not be resolved, even if they have the correct syntax bus the outer bracket is escaped?
RegEx.Execute("This should not match '\{&var:DD.\{MM\}.YYYY\}' but this one '{&var:DD.\{MM\}.YYYY}'")
I hope my question is not confusing and someone can help me
Update 05.07.19 at 12:50
After the great help of #wiktor-stribiżew the result is completed.
As requested i provide some example code:
Sub testRegEx()
Debug.Print FillVariablesInText(Nothing, "Date\\\\{$var01:DD.\{MM\}.YYYY}\\\\ Var:\{$nomatch\}{$var02} Double: {#const}{$var01} rest of string")
End Sub
Function FillVariablesInText(ByRef dict As Dictionary, ByVal txt As String) As String
Const c_varPattern As String = "(?:(?:^|[^\\\n])(?:\\{2})*)\{([#&\$])([\w.]+)(?:\:([^}\\]*(?:\\.[^\}\\]*)*))?(?=\})"
Dim part As String
Dim snippets As New Collection
Dim allMatches, m
Dim i As Long, j As Long, x As Long, n As Long
' Create a RegEx object and execute pattern
Dim RegEx As Object
Set RegEx = CreateObject("VBScript.RegExp")
RegEx.pattern = c_varPattern
RegEx.MultiLine = True
RegEx.Global = True
Set allMatches = RegEx.Execute(txt)
' Start at position 1 of txt
j = 1
n = 0
For Each m In allMatches
n = n + 1
Debug.Print "(" & n & "):" & m.value
Debug.Print " [0] = " & m.SubMatches(0) ' Type [&$#]
Debug.Print " [1] = " & m.SubMatches(1) ' Name
Debug.Print " [2] = " & m.SubMatches(2) ' Format
part = "{" & m.SubMatches(0)
' Get offset for pre-match-string
x = 1 ' Index to Postion at least +1
Do While Mid(m.value, x, 2) <> part
x = x + 1
Loop
' Postition in txt
i = m.FirstIndex + x
' Anything to add to result?
If i <> j Then
snippets.Add Mid(txt, j, i - j)
End If
' Next start postition (not Index!) + 1 for lookahead-positive "}"
j = m.FirstIndex + m.Length + 2
' Here comes a function get a actual value
' e.g.: snippets.Add dict(m.SubMatches(1))
' or : snippets.Add Format(dict(m.SubMatches(1)), m.SubMatches(2))
snippets.Add "<<" & m.SubMatches(0) & m.SubMatches(1) & ">>"
Next m
' Any text at the end?
If j < Len(txt) Then
snippets.Add Mid(txt, j)
End If
' Join snippets
For i = 1 To snippets.Count
FillVariablesInText = FillVariablesInText & snippets(i)
Next
End Function
The function testRegEx gives me this result and debug print:
(1):e\\\\{$var01:DD.\{MM\}.YYYY(2):}{$var02
[0] = $
[1] = var02
[2] =
(1):e\\\\{$var01:DD.\{MM\}.YYYY
[0] = $
[1] = var01
[2] = DD.\{MM\}.YYYY
(2):}{$var02
[0] = $
[1] = var02
[2] =
(3): {#const
[0] = #
[1] = const
[2] =
(4):}{$var01
[0] = $
[1] = var01
[2] =
Date\\\\<<$var01>>\\\\ Var:\{$nomatch\}<<$var02>> Double: <<#const>><<$var01>> rest of string
You may use
((?:^|[^\\])(?:\\{2})*)\{([#$&])([\w.]+)(?::([^}\\]*(?:\\.[^}\\]*)*))?}
To make sure the consecutive matches are found, too, turn the last } into a lookahead, and when extracting matches just append it to the result, or if you need the indices increment the match length by 1:
((?:^|[^\\])(?:\\{2})*)\{([#$&])([\w.]+)(?::([^}\\]*(?:\\.[^}\\]*)*))?(?=})
^^^^^
See the regex demo and regex demo #2.
Details
((?:^|[^\\])(?:\\{2})*) - Group 1 (makes sure the { that comes next is not escaped): start of string or any char but \ followed with 0 or more double backslashes
\{ - a { char
([#$&]) - Group 2: any of the three chars
([\w.]+) - Group 3: 1 or more word or dot chars
(?::([^}\\]*(?:\\.[^}\\]*)*))? - an optional sequence of : and then Group 4:
[^}\\]* - 0 or more chars other than } and \
(?:\\.[^}\\]*)* - zero or more reptitions of a \-escaped char and then 0 or more chars other than } and \
} - a } char
Welcome to the site! If you need to only match balanced escapes, you will need something more powerful. If not --- I haven't tested this, but you could try replacing [^\}]* with [^\{\}]|\\\{|\\\}. That is, match non-braces and escaped brace sequences separately. You may need to change this depending on how you want to handle backslashes in your formatting string.

Regex - Quantifier {x,y} following nothing

I'm creating a basic text editor and I'm using regex to achieve a find and replace function. To do this I've gotten this code:
Private Function GetRegExpression() As Regex
Dim result As Regex
Dim regExString As [String]
' Get what the user entered
If TabControl1.SelectedIndex = 0 Then
regExString = txtbx_Find2.Text
ElseIf TabControl1.SelectedIndex = 1 Then
regExString = txtbx_Find.Text
End If
If chkMatchCase.Checked Then
result = New Regex(regExString)
Else
result = New Regex(regExString, RegexOptions.IgnoreCase)
End If
Return result
End Function
And this is the Find method
Private Sub FindText()
''
Dim WpfTest1 As New Spellpad.Tb
Dim ElementHost1 As System.Windows.Forms.Integration.ElementHost = frm_Menu.Controls("ElementHost1")
Dim TheTextBox As System.Windows.Controls.TextBox = CType(ElementHost1.Child, Tb).ctrl_TextBox
''
' Is this the first time find is called?
' Then make instances of RegEx and Match
If isFirstFind Then
regex = GetRegExpression()
match = regex.Match(TheTextBox.Text)
isFirstFind = False
Else
' match.NextMatch() is also ok, except in Replace
' In replace as text is changing, it is necessary to
' find again
'match = match.NextMatch();
match = regex.Match(TheTextBox.Text, match.Index + 1)
End If
' found a match?
If match.Success Then
' then select it
Dim row As Integer = TheTextBox.GetLineIndexFromCharacterIndex(TheTextBox.CaretIndex)
MoveCaretToLine(TheTextBox, row + 1)
TheTextBox.SelectionStart = match.Index
TheTextBox.SelectionLength = match.Length
Else
If TabControl1.SelectedIndex = 0 Then
MessageBox.Show([String].Format("Cannot find ""{0}"" ", txtbx_Find2.Text), Application.ProductName, MessageBoxButtons.OK, MessageBoxIcon.Information)
ElseIf TabControl1.SelectedIndex = 1 Then
MessageBox.Show([String].Format("Cannot find ""{0}"" ", txtbx_Find.Text), Application.ProductName, MessageBoxButtons.OK, MessageBoxIcon.Information)
End If
isFirstFind = True
End If
End Sub
When I run the program I get errors:
For ?, parsing "?" - Quantifier {x,y} following nothing.; and
For *, parsing "*" - Quantifier {x,y} following nothing.
It's as if I can't use these but I really need to. How can I solve this problem?
? and * are quantifiers in regular expressions:
? is used to specify that something is optional, for instance b?au can match both bau and au.
* means the group with which it binds can be repeated zero, one or multiple times: for instance ba*u can bath bu, bau, baau, baaaaaaaau,...
Now most regular expressions use {l,u} as a third pattern with l the lower bound on the number of times something is repeated, and u the upper bound on the number of occurences. So ? is replaced by {0,1} and * by {0,}.
Now if you provide them without any character before them, evidently, the regex parser doesn't know what you mean. In other words if you do (used csharp, but the ideas are generally applicable):
$ csharp
Mono C# Shell, type "help;" for help
Enter statements below.
csharp> Regex r = new Regex("fo*bar");
csharp> r.Replace("Fooobar fooobar fbar fobar","<MATCH>");
"Fooobar <MATCH> <MATCH> <MATCH>"
csharp> r.Replace("fooobar far qux fooobar quux fbar echo fobar","<MATCH>");
"<MATCH> far qux <MATCH> quux <MATCH> echo <MATCH>"
If you wish to do a "raw text find and replace", you should use string.Replace.
EDIT:
Another way to process them is by escaping special regex characters. Ironically enough, you can do this by replacing them by a regex ;).
Private Function GetRegExpression() As Regex
Dim result As Regex
Dim regExString As [String]
' Get what the user entered
If TabControl1.SelectedIndex = 0 Then
regExString = txtbx_Find2.Text
ElseIf TabControl1.SelectedIndex = 1 Then
regExString = txtbx_Find.Text
End If
'Added code
Dim baseRegex As Regex = new Regex("[\\.$^{\[(|)*+?]")
regExString = baseRegex.Replace(regExString,"\$0")
'End added code
If chkMatchCase.Checked Then
result = New Regex(regExString)
Else
result = New Regex(regExString, RegexOptions.IgnoreCase)
End If
Return result
End Function

VBS script to report AD groups - Regex pattern not working with multiple matches

Having an issue with getting a regex statement to accept two expressions.
The "re.pattern" code here works:
If UserChoice = "" Then WScript.Quit 'Detect Cancel
re.Pattern = "[^(a-z)^(0,4,5,6,7,8,9)]"
re.Global = True
re.IgnoreCase = True
if re.test( UserChoice ) then
Exit Do
End if
MsgBox "Please choose either 1, 2 or 3 ", 48, "Invalid Entry"
While the below "regex.pattern " code does not. I want to use it to format the results of a DSQUERY command where groups are collected, but I don't want any of the info after the ",", nor do i want the leading CN= that is normally collected when the following dsquery is run:
"dsquery.exe user forestroot -samid "& strInput &" | dsget user -memberof")
The string I want to format would look something like this before formatting:
CN=APP_GROUP_123,OU=Global Groups,OU=Accounts,DC=corp,DC=contoso,DC=biz
This is the result I want:
APP_GROUP_123
Set regEx = New RegExp
**regEx.Pattern = "[,.*]["CN=]"**
Result = regEx.Replace(StrLine, "")
I'm only able to get the regex to work when used individually, either
regEx.Pattern = ",."
or
regEx.Pattern = "CN="
code is nested here:
Set InputFile = FSO.OpenTextFile("Temp.txt", 1)
Set InputFile = FSO.OpenTextFile("Temp.txt", 1)
set OutPutFile = FSO.OpenTextFile(StrInput & "-Results.txt", 8, True)
do While InputFile.AtEndOfStream = False
StrLine = InputFile.ReadLine
If inStr(strLine, TaskChoice) then
Set regEx = New RegExp
regEx.Pattern = "[A-Za-z]{2}=(.+?),.*"
Result = regEx.Replace(StrLine, "")
OutputFile.write(Replace(Result,"""","")) & vbCrLf
End if
This should get you started:
str = "CN=APP_GROUP_123,OU=Global Groups,OU=Accounts,DC=corp,DC=contoso,DC=biz"
Set re = New RegExp
re.pattern = "[A-Za-z]{2}=(.+?),.*"
if re.Test(str) then
set matches = re.Execute(str)
matched_str = "Matched: " & matches(0).SubMatches(0)
Wscript.echo matched_str
else
Wscript.echo "Not a match"
end if
Output:Matched: APP_GROUP_123
The regex you need is [A-Za-z]{2}=(.+?),.*
If the match is successful, it captures everything in the parenthesis. .+? means it will match any character non-greedily up until the first comma. The ? in .+? makes the expression non-greedy. If you were to omit it, you would capture everything up to the final comma at ,DC=biz
Your regular expression "[,.*]["CN=]" doesn't work for 2 reasons:
It contains an unescaped double quote. Double quotes inside VBScript strings must be escaped by doubling them, otherwise the interpreter would interpret your expression as a string "[,.*][", followed by an (invalid) variablename CN=] (without an operator too) and the beginning of the next string (the 3rd double quote).
You misunderstand regular expression syntax. Square brackets indicate a character class. An expression [,.*] would match any single comma, period or asterisk, not a comma followed by any number of characters.
What you meant to use was an alternation, which is expressed by a pipe symbol (|), and the beginning of a string is matched by a caret (^):
regEx.Pattern = ",.*|^CN="
With that said, in your case a better approach would be using a group and replacing the whole string with just the group match:
regEx.Pattern = "^cn=(.*?),.*"
regEx.IgnoreCase = True
Result = regEx.Replace(strLine, "$1")

RegEx VBA Excel complex string

I have a function pulled from here. My problem is that I don't know what RegEx pattern I need to use to split out the following data:
+1 vorpal unholy longsword +31/+26/+21/+16 (2d6+13)
+1 vorpal flaming whip +30/+25/+20 (1d4+7 plus 1d6 fire and entangle)
2 slams +31 (1d10+12)
I want it to look like:
+1 vorpal unholy longsword, 31
+1 vorpal flaming whip, 30
2 slams, 31
Here is the VBA code that does the RegExp validation:
Public Function RXGET(ByRef find_pattern As Variant, _
ByRef within_text As Variant, _
Optional ByVal submatch As Long = 0, _
Optional ByVal start_num As Long = 0, _
Optional ByVal case_sensitive As Boolean = True) As Variant
' RXGET - Looks for a match for regular expression pattern find_pattern
' in the string within_text and returns it if found, error otherwise.
' Optional long submatch may be used to return the corresponding submatch
' if specified - otherwise the entire match is returned.
' Optional long start_num specifies the number of the character to start
' searching for in within_text. Default=0.
' Optional boolean case_sensitive makes the regex pattern case sensitive
' if true, insensitive otherwise. Default=true.
Dim objRegex As VBScript_RegExp_55.RegExp
Dim colMatch As VBScript_RegExp_55.MatchCollection
Dim vbsMatch As VBScript_RegExp_55.Match
Dim colSubMatch As VBScript_RegExp_55.SubMatches
Dim sMatchString As String
Set objRegex = New VBScript_RegExp_55.RegExp
' Initialise Regex object
With objRegex
.Global = False
' Default is case sensitive
If case_sensitive Then
.IgnoreCase = False
Else: .IgnoreCase = True
End If
.pattern = find_pattern
End With
' Return out of bounds error
If start_num >= Len(within_text) Then
RXGET = CVErr(xlErrNum)
Exit Function
End If
sMatchString = Right$(within_text, Len(within_text) - start_num)
' Create Match collection
Set colMatch = objRegex.Execute(sMatchString)
If colMatch.Count = 0 Then ' No match
RXGET = CVErr(xlErrNA)
Else
Set vbsMatch = colMatch(0)
If submatch = 0 Then ' Return match value
RXGET = vbsMatch.Value
Else
Set colSubMatch = vbsMatch.SubMatches ' Use the submatch collection
If colSubMatch.Count < submatch Then
RXGET = CVErr(xlErrNum)
Else
RXGET = CStr(colSubMatch(submatch - 1))
End If
End If
End If
End Function
I don't know about Excel but this should get you started on the RegEx:
/(?:^|, |and |or )(\+?\d?\s?[^\+]*?) (?:\+|-)(\d+)/
NOTE: There is a slight caveat here. This will also match if an element begins with + only (not being followed by a digit).
Capture groups 1 and 2 contain the strings that go left and right of your comma (if the whole pattern has index 0). So you can something like capture[1] + ', ' + capture[2] (whatever your syntax for that is).
Here is an explanation of the regex:
/(?:^|, |and |or ) # make sure that we only start looking after
# the beginning of the string, after a comma, after an
# and or after an or; the "?:" makes sure that this
# subpattern is not capturing
(\+? # a literal "+"
\d+ # at least one digit
# a literal space
[^+]*?) # arbitrarily many non-plus characters; the ? makes it
# non-greedy, otherwise it might span multiple lines
# a literal space
\+ # a literal "+"
(\d+)/ # at least one digit (and the brakets are for capturing)

Split a string according to a regexp in VBScript

I would like to split a string into an array according to a regular expression similar to what can be done with preg_split in PHP or VBScript Split function but with a regex in place of delimiter.
Using VBScript Regexp object, I can execute a regex but it returns the matches (so I get a collection of my splitters... that's not what I want)
Is there a way to do so ?
Thank you
If you can reserve a special delimiter string, i.e. a string that you can choose that will never be a part of the real input string (perhaps something like "###"), then you can use regex replacement to replace all matches of your pattern to "###", and then split on "###".
Another possibility is to use a capturing group. If your delimiter regex is, say, \d+, then you search for (.*?)\d+, and then extract what the group captured in each match (see before and after on rubular.com).
You can alway use the returned array of matches as input to the split function. You split the original string using the first match - the first part of the string is the first split, then split the remainder of the string (minus the first part and the first match)... continue until done.
I wrote this for my use. Might be what you're looking for.
Function RegSplit(szPattern, szStr)
Dim oAl, oRe, oMatches
Set oRe = New RegExp
oRe.Pattern = "^(.*)(" & szPattern & ")(.*)$"
oRe.IgnoreCase = True
oRe.Global = True
Set oAl = CreateObject("System.Collections.ArrayList")
Do
Set oMatches = oRe.Execute(szStr)
If oMatches.Count > 0 Then
oAl.Add oMatches(0).SubMatches(2)
szStr = oMatches(0).SubMatches(0)
Else
oAl.Add szStr
Exit Do
End If
Loop
oAl.Reverse
RegSplit = oAl.ToArray
End Function
'**************************************************************
Dim A
A = RegSplit("[,|;|#]", "bob,;joe;tony#bill")
WScript.Echo Join(A, vbCrLf)
Returns:
bob
joe
tony
bill
I think you can achieve this by using Execute to match on the required splitter string, but capturing all the preceding characters (after the previous match) as a group. Here is some code that could do what you want.
'// Function splits a string on matches
'// against a given string
Function SplitText(strInput,sFind)
Dim ArrOut()
'// Don't do anything if no string to be found
If len(sFind) = 0 then
redim ArrOut(0)
ArrOut(0) = strInput
SplitText = ArrOut
Exit Function
end If
'// Define regexp
Dim re
Set re = New RegExp
'// Pattern to be found - i.e. the given
'// match or the end of the string, preceded
'// by any number of characters
re.Pattern="(.*?)(?:" & sFind & "|$)"
re.IgnoreCase = True
re.Global = True
'// find all the matches >> match collection
Dim oMatches: Set oMatches = re.Execute( strInput )
'// Prepare to process
Dim oMatch
Dim ix
Dim iMax
'// Initialize the output array
iMax = oMatches.Count - 1
redim arrOut( iMax)
'// Process each match
For ix = 0 to iMax
'// get the match
Set oMatch = oMatches(ix)
'// Get the captured string that precedes the match
arrOut( ix ) = oMatch.SubMatches(0)
Next
Set re = nothing
'// Check if the last entry was empty - this
'// removes one entry if the string ended on a match
if arrOut(iMax) = "" then Redim Preserve ArrOut(iMax-1)
'// Return the processed output
SplitText = arrOut
End Function