I am looking for a solution to validate and highlight my cell in case false.
I tried the most promising solution: Regex. But still can not find the pattern I need.
My latest attempt was this pattern: "[A-Z-0-9_.]" This works only if the cell contains only a symbol and nothing else, if the symbol is part of a string it does not work.
Problem is that it does not catch cells that have an odd character in a string of text: Example C4UNIT| or B$GROUP.
Specification Cell can contain only capital characters and two allowed symbols Dash - and Underbar _
This is my complete code:
Function ValidateCellContent()
Sheets("MTO DATA").Select
Dim RangeToCheck As Range
Dim CellinRangeToCheck As Range
Dim CollNumberFirst As Integer
Dim CollNumberLast As Integer
Dim RowNumberFirst As Integer
Dim RowNumberLast As Integer
'--Start on Column "1" and Row "3"
CollNumberFirst = 1
RowNumberFirst = 3
'--Find last Column used on row "2" (Write OMI Headings)
CollNumberLast = Cells(2, Columns.count).End(xlToLeft).Column
RowNumberLast = Cells(Rows.count, 1).End(xlUp).Row
'--Set value of the used range of cell addresses like: "A3:K85"
Set RangeToCheck = Range(Chr(64 + CollNumberFirst) & RowNumberFirst & ":" & Chr(64 + CollNumberLast) & RowNumberLast)
Debug.Print "Cells used in active Range = " & (Chr(64 + CollNumberFirst) & RowNumberFirst & ":" & Chr(64 + CollNumberLast) & RowNumberLast)
For Each CellinRangeToCheck In RangeToCheck
Debug.Print "CellinRangeToCheck value = " & CellinRangeToCheck
If Len(CellinRangeToCheck.Text) > 0 Then
'--Non Printables (Space,Line Feed,Carriage Return)
If InStr(CellinRangeToCheck, " ") _
Or InStr(CellinRangeToCheck, Chr(10)) > 0 _
Or InStr(CellinRangeToCheck, Chr(13)) > 0 Then
CellinRangeToCheck.Font.Color = vbRed
CellinRangeToCheck.Font.Bold = True
'--Allowed Characters
ElseIf Not CellinRangeToCheck.Text Like "*[A-Z-0-9_.]*" Then
CellinRangeToCheck.Font.Color = vbRed
CellinRangeToCheck.Font.Bold = True
Else
CellinRangeToCheck.Font.Color = vbBlack
CellinRangeToCheck.Font.Bold = False
End If
End If
Next CellinRangeToCheck
End Function
Try this:
Option Explicit
Private Sub Worksheet_Change(ByVal Target As Range)
'we want only validate when cell content changed, if whole range is involved (i.e. more than 1 cell) then exit sub
If Target.Cells.Count > 1 Then Exit Sub
'if there is error in a cell, also color it red
If IsError(Target) Then
Target.Interior.ColorIndex = 3
Exit Sub
End If
'validate cell with our function, if cell content is valid, it'll return True
'if it i s not valid, then color cell red
If Not ValidateText(Target.Value) Then
Target.Interior.ColorIndex = 3
End If
End Sub
Function ValidateText(ByVal txt As String) As Boolean
Dim i As Long, char As String
'loop through all characters in string
For i = 1 To Len(txt)
char = Mid(txt, i, 1)
If Not ((Asc(char) >= 65 And Asc(char) <= 90) Or char = "-" Or char = "_") Then
'once we come upon invalid character, we can finish the function with False result
ValidateText = False
Exit Function
End If
Next
ValidateText = True
End Function
I've originally assumed you wanted to use RegEx to solve your problem. As per your comment you instead seem to be using the Like operator.
Like operator
While Like accepts character ranges that may resemble regular expressions, there are many differences and few similarities between the two:
Like uses ! to negate a character range instead of the ^ used in RegEx.
Like does not allow/know quantifiers after the closing bracket ] and thus always matches a single character per pair of brackets []. To match multiple characters you need to add multiple copies of your character range brackets.
Like does not understand advanced concepts like capturing groups or lookahead / lookbehind
probably more differences...
The unavailability of quantifiers leaves Like in a really bad spot for your problem. You always need to have one character range to compare to for each character in your cell's text. As such the only way I can see to make use of the Like operator would be as follows:
Private Function IsTextValid(ByVal stringToValidate As String) As Boolean
Dim CharValidationPattern As String
CharValidationPattern = "[A-Z0-9._-]"
Dim StringValidationPattern As String
StringValidationPattern = RepeatString(CharValidationPattern, Len(stringToValidate))
IsTextValid = stringToValidate Like StringValidationPattern
End Function
Private Function RepeatString(ByVal stringToRepeat As String, ByVal repetitions As Long) As String
Dim Result As String
Dim i As Long
For i = 1 To repetitions
Result = Result & stringToRepeat
Next i
RepeatString = Result
End Function
You can then pass the text you want to check to IsTextValid like that:
If IsTextValid("A.ASDZ-054_93") Then Debug.Print "Hurray, it's valid!"
As per your comment, a small Worksheet_Change event to place into the worksheet module of your respective worksheet. (You will also need to place the above two functions there. Alternatively you can make them public and place them in a standard module.):
Private Sub Worksheet_Change(ByVal Target As Range)
Dim ValidationRange As Range
Set ValidationRange = Me.Range("A2:D5")
Dim TargetCell As Range
For Each TargetCell In Target.Cells
' Only work on cells falling into the ValidationRange
If Not Intersect(TargetCell, ValidationRange) Is Nothing Then
If IsTextValid(TargetCell.Text) Then
TargetCell.Font.Color = vbBlack
TargetCell.Font.Bold = False
Else
TargetCell.Font.Color = vbRed
TargetCell.Font.Bold = True
End If
End If
Next TargetCell
End Sub
Regular Expressions
If you want to continue down the RegEx road, try this expression:
[^A-Z0-9_-]+
It will generate a match, whenever a passed-in string contains one or more characters you don't want. All cells with only valid characters should not return a match.
Explanation:
A-Z will match all capital letters,
0-9 will match all numbers,
_- will match underscore and dash symbols.
The preceding ^ will negate the whole character set, meaning the RegEx only matches characters not in the set.
The following + tells the RegEx engine to match one or more characters of the aforementioned set. You only want to match your input, if there is at least one illegal char in there. And if there are more than one, it should still match.
Once in place, adapting the system to changing requirements (different chars considered legal) is as easy as switching out a few characters between the [brackets].
See a live example online.
Related
I would like to convert a range of numbers (and single digits) from a number-only format to alpha-numeric format. Entire statement is in a single, excel cell and would like the converted version to be in a neighboring cell.
As an example:
Assuming 1-24=B1-B24
Assuming 25-48=C1-C24
INPUT—
screen 1-3,5,7-9,11-30,32-37,39-40,41,44-46
DESIRED OUTPUT (all acceptable)
screen B1-B3,B5,B7-B9,B11-C6,C8-C13,C15-C16,C17,C20-C22
OR
screen B1-B3,B5,B7-B9,B11-B24,C1-C6,C8-C13,C15-C16,C17,C20-C22
OR
screen B1-B3,B5,B7-B9,B11-B24
screen C1-C6,C8-C13,C15-C16,C17,C20-C22
Using excel functions is proving quite cumbersome so excel macro would be better. I've looked for examples of requested conversion but haven't found anything.
Any help is greatly appreciated.
Cheers,
Bob
Hey here is a solution that i tested out. Not sure if "screen" needs to be in the string or not. Let me know and I will tweak it if that's the case.
Its a user defined function. So drop this vba in a module and then go to a worksheet and type in "=AlphaConvert(" + the cell reference.
Assumption here is that only one cell will be referenced at a time.
Last this could easily be converted to a sub routine and probably run a bit faster than the function.
Public Function AlphaConvert(TargetCell As Range)
Dim v As Long
Dim vArr() As String
Dim i As Long
Dim iArr() As String
Dim a As String
vArr = Split(TargetCell.Value, ",")
For v = LBound(vArr) To UBound(vArr)
If InStr(vArr(v), "-") > 0 Then
iArr = Split(vArr(v), "-")
For i = LBound(iArr) To UBound(iArr)
If i = LBound(iArr) Then
a = AlphaCode(iArr(i))
Else
a = a & "-" & AlphaCode(iArr(i))
End If
Next i
vArr(v) = a
Else
vArr(v) = AlphaCode(vArr(v))
End If
If v = LBound(vArr) Then
AlphaConvert = vArr(v)
Else
AlphaConvert = AlphaConvert & "," & vArr(v)
End If
Next v
End Function
Private Function AlphaCode(Nbr As Variant)
Select Case Nbr
Case 1 To 24
AlphaCode = "B" & Nbr
Case Else
AlphaCode = "C" & Nbr - 24
End Select
End Function
I am writing a VBA formula to check that all characters in a cell "TestChars" are allowed, where allowed means that each character appears in a list defined by another cell "AllowedChars". To make things even harder, I would like this formula to work on ranges of cells rather than on a single cell.
The current code seems to work:
Option Explicit
Public Function AllCharsValid(InputCells As Range, AllowedChars As String) As Boolean
' Check that all characters in InputCells are among
' the characters in AllowedChars
Dim Char As String
Dim Index As Integer
Dim RangeTestChars As Range
Dim TestChars As String
For Each RangeTestChars In InputCells
TestChars = RangeTestChars.Value
For Index = 1 To Len(TestChars)
Char = Mid(TestChars, Index, 1)
If InStr(AllowedChars, Char) = 0 Then
AllCharsValid = False
Exit Function
End If
Next Index
Next RangeTestChars
AllCharsValid = True
End Function
I have the following questions:
The formula takes a range and returns a single boolean. I would prefer a vectorized function, where, given an input range, you get a corresponding range of booleans. It seems like built-in formulas like 'EXACT' can do this (those formulas where you have to press ctrl-shift-enter to execute them and where you get curly-brackets). Is there a way to do that with user-defined functions?
I am not new to programming, however I am completely new to VBA (I started literally today). Is there any obvious problem, weirdness with the above code?
Are there special characters, extremely long texts or particular input values that would cause the formula to fail?
Is there an easier way to achieve the same effect? Is the code slow?
when you start typing built-in formulas in excel you get suggestions and auto-completion. This doesn't seem to work with my formula, am I asking for too much or is it possible to achieve this?
I realize that this question contains several weakly related sub-questions, so I would be very happy also with sub-answers.
The following code will return a range of boolean values offset one column from the initial input range. Simply create a new tab in Excel and run testAllCharsValid and show the Immediate window in the IDE to see how it works.
Sub testAllCharsValid()
Dim i As Integer
Dim cll As Range, rng As Range
Dim allowedChars As String
' insert test values in sheet: for testing purposes only
With ActiveSheet ' change to Thisworkbook.Sheets("NameOfYourSheet")
Set rng = .Range("A1:A10")
For i = 1 To 10
.Cells(i, 1) = Chr(i + 92)
Next i
End With
' fill allowedChars with letters a to z: for testing purposes only
For i = 97 To 122
allowedChars = allowedChars & Chr(i)
Next i
' get boolean range
Set rng = AllCharsValid(rng, allowedChars)
' check if the returned range contains the expected boolean values
i = 0
For Each cll In rng
i = i + 1
Debug.Print i & " boolean value: " & cll.Value
Next cll
End Sub
' Check that all characters in InputCells are among
' the characters in AllowedChars
Public Function AllCharsValid(InputCells As Range, allowedChars As String) As Range
Dim BoolTest As Boolean
Dim Char As String
Dim Index As Integer
Dim RangeTestChars As Range, RangeBooleans As Range, RangeTemp As Range
Dim TestChars As String
For Each RangeTestChars In InputCells
BoolTest = True
TestChars = RangeTestChars.Value
For Index = 1 To Len(TestChars)
Char = Mid(TestChars, Index, 1)
If InStr(allowedChars, Char) = 0 Then BoolTest = False
Next Index
Set RangeTemp = RangeTestChars.Offset(0, 1) ' change offset to what suits your purpose
RangeTemp.Value = BoolTest
If RangeBooleans Is Nothing Then
Set RangeBooleans = RangeTestChars
Else
Set RangeBooleans = Union(RangeBooleans, RangeTemp)
End If
Next RangeTestChars
Set AllCharsValid = RangeBooleans
End Function
cf 2) If the length of the test string is zero, the function will return True for the cell in question, which may not be desirable.
cf 3) There is a limit to how many characters an Excel cell can contain, read more here. I suppose, if you concatenated some very long strings and sent them to the function, you could reach the integer limit of +32767, which would cause a run-time error due to the integer Index variable. However, since the character limit of Excel cells is exactly +32767, the function should work as is without any problems.
cf 4) None that I know of.
cf 5) This is not the easiest thing to achieve, but there is help to be found here.
For example cell "A1" is linked to cell "B1", so in formula bar for cell "A1" we have:
=B1
How can I check whether value in cell "A1" contains letter B?
I tried the following:
Dim Criteria_3 As Boolean
Dim Value As Range
Set Value = Selection
Dim x As Variant
Set x = Cells
Dim text As String
For Each x In Value
If IsNumeric(x) Then
Criteria_3 = VBA.InStr(1, x.Formula, text) > 0
As soon as value of "Text" is "" it does not work and I really struggle to fined the right solution.
your question is not really conclusive, so here are two options:
To check wheter the value contains B:
blnCheck = 0 < InStr(1, rngCell.Value, "B")
To check wheter the Formula contains B:
blnCheck = 0 < InStr(1, rngCell.Formula, "B")
Regarding your null string problem:
As soon as value of "Text" is "" it does not work and I really struggle to fined the right solution.
That's because you're using VBA.InStr(1, x.Formula, text) and in this case 1 is an invalid index on a string of length 0. You can omit that, or you can code around it like:
If Len(Trim(x.Formula)) = 0 Then
'## Do nothing
Else
Criteria_3 = VBA.InStr(1, x.Formula, text) > 0
End If
To your specific question of identifying when a value contains any alpha character(s):
You can use a function like this to test whether a value contains any letter, by evaluating the Ascii code for each character, and break when True:
Function ContainsAnyLetter(val) As Boolean
Dim ret As Boolean
Dim str$, ch$
Dim i As Long
str = LCase(CStr(val))
For i = 1 To Len(str)
ch = Mid(str, i, 1)
If 97 <= Asc(ch) And Asc(ch) <= 122 Then
ret = True
Exit For
End If
Next
ContainsAnyLetter = ret
End Function
In your code, you could call it like:
Criteria_3 = ContainsAnyLetter(x.Value) '## or x.Formula, depending on your needs
You can use LIKE
https://msdn.microsoft.com/en-us/library/swf8kaxw.aspx
Something like if rngCell.value like "*B*" then
if your goal is to check whether the cell contains any valid range reference, then you could go like this
Option Explicit
Sub main()
Dim cell As Range
For Each cell In Worksheets("Sheet001").Range("A1:A20") '<== jus a test range, set it as per your needs
MsgBox IsCellReference(cell.Formula)
Next cell
End Sub
Function IsCellReference(text As String) As Boolean
On Error Resume Next
IsCellReference = Not Range(Replace(text, "=", "")) Is Nothing
End Function
I've got a column which contains rows that have parameters in them. For example
W2 = [PROD][FO][2.0][Customer]
W3 = [PROD][GD][1.0][P3]
W4 = Issues in production for customer
I have a function that is copying other columns into another sheet, however for this column, I need to do the following
Search the cell and look for [P*]
The asterisk represents a number between 1 and 5
If it finds [P*] then copy P* to the sheet "Calculations" in column 4
Basically, remove everything from the cell except where there is a square bracket, followed by P, a number and a square bracket
Does anyone know how I can do this? Alternatively, it might be easier to copy the column across and then remove everything that doesn't meet the above criteria.
Second Edit:
I edited here to use a regular expression instead of a loop. This may be the most efficient method to achieve your goal. See below and let us know if it works for you:
Function MatchWithRegex(sInput As String) As String
Dim oReg As Object
Dim sOutput As String
Set oReg = CreateObject("VBScript.RegExp")
With oReg
.Pattern = "[[](P[1-5])[]]"
End With
If oReg.test(sInput) Then
sOutput = oReg.Execute(sInput)(0).Submatches(0)
Else
sOutput = ""
End If
MatchWithRegex = sOutput
End Function
Sub test2()
Dim a As String
a = MatchWithRegex(Range("A1").Value)
If a = vbNullString Then
MsgBox "None"
Else
MsgBox MatchWithRegex(Range("A1").Value)
End If
End Sub
First EDIT:
My solution would be something as follows. I'd write a function that first tests if the Pattern exists in the string, then if it does, I'd split it based on brackets, and choose the bracket that matches the pattern. Let me know if that works for you.
Function ExtractPNumber(sInput As String) As String
Dim aValues
Dim sOutput As String
sOutput = ""
If sInput Like "*[[]P[1-5][]]*" Then
aValues = Split(sInput, "[")
For Each aVal In aValues
If aVal Like "P[1-5][]]*" Then
sOutput = aVal
End If
Next aVal
End If
ExtractPNumber = Left(sOutput, 2)
End Function
Sub TestFunction()
Dim sPValue As String
sPValue = ExtractPNumber(Range("A2").Value)
If sPValue = vbNullString Then
'Do nothing or input whatever business logic you want
Else
Sheet2.Range("A1").Value = sPValue
End If
End Sub
OLD POST:
In VBA, you can use the Like Operator with a Pattern to represent an Open Bracket, the letter P, any number from 1-5, then a Closed Bracket using the below syntax:
Range("A1").Value LIke "*[[]P[1-5][]]*"
EDIT: Fixed faulty solution
If you're ok with blanks and don't care if *>5, I would do this and copy down column 4:
=IF(ISNUMBER(SEARCH("[P?]",FirstSheet!$W2)), FirstSheet!$W2, "")
Important things to note:
? is the wildcard symbol for a single character; you can use * if you're ok with multiple characters at that location
will display cell's original value if found, leave blank otherwise
Afterwards, you can highlight the column and remove blanks if needed. Alternatively, you can replace the blank with a placeholder string.
If * must be 1-5, use two columns, E and D, respectively:
=MID(FirstSheet!$W2,SEARCH("[P",FirstSheet!$W2)+2,1)
=IF(AND(ISNUMBER($E2),$E2>0,$E2<=5,MID($W2,SEARCH("[P",FirstSheet!$W2)+3,1))), FirstSheet!$W2, "")
where FirstSheet is the name of your initial sheet.
Let's assume that we have one module with only one Sub in it, and there are no comments. How to identify all variable names ? Is it possible to identify names of variables which are not defined using Dim ? I would like to identify them and replace each with some random name to obfuscate my code (O0011011010100101 for example), replace part is much easier.
List of characters which could be use in names of macros, functions and variables :
ABCDEFGHIJKLMNOPQRSTUVWXYZdefghijklmnopqrstuvwxyzg€‚„…†‡‰Š‹ŚŤŽŹ‘’“”•–—™š›śťžź ˇ˘Ł¤Ą¦§¨©Ş«¬®Ż°±˛ł´µ¶·¸ąş»Ľ˝ľżŔÁÂĂÄĹĆÇČÉĘËĚÍÎĎĐŃŇÓÔŐÖ×ŘŮÚŰÜÝŢßŕáâăäĺćçčéęëěíîďđńňóôőö÷řůúűüýţ˙ÉĘËĚÍÎĎĐŃŇÓÔŐÖ×ŘŮÚŰÜÝŢßŕáâăäĺćçčéęëěíîďđńňóôőö÷řůúűüýţ˙
Below are my function I've wrote recenlty :
Function randomName(n as integer) as string
y="O"
For i = 2 To n:
If Rnd() > 0.5 Then
y = y & "0"
Else
y = y & "1"
End If
Next i
randomName=y
End Function
In goal to replace given strings in another string which represent the code of module I use below sub :
Sub substituteNames()
'count lines in "Module1" which is part of current workbook
linesCount = ActiveWorkbook.VBProject.VBComponents("Module1").CodeModule.CountOfLines
'read code from module
code = ActiveWorkbook.VBProject.VBComponents("Module1").CodeModule.Lines(StartLine:=1, Count:=linesCount)
inputStr = Array("name1", "name2", "name2") 'some hardwritten array with string to replace
namesLength = 20 'length of new variables names
For i = LBound(inputStr) To UBound(inputStr)
outputString = randomName(namesLength-1)
code = Replace(code, inputStr(i), outputString)
Next i
Debug.Print code 'view code
End Sub
then we simply substitute old code with new one, but how to identify strings with names of variables ?
Edition
Using **Option Explicit ** decrease safety of my simple method of obfuscation, because to reverse changes you only have to follow Dim statements and replace ugly names with something normal. Except that to make such substitution harder, I think it's good idea to break the line in the middle of variable name :
O0O000O0OO0O0000 _
0O00000O0OO0
the simple method is also replacing some strings with chains based on chr functions chr(104)&chr(101)&chr(108)&chr(108)&chr(111) :
Sub stringIntoChrChain()
strInput = "hello"
strOutput = ""
For i = 1 To Len(strInput)
strOutput = strOutput & "chr(" & Asc(Mid(strInput, i, 1)) & ")&"
Next i
Debug.Print Mid(strOutput, 1, Len(strOutput) - 1)
End Sub
comments like below could make impression on user and make him think that he does not poses right tool to deal with macro etc.:
'(k=Äó¬)w}ż^¦ů‡ÜOyúm=ěËnóÚŽb W™ÄQó’ (—*-ĹTIäb
'R“ąNPÔKZMţ†üÍQ‡
'y6ű˛Š˛ŁŽ¬=iýQ|˛^˙ ‡ńb ¬ĂÇr'ń‡e˘źäžŇ/âéç;1qýěĂj$&E!V?¶ßšÍ´cĆ$Âű׺Ůî’ﲦŔ?TáÄu[nG¦•¸î»éüĽ˙xVPĚ.|
'ÖĚ/łó®Üă9Ę]ż/ĹÍT¶Mµę¶mÍ
'q[—qëýY~Pc©=jÍ8˘‡,Ú+ń8ŐűŻEüńWü1ďëDZ†ć}ęńwŠbŢ,>ó’Űçµ™Š_…qÝăt±+‡ĽČgřÍ!·eŠP âńđ:ŶOážű?őë®ÁšńýĎáËTbž}|Ö…ăË[®™
You can use a regular expression to find variable assignments by looking for the equals sign. You'll need to add a reference to the Microsoft VBScript Regular Expressions 5.5 and Microsoft Visual Basic for Applications Extensibility 5.3 libraries as I've used early binding.
Please be sure to back up your work and test this before using it. I could have gotten the regex wrong.
UPDATE:
I've refined the regular expressions so that it no longer catches datatypes of strongly typed constants (Const ImAConstant As String = "Oh Noes!" previously returned String). I've also added another regex to return those constants as well. The last version of the regex also mistakenly caught things like .Global = true. That was corrected. The code below should return all variable and constant names for a given code module. The regular expressions still aren't perfect, as you'll note that I was unable to stop false positives on double quotes. Also, my array handling could be done better.
Sub printVars()
Dim linesCount As Long
Dim code As String
Dim vbPrj As VBIDE.VBProject
Dim codeMod As VBIDE.CodeModule
Dim regex As VBScript_RegExp_55.RegExp
Dim m As VBScript_RegExp_55.match
Dim matches As VBScript_RegExp_55.MatchCollection
Dim i As Long
Dim j As Long
Dim isInDatatypes As Boolean
Dim isInVariables As Boolean
Dim datatypes() As String
Dim variables() As String
Set vbPrj = VBE.ActiveVBProject
Set codeMod = vbPrj.VBComponents("Module1").CodeModule
code = codeMod.Lines(1, codeMod.CountOfLines)
Set regex = New RegExp
With regex
.Global = True ' match all instances
.IgnoreCase = True
.MultiLine = True ' "code" var contains multiple lines
.Pattern = "(\sAs\s)([\w]*)(?=\s)" ' get list of datatypes we've used
' match any whole word after the word " As "
Set matches = .Execute(code)
End With
ReDim datatypes(matches.count - 1)
For i = 0 To matches.count - 1
datatypes(i) = matches(i).SubMatches(1) ' return second submatch so we don't get the word " As " in our array
Next i
With regex
.Pattern = "(\s)([^\.\s][\w]*)(?=\s\=)" ' list of variables
' begins with a space; next character is not a period (handles "with" assignments) or space; any alphanumeric character; repeat until... space
Set matches = .Execute(code)
End With
ReDim variables(matches.count - 1)
For i = 0 To matches.count - 1
isInDatatypes = False
isInVariables = False
' check to see if current match is a datatype
For j = LBound(datatypes) To UBound(datatypes)
If matches(i).SubMatches(1) = datatypes(j) Then
isInDatatypes = True
Exit For
End If
'Debug.Print matches(i).SubMatches(1)
Next j
' check to see if we already have this variable
For j = LBound(variables) To i
If matches(i).SubMatches(1) = variables(j) Then
isInVariables = True
Exit For
End If
Next j
' add to variables array
If Not isInDatatypes And Not isInVariables Then
variables(i) = matches(i).SubMatches(1)
End If
Next i
With regex
.Pattern = "(\sConst\s)(.*)(?=\sAs\s)" 'strongly typed constants
' match anything between the words " Const " and " As "
Set matches = .Execute(code)
End With
For i = 0 To matches.count - 1
'add one slot to end of array
j = UBound(variables) + 1
ReDim Preserve variables(j)
variables(j) = matches(i).SubMatches(1) ' again, return the second submatch
Next i
' print variables to immediate window
For i = LBound(variables) To UBound(variables)
If variables(i) <> "" And variables(i) <> Chr(34) Then ' for the life of me I just can't get the regex to not match doublequotes
Debug.Print variables(i)
End If
Next i
End Sub