Changing formulas on the fly with VBA RegEx - regex

Question

i'm trying to change formulas in excel, i need to change the row number of the formulas.
I'm trying do use replace regex to do this. I use an loop to iterate through the rows of the excel and need to change the formula for the row that is iterating at the time. Here is an exemple of the code:
For i = 2 To rows_aux
DoEvents
Formula_string= "=IFS(N19='Z001';'xxxxxx';N19='Z007';'xxxxxx';0=0;'xxxxxxx')"
Formula_string_new = regEx.Replace(Formula_string, "$1" & i)
wb.Cells(i, 33) = ""
wb.Cells(i, 33).Formula = Formula_string_new
.
.
.
Next i
I need to replace rows references but not the ones in quotes or double quotes. Example:
If i = 2 i want the new string to be this:
"=IFS(N2='Z001';'xxxxxx';N2='Z007';'xxxxxx';0=0;'xxxxxxx')"
I'm trying to use this regex:
([a-zA-Z]+)(\d+)
But its changing everything in quotes too. Like this:
If i = 2:
"=IFS(N2='Z2';'xxxxxx';N2='Z2';'xxxxxx';0=0;'xxxxxxx')"
If anyone can help me i will be very grateful!
Thanks in advance.

Answer 1

As others have written, there are probably better ways to write this code. But for a regex that will capture just the Column letter in capturing group #1, try:
\$?\b(XF[A-D]|X[A-E][A-Z]|[A-W][A-Z]{2}|[A-Z]{2}|[A-Z])\$?(?:104857[0-6]|10485[0-6]\d|1048[0-4]\d{2}|104[0-7]\d{3}|10[0-3]\d{4}|[1-9]\d{1,5}|[1-9])d?
Note that is will NOT include the $ absolute addressing token, but could be altered if that were necessary.
Note that you can avoid the loop completely with:
Formula_string = "=IFS(N19=""Z001"",""xxxxxx"",N$19=""Z007"",""xxxxxx"",0=0,""xxxxxxx"")"
Formula_string_new = regEx.Replace(Formula_string, "$1" & firstRow)
With Range(wb.Cells(firstRow, 33), wb.Cells(lastRow, 33))
.Clear
.Formula = Formula_string_new
End With
When we write a formula to a range like this, the references will automatically adjust the way you were doing in your loop.
Depending on unstated factors, you may want to use the FormulaLocal property vice the Formula property.
Edit:
To make this a little more robust, in case there happens to be, within the quote marks, a string that exactly mimics a valid address, you can try checking to be certain that a quote (single or double) neither precedes nor follows the target.
Pattern: ([^"'])\$?\b(XF[A-D]|X[A-E][A-Z]|[A-W][A-Z]{2}|[A-Z]{2}|[A-Z])\$?(?:104857[0-6]|10485[0-6]\d|1048[0-4]\d{2}|104[0-7]\d{3}|10[0-3]\d{4}|[1-9]\d{1,5}|[1-9])d?\b(?!['"])
Replace: "$1$2" & i
However, this is not "bulletproof" as various combinations of included data might match. If it is a problem, let me know and I'll come up with something more robust.

Answer 2

If you can identify some unique features like in the example preceding bracket ( or colon ; and trailing equal = then this might work
Sub test()
Dim s As String, sNew As String, i As Long
Dim Regex As Object
Set Regex = CreateObject("vbscript.regexp")
With Regex
.Global = True
.MultiLine = False
.IgnoreCase = True
.Pattern = "([(;][a-zA-Z]{1,3})(\d+)="
End With
i = 1
s = "=IFS(NANA19='Z001';'xxxxxx';NA19='Z007';'xxxxxx';0=0;'xxxxxxx')"
sNew = Regex.Replace(s, "$1" & i & "=")
Debug.Print s & vbCr & sNew
End Sub