Regex-pattern

gmaxey · #1 01-06-2025, 09:27 AM

Quote:

Originally Posted by batman1

That's right. For example, can you provide a simple pattern in Word Find with the following task: "Find in the sequence all numbers consisting of 4 or 5 digits, whose first 4 digits must have at least 1 digit 2"

batman1,

I suppose the skills that someone is familiar with has bearing on what is simple or not. A simple find pattern? No, and one may not exists. However, with VBA your task is easy enough:

Code:

Sub ScratchMacro()
'A basic Word Macro coded by Gregory K. Maxey
Dim oRng As Range
  Set oRng = ActiveDocument.Range
  With oRng.Find
    .Text = "[0-9]{1,}"
    .MatchWildcards = True
    While .Execute
      Select Case Len(oRng.Text)
        Case 4
          If InStr(oRng.Text, "2") > 0 Then Debug.Print oRng.Text
        Case 5
          If InStr(Left(oRng.Text, 4), "2") > 0 Then Debug.Print oRng.Text
      End Select
    Wend
  End With
lbl_Exit:
  Exit Sub
End Sub

Sample Numbers:
12345 – Match

56789 – No Match
123 – No Match
1234 – Match
4321 – Match
8945 – No Match
98765 – No Match
124567 – No match
54312 – No match

batman1 · #2 01-06-2025, 11:04 AM

Quote:

Originally Posted by gmaxey

batman1,

I suppose the skills that someone is familiar with has bearing on what is simple or not. A simple find pattern? No, and one may not exists. However, with VBA your task is easy enough:

1. There is nothing that says that a number that has 4 or 5 digits has to be a word. So it can be part of a word. Besides, your code will find 3452 in the string "a3452b" (and correctly), so it should find 12456 (1245 also meets the criteria, but we want the longest result possible) in the string "124567".

2. Every case, well almost every case, can be solved in VBA. The only thing is that the solution is relatively short. In the current code, after .Execute you have to use Select.

Besides, if data = "13452234567" the code will not find 34522

vivka · #3 01-06-2025, 11:16 AM

Or another presentation of Greg's idea:

Code:

Sub ScratchMacro()
'Coded by Gregory K. Maxey
'https://www.msofficeforums.com/184580-post31.html

 Dim oRng As range
  Set oRng = ActiveDocument.range
  With oRng.Find
    .text = "<[0-9]{4,5}>"
    .MatchWildcards = True
    While .Execute
        If InStr(oRng.text, "2") > 0 Then MsgBox oRng.text
    Wend
  End With
lbl_Exit:
  Exit Sub
  End Sub

However I had situations when regex helped me out, unlike VBA. Maybe it was because of my poor VBA knowledge.
Greg and Batman1, thank you for another good lesson!

batman1 · #4 01-06-2025, 11:33 AM

Quote:

Originally Posted by vivka

Or another presentation of Greg's idea:

Code:

Sub ScratchMacro()
'Coded by Gregory K. Maxey
'https://www.msofficeforums.com/184580-post31.html

 Dim oRng As range
  Set oRng = ActiveDocument.range
  With oRng.Find
    .text = "<[0-9]{4,5}>"
    .MatchWildcards = True
    While .Execute
        If InStr(oRng.text, "2") > 0 Then MsgBox oRng.text
    Wend
  End With
lbl_Exit:
  Exit Sub
   End Sub

As I explained the number DOES NOT HAVE TO be a word, it can be part of a word.

If the data is:
a3452b
13452
then the code with .text = "[0-9]{4,5}" will find both 3452 and 13452. Meanwhile, the result 13452 is not good because 2 appears in the fifth place and not in 1-4

vivka · #5 01-06-2025, 12:19 PM

In the code

Code:

<[0-9]{4,5}>

< and > identify a 4- or 5-digit 'word' without any letters. You can test it. As for finding 2 in all positions but fifth, Greg's 2nd condition can be used. So, the final code is:

Code:

Sub ScratchMacro() 
 'Coded by Gregory K. Maxey 
'https://www.msofficeforums.com/184580-post31.html  
 Dim oRng As range   
   Set oRng = ActiveDocument.range   
   With oRng.Find     
     .text = "<[0-9]{4,5}>"     
     .MatchWildcards = True     
     While .Execute         
       If InStr(left(oRng.text, 4), "2") > 0 Then MsgBox oRng.text     
     Wend   
   End With 
lbl_Exit:   
   Exit Sub    
End Sub

However, I repeat that my (not rich) experience shows that there are situations when only regex can be helpful.

batman1 · #6 01-06-2025, 01:15 PM

Quote:

Originally Posted by vivka

In the code

Code:

<[0-9]{4,5}>

< and > identify a 4- or 5-digit 'word' without any letters. You can test it. As for finding 2 in all positions but fifth, Greg's 2nd condition can be used. So, the final code is:

Code:

Sub ScratchMacro() 
 'Coded by Gregory K. Maxey 
'https://www.msofficeforums.com/184580-post31.html  
 Dim oRng As range   
   Set oRng = ActiveDocument.range   
   With oRng.Find     
     .text = "<[0-9]{4,5}>"     
     .MatchWildcards = True     
     While .Execute         
       If InStr(left(oRng.text, 4), "2") > 0 Then MsgBox oRng.text     
     Wend   
   End With 
lbl_Exit:   
   Exit Sub    
 End Sub

I have already explained that the task does not require the number to be a word - it can be part of a word.

With data:
13452234567
a3452b
67892
123456 23556789 3456789 a234667 13452234567 a3452b 1342 222

podany kod znajdzie tylko 1342 . Tymczasem wyniki =
34522
3452
7892
12345
23556
23466
34522
3452
1342

vivka · #7 01-06-2025, 01:24 PM

That's clear as mud. In 13452234567 should the code find 34522 and in 123456 12345 ? Then what is the rule/principle of the search? If "Find in the sequence all numbers consisting of 4 or 5 digits, whose first 4 digits have at least 1 digit 2", then 13452234567 contains 3452, 2345, 23456, correct? Ok, never mind! I'm getting too obsessed with the topic and stressing you with my questions.

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