How to return the full string (entire word) when searching with wildcards in word vba

scienceguy · #1 02-13-2019, 06:44 AM

I need to search a document for "words" that start with "32P". However, the numbering after the "32P" will be different (32P1, 32P2, 32P21, etc.). I can write a macro that finds all the "32P" occurrences by using a wildcard ("32P*"). However, I can't figure out how to get the full string (or full "word"). This is what I have so far:

Code:

Sub FindWords2()
        sResponse = "32P*"
        
        If sResponse <> "" Then
            Application.ScreenUpdating = False
            With Selection
                .HomeKey Unit:=wdStory
                With .Find
                    .ClearFormatting
                    .Text = sResponse
                    .MatchWildcards = True
                    ' Loop until Word can no longer
                    ' find the search string and
                    Do While .Execute
                        MsgBox Selection.Range.Text 'This only returns "32P" and not the whole string
                        Selection.MoveRight
                    Loop
                End With
            End With
            Application.ScreenUpdating = True
        End If
End Sub

Any coaching would be greatly appreciated! Thanks in advance!

Guessed · #2 02-13-2019, 03:55 PM

Try

Code:

Sub FindWords2()
  Dim aRng As Range, sResponse As String
  sResponse = "<32P[0-9]{1,}"
  Set aRng = ActiveDocument.Range
  If sResponse <> "" Then
    With aRng.Find
      .ClearFormatting
      .Text = sResponse
      .MatchWildcards = True
      Do While .Execute  ' Loop until Word can no longer find the search string
        MsgBox aRng.Text
      Loop
    End With
    Application.ScreenUpdating = True
  End If
End Sub

scienceguy · #3 02-14-2019, 04:34 AM

Thank you, Andrew! This is great! I really appreciate your time and efforts! This is such a great service.

scienceguy · #4 02-14-2019, 10:49 AM

Hi Andrew, I discovered today that the section numbering has changed. Instead of 32P123, it is now 3.2.P.1.2.3, where the (period + number) after the "P" may repeat one or more times. In addition, the "P" may be a "P" or "S". I was trying to adapt your regular expression to the change. I thought the revised expression would be "<3.2.[PS](\.[0-9])@" but the output only shows the first period and number after the letter (P or S). In other words, 3.2.P.1.2.3 only shows 3.2.P.1. Can you tell me what my regular expression should be, please? Sorry for the trouble!

macropod · #5 02-14-2019, 01:55 PM

Change your Find expression to:
"<3.2.[PS].[.0-9]{1,}"
Do note that this will pick up terminating periods (e.g. where your string is at the end of a sentence), so you'll need to code for that. See, for example: http://www.vbaexpress.com/forum/show...l=1#post388141

scienceguy · #6 02-14-2019, 03:13 PM

Thanks, Paul, but unfortunately, it's still showing the same issue (3.2.P.1.2.3 only shows 3.2.P.1).

Guessed · #7 02-14-2019, 03:24 PM

scienceguy
Perhaps you haven't implemented Paul's suggestion correctly. Note that there is a dot inside the bracket that says [.0-9]. If you didn't include that dot then you would be getting the result you describe.

scienceguy · #8 02-14-2019, 03:33 PM

So sorry! Yes, you're right. I did mistype. Many thanks again!

macropod · #9 02-14-2019, 07:14 PM

Cross-posted at: https://stackoverflow.com/questions/...vba-assistance
For cross-posting etiquette, please read: http://www.excelguru.ca/content.php?184

scienceguy · #10 03-20-2019, 05:42 PM

Hi All,

My application has evolved. Earlier I was searching for a string, such as 3.2.P.1.2.3, where the (period + number) after the "P" may repeat one or more times. Now, it can be:
[2 or 3] + [period or no period] + [1 or 2 or 3] + [period or no period] + [P or S or A] + [period or no period] + [number], where the period (or no period) + number after the P,S or A may repeat any number of times. Some examples would be:

3.2.P.1
2.3.S.1.2
3.2.A.1.2.3.4.5
32P1
23S12
32A12345

I thought the expression would be "<[32].[231].[PSA].[0-9]{1,}"; however, this is only picking up strings that have periods in them. I thought the periods in this expression would represent any (or no) character. Would you help me with the expression, please?

Thanks again,
Roy

macropod · #11 03-20-2019, 05:59 PM

Change your Find expression to:
"<[23][0-9APS.]@[0-9]>"

scienceguy · #12 03-20-2019, 06:11 PM

Hi Paul,

Thanks! It got close. Here are the sample results:

3.2.S.3.1 resulted in 3.2
3.2.S.5 resulted in 3.2
32S1 gave 32S1 (worked)

I also just realized it pulled out 3.2 and 3.1 for 3.2.S.3.1.

Can we get the rest that the first two truncated, please?

Thanks,
Roy

macropod · #13 03-20-2019, 06:29 PM

Try:
"<[23][0-9APS.]{1,}"

scienceguy · #14 03-23-2019, 03:27 AM

Thank you, Paul, for the continued help. After a couple of days of testing, your solution did find the examples I mentioned above. Unfortunately, it found others, too (e.g. 3.5.P). I simply wrote a routine to ignore the ones that we don't want.

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