Regex in Word VBA

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What I am trying to do is find text that matches this regex:

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(The string ARTICLE followed by a space, followed by a digit, followed by a space, followed by an em dash, followed by a space, followed by any number of words, then a newline. An example is "ARTICLE 5 – DESCRIPTION...")

And this regex:

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(An integer, followed by a period, followed by another integer with an optional period and integer, followed by any text, and a newline. An example is "2.1.3 Subsection 2...")

If text matches, the macro should apply a style that includes heading level so I can generate a Table of Contents and populate the navigation pane. If my regex is incorrect, please correct me.

Thank you!
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To give you an idea of the syntax, this will find up to Description...
(ARTICLE )[0-9]( )^=( )<[0-9A-z ]*>

Need to turn Wild Cards on in the search
And this:

(ARTICLE )[0-9]( )^+( )<[0-9A-z ]*[^13]

will find the first paragraph (including marker) starting with "Article #"

em dash is ^+
en dash is ^=

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And this should do the second part...
[0-9]{1,}[.][0-9]{1,}[*][! ]( )<[0-9A-z ]*[^13]
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Patrick MatthewsCommented:
You mention VBA, so I assume that you are implementing the VBScript regex objects.  In that case, your pattern string for the 2nd example should be:


Not that DrTribos's recommendation is for using the search function built into Word, which is accessible via VBA.
Ahhhh... I was wondering.  I've never used RegEx.
indigo6Author Commented:
DrTribos, actually that's fine, I was wondering about what the wildcard syntax would be! I tweaked the second one to be (ARTICLE )[0-9]( )^=( )<[0-9A-z ]*[^13] since it was an En dash, and I had to run a second round with (ARTICLE )[0-9][0-9]( )^=( )<[0-9A-z ]*[^13] to detect article numbers 10 and larger. But that worked great!

     The second one has me perplexed though. It detected strings like the following:
2.2 Access
2.7 Employee Time Off
21.11.3 Class Size (Unit 3)

But not:

5.2 No Lockout
5.4 Savings Clause

From what I can tell, they are formatted the same way. I'm not sure why it does not find these.

matthewspatrick, thanks for the VBA regex! How would I write a macro that changes the style of strings that match that regex?

Sorry I thought I posted this yesterday - apparently it failed.  Here is some code to make the changes using the wild card...
Sub UpdateSections()

Dim para As Paragraph
Dim str As String

str = "(ARTICLE )[0-9]( )^=( )<[0-9A-z ]*[^13]"

    With Selection.Find
        .Text = str
        .Replacement.Text = ""
        .Forward = True
        .Font.Name = "Arial Narrow"
        .Font.Size = 24
        .MatchWildcards = True
        .Forward = True
        .Replacement.Style = "Heading 1"
        .Execute Replace:=wdReplaceAll
    End With

End Sub

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You'll have to change the font names and sizes to suit... :-)
Gah! My other comment is also missing - regarding the strange results... perhaps there is a double space or something confusing the issue.  Can you copy a few of the offending lines and paste them into a sample document?

Also, not sure if you are in the habit of using show - hide... the backward P thing on the home tab of the ribbon (paragraph marker in the paragraph section)... this will show spaces and might make diagnosis easier :-)
Another possible reason that the text is not detected is because there is a soft return instead of a hard return... soft return looks like the symbol on the keyboard for the enter button. Hard return is the 'backward P' paragraph marker....  The wild card search is explicitly looking for a hard return (^13), not sure how other characters would impact...
indigo6Author Commented:
They seem to all be hard returns. Any other reasons why this could be?
Pls copy the offending paragraphs to a sample document and upload.
Sorry for the brevity - on phone
indigo6Author Commented:
I would have to strip too much of the info out, plus the requirements changed, so I'll close this question. It got me moving in the right direction though. Thanks!
No worries - glad to help :-)
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