How do I isolate a space using RegExp in VBA (\\s vs. \\p{Zs})?

Introduction/Question:

I have been studying the use of Regular Expressions (using VBA/Excel), and so far I cannot understand how I would isolate a <space> (or " ") using regexp from other white space characters that are included in \s. I thought that I would be able to use \p{Zs}, but in my testing so far, it has not worked out. Could someone please correct my misunderstanding? I appreciate any helpful input.

To offer proper credit, I modified some code that started off as a very helpful post by @Portland Runner that is found here: How to use Regular Expressions (Regex) in Microsoft Excel both in-cell and loops

This has been my approach/study so far:

Using the string "14z-16z Flavored Peanuts", I've been trying to write a RegExp which removes "14z-16z " and leaves only "Flavored Peanuts". I initially used ^[0-9](\S)+ as strPattern and a sub procedure with following snippets:

Sub REGEXP_TEST_SPACE()

Dim strPattern As String
Dim strReplace As String
Dim strInput As String
Dim regEx As New RegExp

strInput = "14z-16z Flavored Peanuts"
strPattern = "^[0-9](\S)+"
strReplace = ""

With regEx
    .Global = True
    .MultiLine = True
    .IgnoreCase = True
    .pattern = strPattern
End With

If regEx.Test(strInput) Then
    Range("A1").Value = regEx.Replace(strInput, strReplace)
End If

End Sub

This approach gave me an A1 value of " Flavored Peanuts" (note the leading <space> in that string).

I then changed strPattern = "^[0-9](\S)+(\s)" (added the (\s)), which gave me the desired A1 value of "Flavored Peanuts". Great!!! I got the desired output!

But as I understand it, \s represents all white-space characters, equal to [ \f\n\r\t\v]. In this case, I know that the character is just a normal, single space -- I don't need carriage return, horizontal tab, etc. So I tried to see if I could just isolate the <space> character in regex (unicode separator: space), which I believe is \p{Zs} (e.g., strPattern = "^[0-9](\S)+(\p{Zs})"). Using this pattern, however, doesn't return a match whatsoever, nevermind removing the leading space. I also tried the more general \p{Z} (all unicode separators), but that didn't work either.

Clearly I have missed something in my study. Help is both desired and appreciated. Thank you.

Since you are trying to find a correspondence with the \p{Zs} Unicode category class, you might want to also handle all hard spaces. This code will be helpful:

strPattern = "^[0-9](\S)+[ " & ChrW(160) & "]"

Or,

strPattern = "^[0-9](\S+)[ \x0A]"

The [ \x0A] character class will match either a regular space or a hard, non-breaking space.

If you need to match all kinds of spaces, you can use this regex pattern taken based on the information on https://www.cs.tut.fi/~jkorpela/chars/spaces.html:

strPattern = "^[0-9](\S)+[ \xA0\u1680\u180E\u2000-\u200B\u202F\u205F\u3000\uFEFF]"

This is the table with code point explanations:

U+0020  32  SPACE   foo bar Depends on font, typically 1/4 em, often adjusted
U+00A0  160 NO-BREAK SPACE  foo bar As a space, but often not adjusted
U+1680  5760    OGHAM SPACE MARK    foo bar Unspecified; usually not really a space but a dash
U+180E  6158    MONGOLIAN VOWEL SEPARATOR   foo᠎bar No width
U+2000  8192    EN QUAD foo bar 1 en (= 1/2 em)
U+2001  8193    EM QUAD foo bar 1 em (nominally, the height of the font)
U+2002  8194    EN SPACE    foo bar 1 en (= 1/2 em)
U+2003  8195    EM SPACE    foo bar 1 em
U+2004  8196    THREE-PER-EM SPACE  foo bar 1/3 em
U+2005  8197    FOUR-PER-EM SPACE   foo bar 1/4 em
U+2006  8198    SIX-PER-EM SPACE    foo bar 1/6 em
U+2007  8199    FIGURE SPACE    foo bar “Tabular width”, the width of digits
U+2008  8200    PUNCTUATION SPACE   foo bar The width of a period “.”
U+2009  8201    THIN SPACE  foo bar 1/5 em (or sometimes 1/6 em)
U+200A  8202    HAIR SPACE  foo bar Narrower than THIN SPACE
U+200B  8203    ZERO WIDTH SPACE    foobar Nominally no width, but may expand
U+202F  8239    NARROW NO-BREAK SPACE   foo bar Narrower than NO-BREAK SPACE (or SPACE)
U+205F  8287    MEDIUM MATHEMATICAL SPACE   foo bar 4/18 em
U+3000  12288   IDEOGRAPHIC SPACE   foo　bar The width of ideographic (CJK) characters.
U+FEFF  65279   ZERO WIDTH NO-BREAK SPACE

Best regards.

You can explicitly include a white space in your RegEx pattern. The following pattern works just fine