Regex: how to match an word that doesn't end with a specific character

Question

I would like to match the whole "word"—one that starts with a number character and that may include special characters but does not end with a '%'.

Match these:

• 112 (whole numbers)
• 10-12 (ranges)
• 11/2 (fractions)
• 11.2 (decimal numbers)
• 1,200 (thousand separator)

but not

• 12% (percentages)
• A38 (words starting with a alphabetic character)

I've tried these regular expressions:

(\b\p{N}\S)*)

but that returns '12%' in '12%'

(\b\p{N}(?:(?!%)\S)*)

but that returns '12' in '12%'

Can I make an exception to the \S term that disregards %? Or will have to do something else?

I'll be using it in PHP, but just write as you would like and I'll convert it to PHP.

Answer 1

This matches your specification:

\b\p{N}\S*+(?<!%)

Explanation:

\b       # Start of number
\p{N}    # One Digit
\S*+     # Any number of non-space characters, match possessively
(?<!%)   # Last character must not be a %

The possessive quantifier \S*+ makes sure that the regex engine will not backtrack into a string of non-space characters it has already matched. Therefore, it will not "give back" a % to match 12 within 12%.

Of course, that will also match 1!abc, so you might want to be more specific than \S which matches anything that's not a whitespace character.

Answer 2

Can i make an exception to the \S term that disregards %

Yes you can:

[^%\s]

See this expression \b\d[^%\s]* here on Regexr