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.

Undersheriff

Do these numbers appear in some context? Spaces surrounding etc.? Specifically why did you use the trailing \S (which means non-space)? – Lely

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.

Demob

+1, but if you make the \S* possessive - \S*+ - you can get rid of the (?=\s|$). – Biradial

Thanks. This works perfectly. I've extended it a bit to fit some other cases that i discovered. (\b\p{N}\S*+(?<!%|%\p{P}|%\p{P}\p{P})) So it disregards '12%.', '11%-12%', '( 12%)', and '( 12%).' In my data % is followed by a max of two punctations, but i sure the code above can be converted to something more general if needed. – Undersheriff

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

Ocker

But he said that only the last character must not be a %. – Demob

@TimPietzcker thats true, then I showed only the way to exclude a specific character from a predefined class, because the correct answer is already there from you (+1) – Ocker

Yeah, he asked for that (exclusion of a single character), too. So we don't really know what he actually needs :) +1! – Demob

Thanks. I will definitely be using this a lot more. – Undersheriff

Answer 3

KISS (restrictive):

/[0-9][0-9.,-/]*\s/

Skycap

That will match 12 in 12%. – Demob

I assume the OP is looking for "numbers" so "7ISNOTANUMBER" shouldn't match, ergo there should be some kind of separator (that I assumed to be whitespace). – Skycap

Answer 4

\d+([-/\.,]\d+)?(?!%)

Explanation:

\d+        one or more digits
(
   [-/\.,]     one "-", "/", "." or ","
   \d+         one or more digits
)?         the group above zero or one times
(?!%)      not followed by a "%" (negative lookahead)

Vallombrosa

+1. Judging by the examples provided, I suspect this is the answer to the question the OP should have asked. ;) – Biradial

@AlanMoore: Careful, this will match 1 in 12%. The quantifiers should also be made possessive. Then it will work. – Demob

@Tim: Can I recover from this gaffe by saying I assumed there would be word boundaries or anchors around the regex? :P – Biradial

Answer 5

try this one

preg_match("/^[0-9].*[^%]$/", $string);

Enfield

That fails to match single-digit numbers. – Demob

Answer 6

Try this PCRE regex:

/^(\d[^%]+)$/

It should give you what you need.

Antiperistalsis

Only the last character cannot be a percent. – Orthogenesis

Answer 7

I would suggest just:

(\b[\p{N},.-]++(?!%))

That's not very exact regarding decimal delimiters or ranges. (As example). But the ++ possessive quantifier will eat up as many decimals as it can. So that you really just need to check the following character with a simple assertion. Did work for your examples.

Lely

OP did say the first character had to be a number, and you're not enforcing that. The leading \b doesn't do the trick; it will match (e.g.) the comma in foo,bar. – Biradial

Like I said, not very exact. Correct range and fractions treatment has been answered by etuardu. – Lely