CODEWITHSUNDEEP
phpregexcolors
Mav
Mav

Reputation: 1190

Match a number only when it is not part of a hex color code PHP

I'm trying to match numbers and replace them with (matched number)px

The thing is that I want to match them ONLY when they're not part of a hex color code. My input can contain hex color codes in two ways: #xxx or #xxxxxx where the x can be a letter from a-f or a number from 0-9.

The regex I currently have is this:

$input = preg_replace('/(?<!#..)(\d)(?!px)/i', '$1px', $input);

This works only with a 3 digit hex code and that too only when the digit is at the third place.

I want something applicable in all situations. This should replace only those numbers that are not part of a hex code and don't already have a px after them. Thanks!

EDIT: since negative lookbehind can't contain an indefinite number of characters (no quantifiers that is) I have no idea what to do.

The input and output should be like:

input: #da4 10 output: #da4 10px

input: #122222 10 output: #122222 10px

input: #4444dd 20px output: #4444dd 20px

input 30 10 20 20 #414 20 99 #da4 output: 30px 10px 20px 20px #414 20px 99px #da4

Upvotes: 0

Views: 68

Answers (2)

Lucas Araujo
Lucas Araujo

Reputation: 1688

Regex: 

\b(?<!#)\d+\b
# \b     Assert position at a word boundary 
# (?<!#) Negative Lookbehind
# \d+    Match a number with 1 to ilimited digits 
# \b     Assert position at a word boundary 

$input = '30 10 20 20 #414 20 99 #da4 #122222 10 #4444dd 20px';
$input = preg_replace('/\b(?<!#)\d+\b/', '$0px', $input);
print($input);

Code example

Upvotes: 1

Aran-Fey
Aran-Fey

Reputation: 43166

You can use (?<!#)\b(\d+)\b(?!px) (replace with $1px). Demo.

Explanation:

(?<!#)   make sure this isn't hex
\b       make sure we're matching the whole number, not just a part of it
(\d+)    capture the number
\b       again, make sure we've captured the whole number
(?!px)   make sure there's no px

Upvotes: 1

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