Question

Hi all im trying to build a regular expression which will sustain the following criteria.

Word to be censored in this example “view”.
Character to be used after censor: “%”, since “*” mess up my post formatting.

Examples of word use:
view

views
preview
I went to see the great view
The view was great wasn’t it.

Example after word censor:
%%%%
views
preview
I went to see the great %%%%
The %%%% was great wasn’t it.

Here is some code I have:

$string = preg_replace_callback('/\s*'. preg_quote($word, '\\') .'\s*/is', 'bbcode_callback_censored', $string);

Trouble is this matches everything right now since i use “*” in the regex ater “\s”. Any ideas what I could do to fulfill my criteria?

Answer 1

Don’t match for whitespace, use a word boundary

Try

$string = preg_replace_callback('/\b'. preg_quote($word, '\\') .'\b/is', 'bbcode_callback_censored', $string);

See it here on Regexr

You just need to make sure that the content of $word does not start or end with a non word character, then the word boundary will not work.

\b is a word boundary. It matches on a change from a word character (as defined in \w) to a non word character as defined in \W, or the other way round.

Alternative: whitespace boundary

If you don’t like the word boundary because it is possible that your word to replace starts or end with non word characters like “#view”, define your own “whitespace boundary”, e.g. like this:

(?<=^|\s)#view(?=$|\s)

See it here on Regexr

Would look in your code like this

$string = preg_replace_callback('/(?<=^|\s)'. preg_quote($word, '\\') .'(?=$|\s)/is', 'bbcode_callback_censored', $string);

(?<=^|\s) will match if there is the Start of the String or whitespace before

(?=$|\s) will match if there is the end of the String or whitespace ahead