Regex - How to replace the last 3 words of a string with PHP

Question

Trying to wrap the last 3 words in a tag

$str = \'Lorem ipsum dolor sit amet\';
$h2 = preg_replace(\'/^(?:\\w+\\s\\w+)(\\s\\w+)+/\', \'         


        

Answer 1

There is no reason to use regex here at all if you define words as being bounded by a single space. Instead you can use basic string manipulation to get the desired result.

$str = ...; // your input string
$words_with_offsets_in_key = str_word_count($str, 2);
$word_count = count($word_offsets);
if($word_count >= 3) {
    // we have at least 3 words
    // find offset of word three from end of array of words
    // grab third item from end of array
    $third_word_from_end = array_slice($words_with_offsets_in_key, $word_count - 3, 1);
    // inspect its key for offset value in original string
    $offset = key($third_word_from_end);
    // insert span into string
    $str = substr_replace ( $str , '<span>' , $offset, 0) . '</span>';
}

Answer 2

Here it is:

$h2 = preg_replace('/(\w+\s\w+\s\w+)$/', '<span>$1</span>', $str);

Since its last three words, so make the left side(from begin) as open to have the match.

Answer 3

Sabuj Hassan's treats numbers and _ as being part of a word as well, so use that if it makes sense.

Assuming "words" are letters delimited by a space:

$str = 'Lorem ipsum dolor sit amet';
$h2 = preg_replace('/([a-z]+ [a-z]+ [a-z]+)$/i', '<span>$1</span>', $str);

echo $h2;

If you want any non-whitespace considered a word then:

$h2 = preg_replace('/(\S+ \S+ \S+)$/', '<span>$1</span>', $str);