Preg Replace - replace second occurance of a match

Question

I am relatively new to php, and hope someone can help me with a replace regex, or maybe a match replace I am not exactly sure.

I want to automatically bold the (seco

Answer 1

The regular expression itself cannot count, and the preg_ functions provide little help. You need a workaround. If you were to actually search for just a word, you might want to use string functions. Otherwise try:

// just counting
if (7 >= preg_match_all($pattern, $subject, $matches)) {

   $cb_num = 0;
   $subject = preg_replace_callback($pattern, "cb_ibu", $subject);
}

function cb_ibu($match) {
    global $cb_num;

    $match = $match[0];

    switch (++$cb_num) {
        case 2: return "<b>$match</b>";
        case 4: return "<i>$match</i>";
        case 7: return "<u>$match</u>";
        default: return $match;
    }
}

The trick is to have a callback which does the accounting. And there it's quite easy to add any rules.

Answer 2

You could use preg_split to split the text at the matches, apply the modifications, and then put everything back together:

$parts = preg_split('/(example)/', $str, 7, PREG_SPLIT_DELIM_CAPTURE);
if (isset($parts[3])) $parts[3] = '<b>'.$parts[3].'</b>';
if (isset($parts[7])) $parts[7] = '<i>'.$parts[7].'</i>';
if (isset($parts[13])) $parts[13] = '<u>'.$parts[13].'</u>';
$str = implode('', $parts);

The index formula for the i-th match is index = i · 2 - 1.

Answer 3

That's an interesting question. My implementation would be:

function replace_exact($word, $tag, $string, $limit) {
    $tag1 = '<'.$tag.'>';
    $tag2 = '</'.$tag.'>';
    $string = str_replace($word, $tag1.$word.$tag2, $string, 1);
    if ($limit==1) return $string;
    return str_replace($tag1.$word.$tag2,$word,$string,$limit-1);
}

Use it like this:

echo replace_exact('Example', 'b', $source_text, 2);
echo replace_exact('Example', 'i', $source_text, 4);

I don't know about how fast this will work, but it will be faster than preg_replace.