带有if / else条件的php preg_match_all

Question

I'm using the following 'badword filter' and it works -

    $question = Trim(stripslashes($_POST['Message']));     
    $badwords = array("caca","poopoo","pipi");
    $matches = array();
    $matchFound = preg_match_all("/\b(" . implode($badwords,"|") . ")\b/i", $question, $matches);
    if ($matchFound){
        $words = array_unique($matches[0]);
        foreach($words as $word){
            echo '<span class="word">'.$word.'</span>'; //outputs <span class="word">caca</span> <span class="word">poopoo</span> <span class="word">pipi</span>
        }
    }

My question is, I would like to add an if/else condition where if 3+ badwords are found, do this, and if not, do that. Pseudocode:

    if ($severalMatchesFound > 2){
        echo '3 or more badwords found';
    } else {
        echo 'less than 3 badwords found';
    }

Answer 1

$question = Trim(stripslashes($_POST['Message']));     
$badwords = array("caca","poopoo","pipi");
preg_match_all("/\b(" . implode($badwords,"|") . ")\b/i", $question, $matches);
if (count($matches[1]) >= 3){
    // three words or more than three words

    $words = array_unique($matches[1]);
    foreach($words as $word){
        echo '<span class="word">'.$word.'</span>'; //outputs <span class="word">caca</span> <span class="word">poopoo</span> <span class="word">pipi</span>
    }
}elseif(count($matches[1])>=1){
  //more than one but less than three
}else{
  // no words found
}

Sorry forgot [1] on count. I think the elseif would have worked as I wrote it but isset is probably better