句子比較：忽略單詞

Question

我需要一些與句子比較的幫助。

$answer = "This is the (correct and) acceptable answer. Content inside the parenthesis are ignored if not present in the user's answer. If it is present, it should not count against them."; 
    $response = "This is the correct and acceptable answer. Content inside the parenthesis are ignored if not present in the user's answer. If it is present, it should not count against them."; 

    echo "<strong>Acceptable Answer:</strong>"; 
    echo "<pre style='white-space:normal;'>$answer</pre><hr/>"; 
    echo "<strong>User's Answer:</strong>"; 
    echo "<pre>".$response."</pre>"; 

    // strip content in brackets 
    $answer = preg_replace("/\([^)]*\)|[()]/", "", $answer); 

    // strip punctuation 
    $answer = preg_replace("/[^a-zA-Z 0-9]+/", " ", $answer); 
    $response = preg_replace("/[^a-zA-Z 0-9]+/", " ", $response); 

    $common = similar_text($answer, $response, $percent); 
    $orgcount = strlen($answer); 
    printf("The user's response has %d/$orgcount characters in common (%.2f%%).", $common, $percent); 

基本上我想要做的是忽略父母的話。例如，在$ answer字符串中，正確，而位於括號內 - 因此，我不希望這些單詞再次計算用戶的回覆。因此，如果用戶擁有這些詞語，則不會對他們進行計數。如果用戶沒有這些詞，那麼它就不算對他們。

這可能嗎？

Answer 1

感謝評論，我寫了一個解決方案，因爲它是一個「長」的過程，我儘管把它放在一個函數中。 編輯：調試後，它傳出strpos()是造成一些麻煩，如果位置是0，所以我增加了一個OR聲明：

$answer = "(This) is the (correct and) acceptable answer. (random this will not count) Content inside the parenthesis are ignored if not present in the user's answer. If it is present, it should not count against them."; 
$response = "This is the correct and acceptable answer. Content inside the parenthesis are ignored if not present in the user's answer. If it is present, it should not count against them."; 

echo 'The user\'s response has '.round(compare($answer, $response),2).'% characters in common'; // The user's response has 100% characters in common 

function compare($answer, $response){ 
    preg_match_all('/\((?P<parenthesis>[^\)]+)\)/', $answer, $parenthesis); 

    $catch = $parenthesis['parenthesis']; 
    foreach($catch as $words){ 
     if(!strpos($response, $words) === false || strpos($response, $words) === 0){ // if it does exist then remove brackets 
      $answer = str_replace('('.$words.')', $words, $answer); 
     }else{ //if it does not exist remove the brackets with the words 
      $answer = str_replace('('.$words.')', '', $answer); 
     } 
    } 
    /* To sanitize */ 
    $answer = preg_replace(array('/[^a-zA-Z0-9]+/', '/ +/'), array(' ', ' '), $answer); 
    $response = preg_replace(array('/[^a-zA-Z 0-9]+/', '/ +/'), array(' ', ' '), $response); 
    $common = similar_text($answer, $response, $percent); 
    return($percent); 
}