Вграждане на код в друг код

[Blagoicho]

Здравейте.
Може ли някой да вгради този код:

<form action='' method='post'> 
Име:<br> 
<input type='text' name='user'><br>
Коментар:<br> 
<textarea name='answers' cols='20' rows='3'></textarea>
<br> 
<input name='submit' type='submit' value='Изпрати'> 
</form>
<?php 
$id = $_GET['id']; 
if(isset($_POST['submit'])){ 
$date = date("d.m.Y"); 
$sql = mysql_query("INSERT INTO answers (id, user, answers, date) VALUES (NULL, '$id', '$_POST[user]', '$_POST[answers]', '$date')") or die(mysql_error()); 
} 
else { 
echo ""; 
} 
?>
<?php 
$query = mysql_query("SELECT * FROM `answers` WHERE `user`='$id' ORDER BY id DESC") or die(mysql_error()); 
while($row = mysql_fetch_array($query)){ 
echo " 
Отговорил: $row[user] на $row[date]<br> 
$row[answers]<br> <br>"; 
} 
?>

в този

<?php 
$connect = mysql_connect("localhost", "root", ""); 
$select_db = mysql_select_db("db-to"); 
mysql_query('set names cp1251',$connect); 
$id=(int)$_GET['id']; 
$id=abs($id); 
$id=htmlspecialchars(addslashes(trim($id))); 

if ($id < 1) 
{ 
$query = "SELECT * FROM `topics` ORDER BY id DESC"; 
$result=mysql_query($query); 
while($r = mysql_fetch_array($result)) { 
extract($r); 
echo "<a href='viewtopic.php?id=$id'>$name</a><br>"; 
}
} 
else 
{ 
if($_GET['id']){ 
$topicid = $_GET['id']; 
$query1 = "SELECT * FROM `topics` where id = '$topicid'"; 
$result1=mysql_query($query1); 
while($r = mysql_fetch_array($result1)) { 
extract($r); 
$post = nl2br("$post"); 
echo "
Заглавие: <b>$name</b> <br>
Пуснал: <b><i>$user</i></b> <br>
Дата: $date <br>
Тема:<br> $post
";
}
} 
}  
?>

Благодаря