This is the link in order to avoid to be marked as duplicate
I haven't been able to solve the issue following the solutions provided by the developers. Within this snippet of code there must be certainly a syntax error but I can't find where it is, as a matter of fact the below Heredoc statement does not work and prevents the whole code from working, furthermore if I try to run the code on my web server I have a 500 server error. I have modified my question implementing the answers into it.
Before editing this question I've tried to solve the problem on my own, but I've gotten to a blind alley. I've just added the error reporting in the beginning of the code, even if it is not very correct to reopen old questions.
   
<?php error_reporting(E_ALL); ini_set('display_errors', 1);?>
<?php
// take in the id of a director and return his/her full name
function get_director($director_id) {
 
    global $db;
 
    $query = 'SELECT 
            people_fullname 
       FROM
           people
       WHERE
           people_id = ' . $director_id;
    $result = mysql_query($query, $db) or die(mysql_error($db));
 
    $row = mysql_fetch_assoc($result);
    extract($row);
 
    return $people_fullname;
}
// take in the id of a lead actor and return his/her full name
function get_leadactor($leadactor_id) {
 
    global $db;
 
    $query = 'SELECT
            people_fullname
        FROM
            people 
        WHERE
            people_id = ' . $leadactor_id;
    $result = mysql_query($query, $db) or die(mysql_error($db));
 
    $row = mysql_fetch_assoc($result);
 extract($row);
 
    return $people_fullname;
}
// take in the id of a movie type and return the meaningful textual
// description
function get_movietype($type_id) {
 
    global $db;
 
    $query = 'SELECT 
            movietype_label
       FROM
           movietype
       WHERE
           movietype_id = ' . $type_id;
    $result = mysql_query($query, $db) or die(mysql_error($db));
 
    $row = mysql_fetch_assoc($result);
    extract($row);
 
    return $movietype_label;
}
//connect to MySQL
$db = mysql_connect('localhost', 'root', 'xxxxxxxx') or 
    die ('Unable to connect. Check your connection parameters.');
// make sure you’re using the right database
mysql_select_db('moviesite', $db) or die(mysql_error($db));
// retrieve information
$query = 'SELECT
        movie_name, movie_year, movie_director, movie_leadactor,
        movie_type
    FROM
        movie
    ORDER BY
        movie_name ASC,
        movie_year DESC';
$result = mysql_query($query, $db) or die(mysql_error($db));
// determine number of rows in returned result
$num_movies = mysql_num_rows($result);
 $table = <<<ENDHTML
 <div style="text-align: center;"> 
  <h2>Movie Review Database</h2> 
  <table border="1" cellpadding="2" cellspacing="2" 
  style="width: 70%; margin-left: auto; margin-right: auto;"> 
   <tr> 
    <th>Movie Title</th> 
    <th>Year of Release</th> 
    <th>Movie Director</th> 
    <th>Movie Lead Actor</th> 
    <th>Movie Type</th> 
   </tr>
  ENDHTML; 
   /* loop through the results */
    while ($row = mysql_fetch_assoc($result)) {
    extract($row);
    $director = get_director($movie_director);
    $leadactor = get_leadactor($movie_leadactor);
    $movietype = get_movietype($movie_type);
 $table .= <<<ENDHTML
  <tr>
    <td>$movie_name</td>
    <td>$movie_year</td>
    <td>$director</td>
    <td>$leadactor</td>
    <td>$movietype</td>
  </tr>
  ENDHTML;    
 }
$table.= <<<ENDHTML
    </table>    
  <p>$num_movies Movies</p> 
  </div> 
  ENDHTML;
echo $table;
  
?>
> this is the error that I receive
ENDHTML; /* loop through the results */ while ( = mysql_fetch_assoc(Resource id #3)) { extract(); = get_director(); = get_leadactor(); = get_movietype(); 
.= << ENDHTML; } 
     
     
    