Reputation: 66
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; }
Upvotes: 0
Views: 131
Reputation: 177701
How about
$num_movies = mysql_num_rows($result);
?><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>
<?php /* 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);
echo "<tr><td>$movie_name</td><td>$movie_year</td><td>$director</td><td>$leadactor</td><td>$movietype</td></tr>";
}
echo "</table><p>$num_movies Movies</p></div>";
?>
Upvotes: 1
Reputation: 74217
There should not be any spaces before your closing identifier ENDHTML;
which clearly contains 2 spaces for each of them.
Error reporting would have caught that syntax error.
Read up on heredoc:
Warning It is very important to note that the line with the closing identifier must contain no other characters, except a semicolon (;). That means especially that the identifier may not be indented, and there may not be any spaces or tabs before or after the semicolon. It's also important to realize that the first character before the closing identifier must be a newline as defined by the local operating system. This is \n on UNIX systems, including Mac OS X. The closing delimiter must also be followed by a newline.
Footnotes:
mysql_*
functions deprecation notice:
http://www.php.net/manual/en/intro.mysql.php
This extension is deprecated as of PHP 5.5.0, and is not recommended for writing new code as it will be removed in the future. Instead, either the mysqli or PDO_MySQL extension should be used. See also the MySQL API Overview for further help while choosing a MySQL API.
These functions allow you to access MySQL database servers. More information about MySQL can be found at » http://www.mysql.com/.
Documentation for MySQL can be found at » http://dev.mysql.com/doc/.
Upvotes: 1