Hi, i em a newbie just started whit php and MySQL.
My problem is whit code from book.
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="content-type" content="text/html; charset=windows-1251" />
<title>PHP</title>
</head>
<body>
<?php
$link = mysql_connect("*", "*", "*") or die("Server connection");
mysql_select_db("moviesite");
$query = "SELECT movie_name, movie_director, movie_leadactor " .
"FROM movie";
$results = mysql_query($query, $link) or die(mysql_error());
$num_movies = mysql_num_rows($results);
$movie_header =<<<EOD
<h2><center>Movie Review Database</center></h2>
<table width="70%" border="1" cellpadding="2" cellspasing="2" align="center">
<tr>
<th>Movie Title</th>
<th>Year of Release</th>
<th>Movie Director</th>
<th>Movie Lead Actor</th>
<th>Movie Type</th>
</tr>
EOD;
$movie_details = " ";
while($row = mysql_fetch_array($results)) {
$movie_name = $row["movie_name"];
$movie_director = $row["movie_director"];
$movie_leadactor = $row["movie_leadactor"];
$movie_details .=<<<EOD
<tr>
<td>$movie_name</td>
<td>$movie_director</td>
<td>$movie_leadactor</td>
</tr>
EOD;
}
$movie_details .=<<<EOD
<tr>
<td> </td>
</tr>
<tr>
<td>Total : $num_movies Movies</td>
</tr>
EOD;
$movie_footer = "</table>";
$movie =<<<MOVIE
$movie_header;
$movie_details;
$movie_footer;
MOVIE;
echo "Има $num_movies филма в нашата базадани";
echo $movie;
echo "<br />";
echo "<br />";
echo "Успешно извеждане на информация !<br />";
echo "<a href='index.php'>Начало >></a>";
?>
</body>
</html>
When i view the page show :
Parse error: syntax error, unexpected $end in C:\Programing\Web\wamp\wamp\www\PHP\proba_1\table2.php on line 85
..... line 85 is '</html>' the last line of the code !
What can be my problem ?
PHP 5 and MySQL 5.0.27-community-nt