<?
header("Content-Type: text/html; charset=UTF-8");
$db = mysql_connect("localhost","db79","password");
mysql_select_db("d_lx",$db);
mysql_query("SET NAMES utf8");
$result = mysql_query("SELECT * FROM videoByMe", $db);
$num_rows = mysql_num_rows($result);
$idVideo1 = rand (3,$num_rows);
?>
<!DOCTYPE HTML>
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
<html lang="ro">
<title>All Video</title>
<link rel="stylesheet" href="http://site.com/style/bats.css">
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1/jquery.min.js"></script>
<script src="http://www.site.ru/player6/jwplayer.js"></script>
</head>
<body>
<div class="header">
<div class="inner-header">
<a href="#"><img src="http://site.com/image/logo.png" style="width: 70px; margin-bottom: 5px; float: left;">
<a href="#" class="menu">
Home
</a>
<div class="vertical-line"></div>
<a href="#" class="menu">
Reg
</a>
<div class="vertical-line"></div>
<a href="#" class="menu">
Add video
</a>
</div>
</div>
<?
for ($i = 3; $i <= $num_rows; $i++) {
$resultat = mysql_query("SELECT * FROM `videoByMe` WHERE `id` = $i",$db);
$data = mysql_fetch_array($resultat);
$nameVideo = $data['DenumireaVideo'];
$IdvideoMIX = $data['VideoMixID'];
$linkToImage = $data['linkPoza'];
echo '<a href="index.php?id='.$i.'" style="color: #333; text-decoration: none;">
<div class="videos">
<img src="'.$linkToImage.'" style="width: 285px; height: 150px;">
<div style="font-size: 16px; margin-top: 5px; font-weight: bolder;">'.$nameVideo.'</div>
</div>
</a>';
}
mysql_close($db);
?>
</div>
</div>
</div>
</body>
</html>
некоторый фото не показывает .
Средний
Простой
