talevtehnology
Registered
Здравейте как мога да направя този скрипт като се щракне на свали да отваря клипа ето кода:
Код:
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<title>Сваляне от Vbox7</title>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8">
</head>
<body>
<?php
error_reporting(E_ALL ^ E_WARNING ^ E_NOTICE);
if($_POST['submit'] and strlen($_POST['url'])>0){
$file = $_POST['url'];
$number = explode(":", $file);
$i = substr($number[2],0 ,2);
if(fopen("http://media.vbox7.com/s/$i/$number[2].flv","r")){
$url = "http://media.vbox7.com/s/$i/$number[2].flv";
} else {
for($ii = 2; $ii <= 21; $ii++)
{
if($ii < 10)
{
if(fopen("http://media0$ii.vbox7.com/s/$i/$number[2].flv", "r"))
{
$url = "http://media0$ii.vbox7.com/s/$i/$number[2].flv";
}
} else {
if(fopen("http://media$ii.vbox7.com/s/$i/$number[2].flv", "r"))
{
$url = "http://media$ii.vbox7.com/s/$i/$number[2].flv";
}
}
}
}
echo "<center><b><font size='6'><a style='text-decoration:none; color:#aaa;' href='".$url."'>Свали видеото!</a></font></center>";
} else {
echo '<center><form method="post" action="">
<b>Линк към клипа (http://vbox7.com/play:ed7caef8)</b><br>
<input type="text" name="url" size="35" value=""><br>
<input type="submit" name="submit" style="font-style: bold;" value="Download">
</form></center><br />';
}
?>
</body>
</html>