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o kaip man padaryti kad is mysql bazes i sita vieta ikeltu link


pvz kad butu taip

$url = zippy("zippyshare nuoroda is mysql");


cia visas kodas
function zippy($url) {
$id = explode("v/",$url);
if(@$id[1]) {
$id = explode("/file",$id[1]);
$id = $id[0];
$server = explode("//www",$url);
$server = explode(".zipp",$server[1]);
$server = $server[0];
}else{
$id = explode("key%3D",$url);
$id = $id[1];
$server = explode("//www",$url);
$server = explode(".zipp",$server[1]);
$server = $server[0];
}
return array($server,$id);
}

$file = $url[1];
$www = $url[0];