我有像Tv.Show.Name.S01E02.somthing.not.needed.avi这样的文件名
如何提取它以获得单独的变量,如:
答案 0 :(得分:7)
我会这样做:
if (preg_match("'^(.+)\.S([0-9]+)E([0-9]+).*$'i",$filename,$n))
{
$name = preg_replace("'\.'"," ",$n[1]);
$season = intval($n[2],10);
$episode = intval($n[3],10);
}