提取文件名，並重定向URL

Question

我有名字2011-08-21-filename.php

我想提取的日期和標題的文件，所以永久會像/2011/08/21/filename

Answer 1

$filename = "2011-08-21-filename.php"; 
preg_match('/(\d{4})-(\d{2})-(\d{2})-(\w*).php/', $filename, $matches); 
$path = implode("/", array_slice($matches, 1)); // => string(19) "2011/08/21/filename" 

Answer 2

$filename = str_replace(".php", "", preg_replace("/[0-9-]/", "", $string)); 
$directory = $dir.$filename; 

快速的解決方案，對不起，我不是那麼好用正則表達式