如何在PHP中以最佳方式从URL获取带扩展名的文件名？

Question

假设，我遵循以下网址：http://localhost/project-folder/sub-dir-1/images/products/Moulin_1407822344.jpg

现在我只想从上面的URL获得Moulin_1407822344.jpg。

我应该如何在PHP中以简短，甜蜜和最佳的方式获得这个？

有人可以帮我这方面吗？提前致谢。

Answer 1

只需使用basename函数。

basename('http://localhost/project-folder/sub-dir-1/images/products/Moulin_1407822344.jpg');

Answer 2

$link = "http://localhost/project-folder/sub-dir-1/images/products/Moulin_1407822344.jpg";
$parts = explode('/', $link);

$filename = end($parts);

Answer 3

使用pathinfo php函数从URL获取文件名

$ fileName = pathinfo（'path / products / Moulin_1407822344.jpg'）;

echo $ fileName ['basename'];

Answer 4

使用basename（）应该工作：

$url = "http://localhost/project-folder/sub-dir-1/images/products/Moulin_1407822344.jpg";

echo basename($url); 

Live Demo

Answer 5

你需要basename函数

echo basename("http://localhost/project-folder/sub-dir-1/images/products/Moulin_1407822344.jpg") ;

Answer 6

$reg = '/(?<=jpg|png|gif|jpeg).*/';
$url = 'https://example.com/2012/12/zoe.jpg?w=600&#038;h=400&#038;crop=1,"; "=""';
$rep = '';
$result = preg_replace($reg, $rep, $url);
$cleanfile = basename($result);
echo $cleanfile;

http://devcodepro.com/view/58/14/How-to-get-clean-file-image-name-with-extension-from-the-URL-in-PHP-