这是一个问题，你可以在网络上的任何地方看到各种答案：
$ext = end(explode('.', $filename));
$ext = substr(strrchr($filename, '.'), 1);
$ext = substr($filename, strrpos($filename, '.') + 1);
$ext = preg_replace('/^.*\.([^.]+)$/D', '$1', $filename);
$exts = split("[/\\.]", $filename);
$n = count($exts)-1;
$ext = $exts[$n];
etc.
然而，总是有“最好的方法”，它应该是堆栈溢出。
当前回答
IMO，如果你有像name.name.name.ext这样的文件名（很难看，但有时会发生这种情况），这是最好的方法：
$ext = explode('.', $filename); // Explode the string
$my_ext = end($ext); // Get the last entry of the array
echo $my_ext;
其他回答
很抱歉“简短的问题；但不是简短的回答”
PATH示例1
$path = "/home/ali/public_html/wp-content/themes/chicken/css/base.min.css";
$name = pathinfo($path, PATHINFO_FILENAME);
$ext = pathinfo($path, PATHINFO_EXTENSION);
printf('<hr> Name: %s <br> Extension: %s', $name, $ext);
URL示例2
$url = "//www.example.com/dir/file.bak.php?Something+is+wrong=hello";
$url = parse_url($url);
$name = pathinfo($url['path'], PATHINFO_FILENAME);
$ext = pathinfo($url['path'], PATHINFO_EXTENSION);
printf('<hr> Name: %s <br> Extension: %s', $name, $ext);
示例1的输出：
Name: base.min
Extension: css
示例2的输出：
Name: file.bak
Extension: php
工具书类
https://www.php.net/manual/en/function.pathinfo.phphttps://www.php.net/manual/en/function.realpath.phphttps://www.php.net/manual/en/function.parse-url.php
我尝试了一个简单的解决方案，它可能会帮助其他人从具有get参数的URL中获取文件名
<?php
$path = "URL will be here";
echo basename(parse_url($path)['path']);
?>
谢谢
快速修复可能是这样的。
// Exploding the file based on the . operator
$file_ext = explode('.', $filename);
// Count taken (if more than one . exist; files like abc.fff.2013.pdf
$file_ext_count = count($file_ext);
// Minus 1 to make the offset correct
$cnt = $file_ext_count - 1;
// The variable will have a value pdf as per the sample file name mentioned above.
$file_extension = $file_ext[$cnt];
substr($path, strrpos($path, '.') + 1);
事实上，我在找这个。
<?php
$url = 'http://example.com/myfolder/sympony.mp3?a=1&b=2#XYZ';
$tmp = @parse_url($url)['path'];
$ext = pathinfo($tmp, PATHINFO_EXTENSION);
var_dump($ext);
