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

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

其他回答

E-satis的响应是确定文件扩展名的正确方法。

或者,您可以使用fileinfo来确定文件的MIME类型,而不是依赖文件扩展名。

下面是处理用户上传的图像的简化示例:

// Code assumes necessary extensions are installed and a successful file upload has already occurred

// Create a FileInfo object
$finfo = new FileInfo(null, '/path/to/magic/file');

// Determine the MIME type of the uploaded file
switch ($finfo->file($_FILES['image']['tmp_name'], FILEINFO_MIME)) {        
    case 'image/jpg':
        $im = imagecreatefromjpeg($_FILES['image']['tmp_name']);
    break;

    case 'image/png':
        $im = imagecreatefrompng($_FILES['image']['tmp_name']);
    break;

    case 'image/gif':
        $im = imagecreatefromgif($_FILES['image']['tmp_name']);
    break;
}

在PHP中获取文件扩展名的最简单方法是使用PHP的内置函数pathinfo。

$file_ext = pathinfo('your_file_name_here', PATHINFO_EXTENSION);
echo ($file_ext); // The output should be the extension of the file e.g., png, gif, or html

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];

只要它不包含路径,您也可以使用:

array_pop(explode('.', $fname))

其中$fname是文件的名称,例如:my_picture.jpg。结果将是:jpg

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