如何在 PHP 中获取文件扩展名？
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How do I get a file extension in PHP?
提问by e-satis
This is a question you can read everywhere on the web with various answers:
这是一个您可以在网络上随处阅读的问题，并提供各种答案：
$ext = end(explode('.', $filename));
$ext = substr(strrchr($filename, '.'), 1);
$ext = substr($filename, strrpos($filename, '.') + 1);
$ext = preg_replace('/^.*\.([^.]+)$/D', '', $filename);
$exts = split("[/\.]", $filename);
$n = count($exts)-1;
$ext = $exts[$n];
etc.
等等。
However, there is always "the best way" and it should be on Stack Overflow.
然而，总是有“最好的方法”，它应该在 Stack Overflow 上。
回答by e-satis
People from other scripting languages always think theirs is better because they have a built-in function to do that and not PHP (I am looking at Pythonistas right now :-)).
来自其他脚本语言的人总是认为他们的更好，因为他们有一个内置的函数来做到这一点，而不是 PHP（我现在正在看 Pythonistas :-)）。
In fact, it does exist, but few people know it. Meet pathinfo():
事实上，它确实存在，但很少有人知道。见面pathinfo()：
$ext = pathinfo($filename, PATHINFO_EXTENSION);
This is fast and built-in. pathinfo()can give you other information, such as canonical path, depending on the constant you pass to it.
这是快速和内置的。pathinfo()可以为您提供其他信息，例如规范路径，具体取决于您传递给它的常量。
Remember that if you want to be able to deal with non ASCII characters, you need to set the locale first. E.G:
请记住，如果您希望能够处理非 ASCII 字符，则需要先设置语言环境。例如：
setlocale(LC_ALL,'en_US.UTF-8');
Also, note this doesn't take into consideration the file content or mime-type, you only get the extension. But it's what you asked for.
另外，请注意这不考虑文件内容或 mime 类型，您只能获得扩展名。但这是你要求的。
Lastly, note that this works only for a file path, not a URL resources path, which is covered using PARSE_URL.
最后，请注意，这仅适用于文件路径，而不适用于使用 PARSE_URL 涵盖的 URL 资源路径。
Enjoy
享受
回答by Adam Wright
$path_info = pathinfo('/foo/bar/baz.bill');
echo $path_info['extension']; // "bill"
回答by T.Todua
Example URL: http://example.com/myfolder/sympony.mp3?a=1&b=2#XYZ
示例网址： http://example.com/myfolder/sympony.mp3?a=1&b=2#XYZ
A) Don't use suggested unsafe PATHINFO:
A)不要使用建议的 unsafePATHINFO：
pathinfo($url)['dirname'] 'http://example.com/myfolder'
pathinfo($url)['basename'] 'sympony.mp3?a=1&b=2#XYZ' // <------- BAD !!
pathinfo($url)['extension'] 'mp3?a=1&b=2#XYZ' // <------- BAD !!
pathinfo($url)['filename'] 'sympony'
B) Use PARSE_URL:
B) 使用PARSE_URL：
parse_url($url)['scheme'] 'http'
parse_url($url)['host'] 'example.com'
parse_url($url)['path'] '/myfolder/sympony.mp3'
parse_url($url)['query'] 'aa=1&bb=2'
parse_url($url)['fragment'] 'XYZ'
BONUS: View all native PHP examples
奖励：查看所有原生 PHP 示例
回答by hakre
There is also SplFileInfo:
还有SplFileInfo：
$file = new SplFileInfo($path);
$ext = $file->getExtension();
Often you can write better code if you pass such an object around instead of a string. Your code is more speaking then. Since PHP 5.4 this is a one-liner:
如果传递这样的对象而不是字符串，通常可以编写更好的代码。你的代码更会说话。从 PHP 5.4 开始，这是一个单行：
$ext = (new SplFileInfo($path))->getExtension();
回答by Toxygene
E-satis's response is the correct way to determine the file extension.
E-satis 的响应是确定文件扩展名的正确方法。
Alternatively, instead of relying on a files extension, you could use the fileinfoto determine the files MIME type.
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。
Here's a simplified example of processing an image uploaded by a user:
这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;
}
回答by Anonymous
As long as it does not contain a path you can also use:
只要它不包含路径，您也可以使用：
array_pop(explode('.', $fname))
Where $fnameis a name of the file, for example: my_picture.jpg.
And the outcome would be: jpg
$fname文件名在哪里，例如：my_picture.jpg. 结果将是：jpg
回答by Alix Axel
Sometimes it's useful to not to use pathinfo($path, PATHINFO_EXTENSION). For example:
有时不使用pathinfo($path, PATHINFO_EXTENSION). 例如：
$path = '/path/to/file.tar.gz';
echo ltrim(strstr($path, '.'), '.'); // tar.gz
echo pathinfo($path, PATHINFO_EXTENSION); // gz
Also note that pathinfofails to handle some non-ASCII characters (usually it just suppresses them from the output). In extensions that usually isn't a problem, but it doesn't hurt to be aware of that caveat.
另请注意，pathinfo无法处理一些非 ASCII 字符（通常它只是从输出中抑制它们）。在通常不是问题的扩展中，但意识到这一警告并没有什么坏处。
回答by Subodh Ghulaxe
1)If you are using (PHP 5 >= 5.3.6)you can use SplFileInfo::getExtension— Gets the file extension
1)如果您正在使用(PHP 5 >= 5.3.6)您可以使用SplFileInfo::getExtension— 获取文件扩展名
Example code
示例代码
<?php
$info = new SplFileInfo('test.png');
var_dump($info->getExtension());
$info = new SplFileInfo('test.tar.gz');
var_dump($info->getExtension());
?>
This will output
这将输出
string(3) "png"
string(2) "gz"
2)Another way of getting the extension if you are using (PHP 4 >= 4.0.3, PHP 5)is pathinfo
2)如果您正在使用(PHP 4 >= 4.0.3, PHP 5)，另一种获取扩展的方法是pathinfo
Example code
示例代码
<?php
$ext = pathinfo('test.png', PATHINFO_EXTENSION);
var_dump($ext);
$ext = pathinfo('test.tar.gz', PATHINFO_EXTENSION);
var_dump($ext);
?>
This will output
这将输出
string(3) "png"
string(2) "gz"
// EDIT: removed a bracket
// 编辑：删除了一个括号
回答by Shahbaz
回答by pooya_sabramooz
You can try also this (it works on PHP 5.* and 7):
你也可以试试这个（它适用于 PHP 5.* 和 7）：
$info = new SplFileInfo('test.zip');
echo $info->getExtension(); // ----- Output -----> zip
Tip: it returns an empty string if the file doesn't have an extension
提示：如果文件没有扩展名，它将返回一个空字符串
