获取当前文件 PHP 的父文件夹
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Get parent folder of current file PHP
提问by Andrey
I've done quite a bit of searching for this, so I'm sorry if this is a dupe.
Anyway, I need to get the name of the folder the current file is in. For example, I want to convert something like example.com/folder/subfolder/file.phpto subfolder.
My current code is dirname($_SERVER['PHP_SELF']), but that returns /folder/subfolderinstead of subfolder. Thank you!
我已经做了很多搜索，所以如果这是一个骗局，我很抱歉。
无论如何，我需要获取当前文件所在文件夹的名称。例如，我想将类似的内容转换example.com/folder/subfolder/file.php为subfolder.
我当前的代码是dirname($_SERVER['PHP_SELF'])，但它返回/folder/subfolder而不是subfolder。谢谢！
回答by Alfonso de la Osa
The simpliest way is:
最简单的方法是：
basename(__DIR__);
回答by DaveRandom
You need to combine your existing code using dirname()with a call to basename():
您需要将现有代码 usingdirname()与调用结合起来basename()：
$parent = basename(dirname($_SERVER['PHP_SELF']));
回答by Luke Pittman
dirname()used with basename()would work ... also this if you want to get them all:
dirname()使用 withbasename()会起作用......如果你想把它们全部都用这个：
$folders = explode ('/', $_SERVER['PHP_SELF']);
Now $folders would contain an array of all of the folder names.
现在 $folders 将包含所有文件夹名称的数组。
Cheers.
干杯。
回答by tribulant
$folder = basename(dirname(__FILE__));
回答by MK2009
str_replace('/','',$_SERVER[REQUEST_URI]);
This works for php7: str_replace('/','',$_SERVER["REQUEST_URI"]);
这适用于 php7： str_replace('/','',$_SERVER["REQUEST_URI"]);
Result:
结果：
lib/**images**/index.php
lib/**images**/index.php
images
图片
