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从PHP中的其他文件夹打开文件

[英]Open file from a different folder in PHP

Let's say I have two folders in my PHP server root directory: x and y . 假设我的PHP服务器根目录中有两个文件夹: xy There are two more directories inside the x folder: a and b . x文件夹中还有两个目录: ab In the /x/a directory I have the index.php file and in the /y folder a stuff.php file. /x/a目录中我有index.php文件,在/y文件夹中有一个stuff.php文件。
From my /x/a/index.php file, I want to include the /y/stuff.php file. 从我的/x/a/index.php文件中,我想要包含/y/stuff.php文件。 How do I do that? 我怎么做?
My server doesn't allow including files from other domains, so adding the full URL doesn't work! 我的服务器不允许包含来自其他域的文件,因此添加完整的URL不起作用! Also, I'd like to know how to start a path from the root in PHP. 另外,我想知道如何从PHP中的根开始路径。 I use ./blabla from the index of my root and ../blabla from directories in the root, but, unfortunately, .../blabla doesn't work from 2nd grade directories. 我用./blabla从我的根和指数../blabla从目录的根,但不幸的是, .../blabla不从二年级目录工作。

To access the root begin your path with "/". 要访问root,请以“/”开头。

If you want to go up one directory from where you are currently, eg from /x/a/ to /x/ you could use "../". 如果你想从你当前所在的目录上去,例如从/ x / a /到/ x /你可以使用“../”。

If you want to go back up two directories (eg from /x/a/ to /) you could use "../../" (rather than ".../" which you mentioned). 如果你想要备份两个目录(例如从/ x / a /到/)你可以使用“../../”(而不是你提到的“... /”)。

尝试:

include "../../y/stuff.php";

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