如何相对要求PHP文件(在不同目录级别)?

问题描述

我具有以下文件结构:

rootDIR
    dir1
        subdir1
           file0.php
           file1.php
    dir2
       file2.php
       file3.php
       file4.php

file1.php从dir2需要file3.php和file4.php像这样:

file1.php requires file3.php and file4.php from dir2 like this :

require('../../dir2/file3.php')

file2.php需要这样的file1.php:

require('../dir1/subdir1/file1.php')

但是随后file1.php中的require无法打开file3.php和file4.php(可能由于路径相对性)

But then require in file1.php fails to open file3.php and file4.php ( maybe due to the path relativeness)

但是，file2.php的原因是什么，我该怎么办，所以file1.php正确地需要file3.php和file4.php?

However, what is the reason and what can I do for file2.php so file1.php properly require file3.php and file4.php?

推荐答案

对于相对路径，您可以直接使用__DIR__而不是dirname(__FILE__)(只要您是使用PHP 5.3.0及更高版本):

For relative paths you can use __DIR__ directly rather than dirname(__FILE__) (as long as you are using PHP 5.3.0 and above):

require(__DIR__.'/../../dir2/file3.php');

请记住在引号内路径的开头添加其他正斜杠.

Remember to add the additional forward slash at the beginning of the path within quotes.

请参阅:

• PHP-与require_once/include/require，路径是相对于什么的?
• PHP-相对路径"require"

• PHP - with require_once/include/require, the path is relative to what?
• PHP - Relative paths "require"

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