需要文件夹中的所有文件

Is there an easy way to require all files in a folder?

是否有一种简单的方法来要求文件夹中的所有文件？

8 个解决方案

#1

No short way of doing it, you'll need to implement it in PHP. Something like this should suffice:

没有简短的方法，你需要在PHP中实现它。这样的东西应该足够了：

foreach (scandir(dirname(__FILE__)) as $filename) {
    $path = dirname(__FILE__) . '/' . $filename;
    if (is_file($path)) {
        require $path;
    }
}

#2

Probably only by doing something like this:

可能只做这样的事情：

$files = glob($dir . '/*.php');

foreach ($files as $file) {
    require($file);   
}

It might be more efficient to use opendir() and readdir() than glob().

使用opendir（）和readdir（）比使用glob（）更有效。

#3

There is no easy way, as in Apache, where you can just Include /path/to/dir, and all the files get included.

没有简单的方法，就像在Apache中一样，你可以只包含/ path / to / dir，并包含所有文件。

A possible way is to use the RecursiveDirectoryIterator from the SPL:

一种可能的方法是使用SPL中的RecursiveDirectoryIterator：

function includeDir($path) {
    $dir      = new RecursiveDirectoryIterator($path);
    $iterator = new RecursiveIteratorIterator($dir);
    foreach ($iterator as $file) {
        $fname = $file->getFilename();
        if (preg_match('%\.php$%', $fname)) {
            include($file->getPathname());
        }
    }
}

This will pull all the .php ending files from $path, no matter how deep they are in the structure.

这将从$ path中提取所有.php结尾文件，无论它们在结构中有多深。

#4

Simple:

简单：

foreach(glob("path/to/my/dir/*.php") as $file){
    require $file;
}

#5

Use a foreach Loop.

foreach (glob("classes/*") as $filename) {
  require $filename;
}

For more details, check out this previously posted question:

有关更多详细信息，请查看此前发布的问题：

#6

As require_all() function:

作为require_all（）函数：

//require all php files from a folder
function require_all ($path) {

    foreach (glob($path.'*.php') as $filename) require_once $filename;

}

#7

recursively all file list and require_once in one directory:

递归地在一个目录中的所有文件列表和require_once：

$files = array();

function require_once_dir($dir){

       global $files;

       $item = glob($dir);

       foreach ($item as $filename) {

             if(is_dir($filename)) {

                  require_once_dir($filename.'/'. "*");

             }elseif(is_file($filename)){

                  $files[] = $filename;
             }
        }
}

$recursive_path = "path/to/dir";

require_once_dir($recursive_path. "/*");

for($f = 0; $f < count($files); $f++){

     $file = $files[$f];

     require_once($file);
}

#8

My way to require all siblings:

我要求所有兄弟姐妹的方式：

<?php
$files = glob(__DIR__ . '/*.php');
foreach ($files as $file) {
    // prevents including file itself
    if ($file != __FILE__) {
        require($file);
    }
}