Is there any elegant way to check if a file was included by using include/include_once/require/require_once or if the page was actually loaded directly? I'm trying to set up a testing file inside class files while I'm creating them.
I'm looking for something similar to Python's if __name__ == "__main__": technique. Without setting globals or constants.
引用自：如何知道是否通过 require_once() 调用了 php 脚本？
我正在寻找一种方法来确定文件是否已被包含或直接调用，所有这些都来自文件中。在我的探索中的某个时刻，我通过了这个线程。从 PHP 手册中检查此站点和其他站点和页面上的各种其他线程，我得到启发并想出了这段代码：
if (basename(__FILE__) == basename($_SERVER["SCRIPT_FILENAME"])) {
echo "called directly";
} else {
echo "included/required";
}
本质上，它比较当前文件的名称（可能包含的文件）是否与正在执行的文件相同。
信用：@Interwebs 牛仔
您可以通过get_included_files执行此操作— 返回一个包含包含或所需文件名称的数组并验证__FILE__
I appreciate all the answers, but I didn't want to use any one's solution here, so I combined your ideas and got this:
<?php
// place this at the top of the file
if (count(get_included_files()) == 1) define ('TEST_SUITE', __FILE__);
// now I can even include bootstrap which will include other
// files with similar setups
require_once '../bootstrap.php'
// code ...
class Bar {
...
}
// code ...
if (defined('TEST_SUITE') && TEST_SUITE == __FILE__) {
// run test suite here
}
?>
if (defined('FLAG_FROM_A_PARENT'))
// Works in all scenarios but I personally dislike this
if (__FILE__ == get_included_files()[0])
// Doesn't work with PHP prepend unless calling [1] instead.
if (__FILE__ == $_SERVER['SCRIPT_FILENAME'])
// May break on Windows due to mixed DIRECTORY_SEPARATOR
if (basename(__FILE__) == basename($_SERVER['SCRIPT_FILENAME']))
// Doesn't work with files with the same basename but different paths
if (realpath(__FILE__) == realpath($_SERVER['SCRIPT_FILENAME']))
// Seems to do the trick as long as document root is properly configured
注意：在 WAMP 服务器上，虚拟主机有时会继承默认的文档根设置，从而导致$_SERVER['DOCUMENT_ROOT']显示错误的路径。
<?php
if (__FILE__ == $_SERVER['SCRIPT_FILENAME'])
{
//file was navigated to directly
}
?>
取自 mgutt 对此处略有不同的问题的回答。重要的是要注意，如果脚本是从命令行运行的，这不起作用，但除此之外它的功能与 python 的完全一样
if __name__ == '__main__':
据我所知
它们无法将它们分开，include/include_once/require/require_once但是 php 具有get_included_files并且get_required_files是相同的，并且只返回所有包含文件的数组。required如果它的或，它不会将它分开included。
例子a.php
include 'b.php';
include_once 'c.php';
require 'd.php';
var_dump(get_required_files());
输出
array
0 => string '..\lab\stockoverflow\a.php' (length=46) <---- Returns current file
1 => string '..\lab\stockoverflow\b.php' (length=46)
2 => string '..\lab\stockoverflow\c.php' (length=46)
3 => string '..\lab\stockoverflow\d.php' (length=46)
但是你可以做类似的事情
$inc = new IncludeManager($file);
var_dump($inc->find("b.php")); // Check if a file is included
var_dump($inc->getFiles("require_once")); // Get All Required Once
使用的类
class IncludeManager {
private $list = array();
private $tokens = array();
private $find;
private $file;
private $type = array(262 => "include",261 => "include_once",259 => "reguire",258 => "require_once");
function __construct($file) {
$this->file = $file;
$this->_parse();
}
private function _parse() {
$tokens = token_get_all(file_get_contents($this->file));
for($i = 0; $i < count($tokens); $i ++) {
if (count($tokens[$i]) == 3) {
if (array_key_exists($tokens[$i][0], $this->type)) {
$f = $tokens[$i + 1][0] == 371 ? $tokens[$i + 2][1] : $tokens[$i + 1][1];
$this->list[] = array("pos" => $i,"type" => $this->type[$tokens[$i][0]],"file" => trim($f, "\"\'"));
}
}
}
}
public function find($find) {
$finds = array_filter($this->list, function ($v) use($find) {
return $v['file'] == $find;
});
return empty($finds) ? false : $finds;
}
public function getList() {
return $this->list;
}
public function getFiles($type = null) {
$finds = array_filter($this->list, function ($v) use($type) {
return is_null($type) ? true : $type == $v['type'];
});
return empty($finds) ? false : $finds;
}
}
get_included_files()返回数组，其中 0 索引表示第一个“包含”文件。因为直接运行在这个术语中意味着“包含”，所以您可以简单地检查第一个索引是否相等__FILE__：
if(get_included_files()[0] == __FILE__){
do_stuff();
}
这不能在 PHP 4 上工作，因为 PHP 4 没有在这个数组中添加运行文件。
$target_file = '/home/path/folder/file.php'; // or use __FILE__
if ($x=function($e){return str_replace(array('\\'), '/', $e);}) if(in_array( $x($target_file), array_map( $x , get_included_files() ) ) )
{
exit("Hello, already included !");
}
这是一个不同的想法。只需在需要时包含该文件。在包含文件中，您可以决定是否需要包含内容：
<?php
if (defined("SOME_UNIQUE_IDENTIFIER_FOR_THIS_FILE"))
return;
define("SOME_UNIQUE_IDENTIFIER_FOR_THIS_FILE", 1);
// Rest of code goes here
我认为这不是get_included_files完美的解决方案，如果您的主脚本在检查之前包含其他一些脚本怎么办？我的建议是检查是否__FILE__等于realpath($argv[1])：
<?php
require('phpunit/Autoload.php');
class MyTests extends PHPUnit_Framework_TestCase
{
// blabla...
}
if (__FILE__ == realpath($argv[0])) {
// run tests.
}
当我遇到这个问题时，我采取了类似的方法。我找到的解决方案是在 include_once 方法中根据需要加载每个文件。希望这可以帮助。
$FILES = get_included_files(); // Retrieves files included as array($FILE)
$FILE = __FILE__; // Set value of current file with absolute path
if(!in_array($FILE, $FILES)){ // Checks if file $FILE is in $FILES
include_once "PATH_TO_FILE"; // Includes file with include_once if $FILE is not found.
}
我建立了以下功能来检查加载的文件：
ARRAY_DUMP($FILES);
function ARRAY_DUMP($array){
echo "
<span style='font-size:12px;'>".date('h:i:s').":</span>
<pre style='font-size:12px;'>", print_r($array, 1), "</pre>
";
}
输出：
currentArray
(
[0] => /home/MY_DOMAIN/hardeen/index.php
[1] => /home/MY_DOMAIN/hardeen/core/construct.php
[2] => /home/MY_DOMAIN/hardeen/core/template.php
[3] => /home/MY_DOMAIN/hardeen/bin/tags.php
[4] => /home/MY_DOMAIN/hardeen/bin/systemFunction.php
)
太简单了..我做了这样的事情：
//code for file.php
if (!isset($file_included)){
echo "It was loaded!";
} else {
echo "It was included!";
}
//code for loader.php
//proves that atleast loader.php has loaded,
//not the file we targeted first..
$file_included = true;
include("../file.php");
就是这样......就像在python中一样简单。