php 如何从继承的方法中获取派生类的路径？

How to get the path of a derived class from an inherited method?

提问by Jacco

How to get the path of the current class, from an inherited method?

如何从继承的方法中获取当前类的路径？

I have the following:

我有以下几点：

<?php // file: /parentDir/class.php
   class Parent  {
      protected function getDir() {
         return dirname(__FILE__);
      }
   }
?>

and

和

<?php // file: /childDir/class.php
   class Child extends Parent {
      public function __construct() {
         echo $this->getDir(); 
      }
   }
   $tmp = new Child(); // output: '/parentDir'
?>

The __FILE__constant always points to the source-file of the file it is in, regardless of inheritance.
I would like to get the name of the path for the derived class.

该__FILE__常数总是指向它在文件的源文件，无论继承。
我想获取派生类的路径名称。

Is there any elegantway of doing this?

有没有优雅的方法来做到这一点？

I could do something along the lines of $this->getDir(__FILE__);but that would mean that I have to repeat myself quite often. I'm looking for a method that puts all the logic in the parent class, if possible.

我可以做一些类似的事情，$this->getDir(__FILE__);但这意味着我必须经常重复自己。如果可能的话，我正在寻找一种将所有逻辑放在父类中的方法。

Update:
Accepted solution (by Palantir):

更新：
接受的解决方案（由Palantir 提供）：

<?php // file: /parentDir/class.php
   class Parent  {
      protected function getDir() {
         $reflector = new ReflectionClass(get_class($this));
         return dirname($reflector->getFileName());
      }
   }
?>

回答by Palantir

Using ReflectionClass::getFileNamewith this will get you the dirname the class Childis defined on.

ReflectionClass::getFileName与此一起使用将为您提供Child定义类的目录名。

$reflector = new ReflectionClass("Child");
$fn = $reflector->getFileName();
return dirname($fn);

You can get the class name of an object with get_class():)

您可以使用get_class():)获取对象的类名

回答by Artefacto

Yes. Building on Palantir's answer:

是的。基于 Palantir 的回答：

   class Parent  {
      protected function getDir() {
         $rc = new ReflectionClass(get_class($this));
         return dirname($rc->getFileName());
      }
   }

回答by Ian Bytchek

Don't forget, since 5.5 you can use classkeyword for the class name resolution, which would be a lot faster than calling get_class($this). The accepted solution would look like this:

不要忘记，从 5.5 开始，您可以使用class关键字来解析类名，这比调用get_class($this). 接受的解决方案如下所示：

protected function getDir() {
    return dirname((new ReflectionClass(static::class))->getFileName());
}

回答by ya.teck

If you are using Composerfor autoloading you can retrieve the directory without reflection.

如果您使用Composer进行自动加载，则无需反射即可检索目录。

$autoloader = require 'project_root/vendor/autoload.php';
// Use get_called_class() for PHP 5.3 and 5.4
$file = $autoloader->findFile(static::class);
$directory = dirname($file);

回答by Adrian

You can also pass the directory as constructor arg. Not super elegant, but at least you don't have to work with reflection or composer.

您还可以将目录作为构造函数 arg 传递。不是超级优雅，但至少您不必使用反射或作曲家。

Parent:

家长：

<?php // file: /parentDir/class.php
   class Parent  {
      private $directory;

      public function __construct($directory) {
          $this->directory = $directory;
      }

      protected function getDir() {
         return $this->directory;
      }
   }
?>

Child:

孩子：

<?php // file: /childDir/class.php
   class Child extends Parent {
      public function __construct() {
        parent::__construct(realpath(__DIR__));
        echo $this->getDir(); 
      }
   }
?>

回答by Remy Mellet

<?php // file: /parentDir/class.php
   class Parent  {
      const FILE = __FILE__;
      protected function getDir() {
         return dirname($this::FILE);
      }
   }
?>


<?php // file: /childDir/class.php
   class Child extends Parent {
      const FILE = __FILE__;
      public function __construct() {
         echo $this->getDir(); 
      }
   }
   $tmp = new Child(); // output: '/childDir'
?>

Please not that if you need to get the dir, directly use __DIR__.

请注意，如果您需要获取目录，请直接使用__DIR__.