孩子的文件路径

发布于 2024-10-14 06:26:46 字数 319 浏览 1 评论 0原文

<代码>

class a1 extends class b{}
class a2 extends class b{}
class a3 extends class b{}
class a4 extends class b{}
class a5 extends class b{}

我需要 $this->filePath 在每个 a1 到 a5 中指向其文件的位置，但是当我在父级中设置 $this->filePath = __FILE__ 时，在子级中 $this->filePath 指向父级位置

原文

class a1 extends class b{}
class a2 extends class b{}
class a3 extends class b{}
class a4 extends class b{}
class a5 extends class b{}

I need $this->filePath in each a1 to a5 that point to location of it's file, but when I set $this->filePath = __FILE__ in parent, in children $this->filePath point to parent location

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鹿! 2024-10-21 06:26:46

由于 __FILE__ 的性质，无法在 b 的构造函数中全局执行此操作 - 它的行为不像函数，而是一个需要处理的魔术常量（并替换为其实际值）当解释文件时。

您必须分别为每个孩子执行此操作。这在 PHP 5 中有效：

class a1 extends b{ private $path = __FILE__;}
class a2 extends b{ private $path = __FILE__;}
class a3 extends b{ private $path = __FILE__;}
class a4 extends b{ private $path = __FILE__;}
class a5 extends b{ private $path = __FILE__;}

我知道在父级中执行此操作的唯一方法是使用 debug_backtrace()，但这不是一个好的做法。

There is no way to do this globally in b's constructor due to the nature of __FILE__ - it does not behave like a function, but is a magic constant that gets processed (and replaced with its actual value) when the file is interpreted.

You will have to do this in each child separately. This works in PHP 5:

class a1 extends b{ private $path = __FILE__;}
class a2 extends b{ private $path = __FILE__;}
class a3 extends b{ private $path = __FILE__;}
class a4 extends b{ private $path = __FILE__;}
class a5 extends b{ private $path = __FILE__;}

The only way I know to do this in the parent is using debug_backtrace(), and that is not a good practice.

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