doudoulb1234
2016-11-17 18:04
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PHP:有没有办法在对象构造函数中使用全局变量?

I have a php script that is intended to be user-editable, with some options set right at the beginning, including allowable extensions for image files.

Later on, I want to use this in an object constructor to set whether a file is an image or not.

$extensions = ["jpg", "jpeg", "gif", "png"];

class directory_entry {

  function __construct($name) {

    $this->name = $name;

    $extension = pathinfo($name, PATHINFO_EXTENSION);
    $this->is_image = in_array(strtoupper($extension), $extensions); // throws an error

    $this->image = $this->is_image($name);

  }
}

This doesn't work because I can't call $extensions from inside an class(?). Is there a way to do it without including $extensions in every object of this class? Seems nuts to declare objects with

$files[] = new directory_entry($value, $extensions);

and have every instance of this class have a copy of the same array?

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2条回答 默认 最新

  • dopt85756 2016-11-18 08:02
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    The solution that seems to work best for what I'm trying to do (bear in mind that I'm a PHP and OOP n00b) is the globals keyword:

    $extensions = ["jpg", "jpeg", "gif", "png"];
    
    class directory_entry {
    
      function __construct($name) {
    
        $this->name = $name;
    
        $extension = pathinfo($name, PATHINFO_EXTENSION);
        global $extensions; //use the global version of $extensions
        $this->is_image = in_array(strtoupper($extension), $extensions); // works!
    
        $this->image = $this->is_image($name);
    
      }
    }
    
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  • douzi1991 2016-11-17 18:07

    Why not create a classe for this?

    class ExtensionsEnum
    {
        const IMAGES = ["jpg", "jpeg", "gif", "png"];
    }
    

    Then

    use ExtensionsEnum;
    
    class directory_entry {
    ...
        $this->is_image = in_array(strtoupper($extension), ExtensionsEnum::IMAGES); // throws an error
    ...
    }
    
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