将对象实例化为具有命名空间的类

Question

I don't know how to explain it, but I will do my best. Ok, I have these three files:

• Theme.php

  path: /shared/models/Theme.php
  class: Theme
  namespace: namespace models;
  

• Custom.php

  path: /themes/default/Custom.php
  class: Custom
  namespace: this class does not use namespace
  

• Settings.php

  path: /controllers/Settings.php
  class: Settings
  namespace: this class does not use namespace
  

In my Settings.php looks like:

<?php
class Settings
{
    public function apply()
    {
        $theme = new \models\Theme();
        $theme->customize(); // in this method the error is triggered
    }
}

Now, look Theme class below:

<?php
namespace models;

class Theme
{
    public function customize()
    {
        $ext = "/themes/default/Custom.php";
        if (file_exists($ext))
        {
            include $ext;
            if (class_exists('Custom'))
            {            
                $custom = new Custom(); 
                //Here, $custom var in null, why???
            }
        }
    }
}

When I execute the code, I get the following error:

Message: require(/shared/models/Custom.php) [function.require]: failed to open stream: No such file or directory
Line Number: 64

Why the interpreter is trying to load the Custom class from another directory instead of specified with $ext var?

Answer 1

When invoking new Custom() inside of a class in \models namespace you're actually trying to instantiate \models\Custom. Since you say your Custom class "has no namespace", try new \Custom() instead.

The error you get seems to be coming from some class autoloader that tries to require class file for \models\Custom and fails.