Variable Class Names ignore “use”

问题

From other posts, it appears that if you have namespaces defined and want to dynamically create an object in another namespace, you have to construct a string and use that in the new call. However, I'm getting a weird behavior. It appears that this method does not work going across namespaces.

User.php:

namespace application\models;

class User {

        public function hello() {
                echo "Hello from User!";
        }
}

Controller.php:

namespace application\controllers;

use application\models;

require('User.php');

$userStr = 'models\\User';
//$userOne = new $userStr();  //Doesn't work. Gets a "Class 'models\User' not found" error
$userOne = new models\User();  //Works fine

$userStr = '\\application\\models\\User';
$userTwo = new $userStr();  //Works fine

$userOne->hello();
$userTwo->hello();

Any idea why when using a variable for the class name, I need to use the fully qualified namespace when it's in a variable, but hard coded, I can leverage the "use" command?

回答1:

You can not import with use into variable classnames. That is a limitation of PHP.

See as well the related questions:

• Expanding PHP namespace alias to full namespace string
• Can't get constant from dynamic class using namespaces

来源：https://stackoverflow.com/questions/13313904/variable-class-names-ignore-use