变量类名忽略"use". [英] Variable Class Names ignore "use"

问题描述

从其他帖子中看来，如果您定义了名称空间，并且想要在另一个名称空间中动态创建对象，则必须构造一个字符串并在新调用中使用它.但是，我的行为很奇怪.似乎该方法无法跨命名空间使用.

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:

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();

有人知道为什么在使用变量作为类名时，需要在变量中使用完全限定的名称空间，但是经过硬编码后，我才能利用"use"命令吗?

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?

推荐答案

您不能使用 转换为变量类名.那是PHP的局限性.

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

也请参阅相关问题:

• 将PHP名称空间别名扩展为完整的名称空间字符串
• 无法使用名称空间从动态类中获取常量

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

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