问题描述

想知道其他人在使用 PHP 5.3 命名类的新功能时是否遇到过这个问题.

Wondering if anyone else has encountered this problem when utilizing the new ability to namespace classes using PHP 5.3.

我正在生成一个动态类调用，该调用利用一个单独的类在我的应用程序中定义用户类型.基本上，类定义器采用类型的整数表示并解释它们，返回一个包含类名的字符串，作为该用户的模型调用.

I am generating a dynamic class call utilizing a separate class for defining user types in my application. Basically the class definer takes an integer representation of types and interprets them, returning a string containing the classname to be called as the model for that user.

我有一个用户类型的对象模型，该名称在全局范围中定义，但我在 Editor 命名空间中有另一个同名的用户编辑器对象.出于某种原因，PHP 不允许我进行如下命名空间的动态调用.

I have an object model for the user's type with that name defined in the global scope, but I have another object with the same name for the user's editor in the Editor namespace. For some reason, PHP won't allow me to make a namespaced dynamic call as follows.

$definition = Definer::defineProfile($_SESSION['user']->UserType);
new Editor$definition();

相同的语法适用于在全局命名空间中调用全局基本对象模型，我在整个应用程序中以这种方式可靠地使用它.

The identical syntax works for calling the global basic object model in the global namespace and I use it this way reliably throughout the application.

$definition = Definer::defineProfile($_SESSION['user']->UserType);
new $definition();

这将正确调用动态所需的类.

This will correctly call the dynamically desired class.

是否有原因两者的行为会有所不同，或者由于这是一项新功能，因此该庄园尚未实现对命名空间的动态调用?是否有另一种方法可以从另一个命名空间动态调用一个类，而无需在代码中显式放置其名称，而是从一个变量中?

Is there a reason the two would behave differently, or has dynamic calling for namespaces not been implemented in this manor yet as this is a new feature? Is there another way to dynamically call a class from another namespace without explicitly placing its name in the code, but from within a variable?

推荐答案

好吧，把字符串中的命名空间拼出来就好了:

Well, just spell out the namespace in the string:

$definition = Definer::defineProfile($_SESSION['user']->UserType);
$class = '\Editor\' . $definition;
$foo = new $class();

如果它是子命名空间(如注释中所示)，只需在命名空间前面加上 __NAMESPACE__:

And if it's a child namespace (as indicated in the comments), simply prepend the namespace with __NAMESPACE__:

$class = __NAMESPACE__ . '\Editor\' . $definition;

因此，如果当前命名空间是 FooBar，并且 $definition 是Baz"，则生成的类将是 FooBarEditor巴兹

So if the current namespace is FooBar, and $definition is "Baz", the resulting class would be FooBarEditorBaz

这篇关于具有动态类名的 PHP 命名空间的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持跟版网！