使用关键字“使用"有条件地导入类 [英] Import class conditionally with the keyword 'use'

问题描述

我从来没有在任何地方看到过这种结构，所以我想知道这样的表达式是否有问题:

I've never seen this structure anywhere, so I wonder if there's something wrong with an expression like this:

if (condition) {

    use SymfonyComponentHttpFoundationResponse;

}

推荐答案

use 唯一要做的就是别名一个类名.而已.仅此而已.
不必在脚本中重复编写完全限定的类名:

The only thing use does is to alias a class name. That's it. Nothing more.
Instead of having to repeatedly write the fully qualified classname in your script:

$q = new FooBarBazQuux;
if ($q instanceof FooBarBazQuux) ...

您可以将其缩短为:

use FooBarBazQuux;

$q = new Quux;
if ($q instanceof Quux) ...

因此，想要有条件地使用 use 绝对没有意义.它只是一个语法助手；如果可以有条件地使用它，您的 脚本语法 会变得模棱两可，这是没有人想要的.

As such, it makes absolutely no sense to want to use use conditionally. It's just a syntactic helper; if it could be used conditionally your script syntax would become ambiguous, which is something nobody wants.

它不会减少代码加载，因为代码只能通过 require/include 调用或通过自动加载显式加载.后者是非常受欢迎的，因为它已经懒惰地只在需要时才开始行动.

It doesn't reduce code loading, because code is only loaded explicitly by require/include calls or via autoloading. The latter one is greatly preferred, since it already lazily springs into action only when needed.

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