如何使PHP库松散耦合? [英] How to make PHP libraries loosely coupled?
问题描述
我的框架"中有一些库，例如路由，配置，记录器，...我希望它们彼此独立，就像一些众所周知的PHP框架使它们独立一样.
I have some libraries in my 'framework' like routing, config, logger,... I want them to be independent of each other, like some of well known PHP frameworks make them.
我了解松耦合的所有原理，但是我不知道如何同时遵循松耦合和DRY原理.如果我将路由库用作配置和记录器，那么我就不重复了，但是如果我想单独使用路由器，它将无法正常工作.同样，如果我将日志记录和配置代码写入路由库，我会重复自己.
I understand all the principles of loose coupling, but I have no clue how to follow both loose coupling and DRY principles. If I make routing library that config and logger, then I don't repeat myself, but if I want to use router on its own it won't work. Similarly if I write logging and config code into my routing library, I would repeat myself.
推荐答案
松耦合通常意味着您的组件不期望一个具体的实例，而只是一个具有兼容接口的实例.
Loose coupling normally means that your components do not expect a concrete instance but just one instance that has a compatible interface.
然后可以将每个协作者替换为相同类型的另一个协作者.该代码不再依赖于其中之一的具体实现.
Each collaborator can be replaced then with a different one of the same type. The code is not dependent on a concrete implementation of one of those any longer.
所以:
-
请勿使用:
Do not use:
-
全局(静态)函数
global (static) functions
Foo:bar();
基于类的编程(传递类名)
class based programming (passing a classname around)
stream_wrapper_register("var", "VariableStream");
全局常量
global constants
if ( !defined('ABSPATH') )
define('ABSPATH', dirname(__FILE__) . '/');
但是:
-
使用对象
Use objects
$foo->bar();
针对接口的程序
Program against interfaces
public function __construct(LoggerInterface $logger) {
带有模拟的单元测试
Unit-test with mocks
$logger = $this->getMock('LoggerInterface', array('log'));
另请参见:
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