英文:
Define the same PHP function in many namespaces
问题
为了在单元测试中创建全局函数的模拟版本，我将其模拟版本定义在SUT类的命名空间中，如此处所解释的那样。
考虑一个全局函数 foo() 在 Bar 命名空间和 Baz 命名空间中都被使用的情况。如果我想在两个地方使用相同的 foo() 模拟版本，似乎我被迫要为 foo() 进行两次声明。
/* bar-namespace-mocks.php */namespace Bar;function foo(){return 'foo mock';}
/* baz-namespace-mocks.php */namespace Baz;function foo(){return 'foo mock';}
这段代码不符合DRY原则。更好的方式是只声明一次foo模拟版本。
/* foo-mock.php */function foo(){return 'foo mock';}
然后按需要导入每个命名空间，如以下伪代码所示：
/* namespace-mocks.php */namespace Bar{import 'foo-mock.php';}namespace Baz{import 'foo-mock.php';}
使用include导入，例如
namespace Baz;include 'foo-mock.php';
不会导致模拟在Baz命名空间中被声明。是否有办法在多个命名空间中声明一个函数而不必多次版本化它？
英文:
In order to create a mock version of a global function for unit testing, I define its mock version in the namespace of the SUT class, as is explained here
Consider a case of global function foo() which is used in namespace Bar and in namespace Baz. If I want to use the same mock version of foo() in both places, it seems I am forced to make two declarations of the mock for foo()
/* bar-namespace-mocks.php */namespace Bar;function foo(){return 'foo mock';}
/* baz-namespace-mocks.php */namespace Baz;function foo(){return 'foo mock';}
This code does not conform to the DRY principal.
It would be preferred to have one declaration of the foo mock
/* foo-mock.php */function foo(){return 'foo mock';}
and then import into each namespace as needed like the following pseudo code:
/* namespace-mocks.php */namespace Bar{import 'foo-mock.php';}namespace Baz{import 'foo-mock.php';}
Importing using include, e.g.
namespace Baz;include 'foo-mock.php'
does not cause the mock to be declared in the Baz namespace. Is there any way to declare a function in more than one namespace without having more than one version of it?
答案1
得分: 1
如果您需要对一个本地函数进行抽象化，那么请为其创建一个契约（contract），以便您可以使用依赖注入来提供其提供的服务：
interface FooInterface{public function get(): string;}
然后提供一个默认的实现，使用本地函数：
class NativeFoo implements FooInterface{public function get(): string{return \foo();}}
然后您的主要类可能看起来像这样：
class A{protected FooInterface $foo;public function __construct(FooInterface $foo = null){$this->foo = $foo ?? new NativeFoo();}public function whatever(){$foo = $this->foo->get();}}
因此，在构造函数中有一个可选参数，如果您不提供值，将会得到本地的foo函数作为默认值。您只需执行以下操作：
$a = new A();$a->whatever();
但如果您想要覆盖它，可以这样做：
$foo = new SomeOtherFoo();$a = new A($foo);$a->whatever();
然后，在您的测试中，您可以构建一个显式的模拟：
class MockFoo implements FooInterface{public function get(): string{return 'some test value';}}
并传递该实例：
$foo = new MockFoo();$a = new A($foo);$a->whatever();
或者使用PHPUnit模拟构建器：
$foo = $this->getMockBuilder(FooInterface::class)->... // 具体操作
如果您想要一个更简单的版本，您可以只使用一个可调用对象：
class A{protected $foo;public function __construct(callable $foo = null){$this->foo = $foo ?? 'foo';}public function whatever(){$foo = call_user_func($this->foo);}}// 默认用法$a = new A();$a->whatever();// 测试用法$a = new A(fn() => 'some other value');$a->whatever();
英文:
If you need to abstract away a native function, then make a contract for it, so that you can use dependency injection for the service it provides:
interface FooInterface{public function get(): string;}
Then provide a default implementation that uses the native function:
class NativeFoo implements FooInterface{public function get(): string{return \foo();}}
Then your main class might look something like:
class A{protected FooInterface $foo;public function __construct(FooInterface $foo = null){$this->foo = $foo ?? new NativeFoo();}public function whatever(){$foo = $this->foo->get();}}
So, you've got an optional argument in the constructor, and if you don't provide a value, you'll get the native foo function as a default. You'd just do:
$a = new A();$a->whatever();
But if you want to override that, you can do:
$foo = new SomeOtherFoo();$a = new A($foo);$a->whatever();
Then, in your test, you can build an explicit mock:
class MockFoo implements FooInterface{public function get(): string{return 'some test value';}}
And pass in instance of that:
$foo = new MockFoo();$a = new A($foo);$a->whatever();
Or use the PHPUnit mock builder:
$foo = $this->getMockBuilder(FooInterface::class)->... // whatever
If you want a simpler version, you could just use a callable:
class A{protected $foo;public function __construct(callable $foo = null){$this->foo = $foo ?? 'foo';}public function whatever(){$foo = call_user_func($this->foo);}}// default usage$a = new A();$a->whatever();// test usage$a = new A(fn() => 'some other value');$a->whatever();
答案2
得分: 0
I'll provide the translation of the code portion you provided:
在我全心全意同意Alex Howansky的答案的同时，如果你真的执意要使用纯函数，那么你总是可以：```lang-phpnamespace Mocks {function foo() {return 'foo';}}namespace Bar {function foo() {return \Mocks\foo();}}namespace Baz {function foo() {return \Mocks\foo();}}namespace asdf {var_dump(\Bar\foo(), \Baz\foo());}
输出：
string(3) "foo"string(3) "foo"
尽管在这一点上，我们实际上只是在重新发明类，但更糟糕。 在某个时候，你会想知道如何给变量命名空间，这是不可能的，所以你将在那一点上将其整合到类中。
类、接口、特性/继承将是最合适的解决这个问题的方式。
<details><summary>英文:</summary>While I whole-heartedly agree with Alex Howansky's answer, if you're truly hell-bent on using plain functions, then you can always:```lang-phpnamespace Mocks {function foo() {return 'foo';}}namespace Bar {function foo() {return \Mocks\foo();}}namespace Baz {function foo() {return \Mocks\foo();}}namespace asdf {var_dump(\Bar\foo(), \Baz\foo());}
Output:
string(3) "foo"string(3) "foo"
Though at this point we're really just on our way to reinventing classes, but worse. At some point you're going to wonder how to namespace a variable, and that is simply not possible, so you'll be rolling this into classes at that point.
Classes, interfaces, and traits/inheritance would be most properly leveraged to solve this problem.
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