php - Php doc:键入一个字符串类名,该类名将始终实现给定的接口
问题描述
我有一个返回类名(字符串)的函数,但返回的类总是实现相同的接口:
<?php
interface BaseInterface {}
class SomeClass implements BaseInterface {}
class AnotherClass implements BaseInterface {}
/**
* @return ....
*/
function getClassName($someCondition): string
{
if ($someCondition === 42) {
return SomeClass::class;
} else {
return AnotherClass::class;
}
}
是否可以在@return
标签中提供此信息?像这样的东西(想象的语法):
/** @return BaseInterface::class */
最终目标是让 IDE 理解这getClassName(...)::myInterfaceMethod()
是有效的。
编辑:问题的具体用例
这是一个可运行的示例,显示了我::class
在实际代码中的使用方式。
我将代码改写得尽可能短,但它实际上是 Symfony 项目的一部分,使用 php 7.4 运行。
这个想法是为每个社交网络实现一个类;所有这些小类都实现SocialNetworkInterface
并且只有静态方法。它们永远不会被实例化,但是通过调用它们的静态方法来查询它们。
注意:变量$user
应该输入为User
,但为了缩短代码,我只是在参数中不输入它们。
我的问题中解释的问题(IDE 抱怨)发生在第 79 行,里面renderLinkForUserNetwork()
:
$class = $networkClass::getFontAwesomeLogoClass();
$url = $networkClass::getUrlForUsername($networkUsername);
<?php
interface SocialNetworkInterface
{
public static function getName(): string;
public static function getBaseUrl(): string;
public static function getUrlForUsername(string $username): string;
public static function getFontAwesomeLogoClass(): string;
}
abstract class AbstractSocialNetwork implements SocialNetworkInterface
{
public static function getUrlForUsername(string $username): string
{
// Most social networks use this kind of urls, but of course, this can
// be overloaded in final classes:
$baseUrl = static::getBaseUrl();
return "$baseUrl/$username";
}
public static function getFontAwesomeLogoClass(): string
{
$name = static::getName();
return "fab fa-$name";
}
}
class Facebook extends AbstractSocialNetwork
{
public static function getName(): string
{
return 'facebook';
}
public static function getBaseUrl(): string
{
return ('https://www.facebook.com');
}
}
class Twitter extends AbstractSocialNetwork
{
public static function getName(): string
{
return 'twitter';
}
public static function getBaseUrl(): string
{
return ('https://twitter.com');
}
}
class SocialNetworkHandler
{
public function getSocialNetworksForUser($user): array {
// Should return all the social network classes for which the user has
// information to give (normally depends on the user, fake data is given
// here for simplicity):
return [
Facebook::class,
Twitter::class,
];
}
public function renderLinkForUserNetwork($user, string $networkClass): ?string
{
// Make it runnable for this example, normally retrieved from $user
// and $networkClass:
$networkUsername = 'example';
// An ugly workaround: of course, $networkClass is not an instance of
// SocialNetworkInterface, but calling static methods work the same
// between class name strings and objects:
/** @var SocialNetworkInterface $networkClass */
// When not using the ugly workaround just above, the IDE is complaining
// here (method not found in string):
$class = $networkClass::getFontAwesomeLogoClass();
$url = $networkClass::getUrlForUsername($networkUsername);
return <<< HTML
<a href="$url"><i class="$class"></i></a>
HTML;
}
public function renderLinksForUser($user): string
{
$links = [];
foreach ($this->getSocialNetworksForUser($user) as $networkClass) {
$links[] = $this->renderLinkForUserNetwork($user, $networkClass);
}
return implode("\n", $links);
}
}
$handler = new SocialNetworkHandler();
$user = null; // Should be a User entity
echo $handler->renderLinksForUser($user);
输出:
<a href="https://www.facebook.com/example"><i class="fab fa-facebook"></i></a>
<a href="https://twitter.com/example"><i class="fab fa-twitter"></i></a>
解决方案
不,不幸的是,它不是这样工作的。您应该在此处指定返回类型,即使 BaseInterface::class 是动态的,它也会是返回值(类型为string
)。
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