如何通过 PHP 中的另一个变量引用对象？

Question

我正在学习 php，但遇到了以下令人头疼的问题：

我有 2 个对象，比方说

class Fruit {

public $weight;

}

$apple = new Fruit();

$apple->weight = 1;

$banana = new Fruit();

$banana->weight = 2;

后来，我从用户那里得到一些输入作为变量（比如，你最喜欢哪种水果？）：

$user_preference =  'apple';

现在，如何动态引用正确的对象？如何获得类似的东西

echo $user_preference->weight; 

?

Answer 1

我会创建一个地图（例如，当从数据库中检索数据时很有用）$apple = new Fruit();$apple->weight = 1;$banana = new Fruit();$banana->weight = 2;$fruitMap = ['apple'=>$apple,'banana'=>$banana];$user_preference =  'apple';echo $fruitMap[$user_preference]->weight;但是检查密钥是否存在

Answer 2

您可以使用变量变量：<?phpclass Fruit {&nbsp; &nbsp; public $weight;}$apple = new Fruit();$apple->weight = 1;$banana = new Fruit();$banana->weight = 2;$user_preference =&nbsp; 'apple';//&nbsp; &nbsp;vv---------------- Check this notationecho $$user_preference->weight;&nbsp; // outputs 1自己测试一下请注意，这可能会导致安全漏洞，因为永远不要相信用户输入。永远不要相信用户输入，尤其是在控制代码执行时。永远不要相信用户输入。想象一下echo $$user_input;，用户输入是database_password为避免这种情况，您需要清理用户输入，例如：<?phpclass Fruit {&nbsp; &nbsp; public $weight;}$apple = new Fruit();$apple->weight = 1;$banana = new Fruit();$banana->weight = 2;$allowed_inputs = ['apple', 'banana'];$user_preference =&nbsp; 'apple';if (in_array($user_preference, $allowed_inputs)){&nbsp; &nbsp; echo $$user_preference->weight;&nbsp; // outputs 1}else{&nbsp; &nbsp; echo "Nope ! You can't do that";}但这是以输入更多代码为代价的。ka_lin 的解决方案更安全，更易于维护