在PHP类中动态创建实例变量 [英] Dynamically creating instance variables in PHP classes

问题描述

我不确定这是否是一个简单的问题，但在PHP类中：

I'm not sure if this is a trivial questions but in a PHP class:

MyClass：
$ b

MyClass:

class MyClass {
   public $var1;
   public $var2;

    constructor() { ... }

    public method1 () {

    // Dynamically create an instance variable
         $this->var3 = "test"; // Public....?


    }
}

主要：

$test = new MyClass();
$test->method1();
echo $test->var3; // Would return "test"

我将如何让这工作？ Ps。我写了这么快，所以请忽略我在设置类或调用方法时出现的任何错误。

Does this work?? How would I get this to work? Ps. I wrote this quickly so please disregard any errors I made with setting up the class or calling methods!

EDIT
这些实例变量，我创建私人??

EDIT What about making these instance variables that I create private??

编辑2
感谢所有回应 - 每个人都是正确的 - 我应该自己测试了，但我第二天早上有一个考试，在学习时有这个想法，我想检查它是否工作。人们不断地暗示其糟糕的OOP - 也许，但它允许一些优雅的代码。让我解释一下，看看你还是这样认为。这是我想出的：

EDIT 2 Thanks all for responding - Everyone is right - I should have just tested it out myself, but I had an exam the next morning and had this thought while studying that I wanted to check to see if it worked. People keep suggesting that its bad OOP - maybe but it does allow for some elegant code. Let me explain it a bit and see if you still think so. Here's what I came up with:

//PHP User Model: 

class User {
    constructor() { ... }

    public static find($uid) {
         $db->connect(); // Connect to the database

         $sql = "SELECT STATEMENT ...WHERE id=$uid LIMIT 1;";
         $result = $db->query($sql); // Returns an associative array

         $user = new User();

         foreach ($result as $key=>$value)
            $user->$$key = $value; //Creates a public variable of the key and sets it to value

         $db->disconnect();
    }
}

//PHP Controller:

function findUser($id) {

    $User = User::find($id);

    echo $User->name;
    echo $User->phone;
    //etc...

}

只是把它放在一个关联数组，但我不能正确地命名该数组有意义的（即$ user-> data ['name'] ...丑陋。）无论哪种方式，你必须知道什么是数据库，所以我真的不明白什么是参数是它的混乱，特别是因为你可以只是var dump对象进行调试。

I could have just put it in an associative array but I can never correctly name that array something meaningful (ie. $user->data['name'] ... ugly.) Either way you have to know what is in the database so I do not really understand what the argument is that its confusing, especially since you can just var dump objects for debugging.

推荐答案

会真的工作。自动创建的实例变量具有公开的可见性。

Yes that will indeed work. Auto-created instance variables are given public visibility.

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