创建的类有错误.帮助找到它，请)

Question

我正在学习php，其中一个任务是创建一个特定的类。我已经尝试了很多，但仍然有错误。

任务：

创建类User，它将具有$name和$age私有属性和公共$site属性，这些值应该在类构造函数中设置，方法getFullInfo()，这将返回(而不是打印！ $this->name at the age of $this->age is a user of $this->site .构造函数应检查输入年龄，如果它大于130或小于0 $age值设置为unset(仅作为字符串(。成功创建实例后，构造函数应打印User was successfully created! .

我使用PHP 5。

我学习的网站不应用我的解决方案版本。希望你能帮助我^^

<?php 
class User{
    private $name;
    private $age;
    public $site;
    function __construct($q,$w,$e){
        echo "User was successfully created!";
        $this->name=$q;
        $this->site=$e;
        if($w>130 || $w<0){
            unset($w);
        };
        $this->age=$w;
    }
    public function getFullInfo(){
        return "$this->name at the age of $this->age is a user of $this->site";
    } 
} 
?>

Answer 1

该任务说，如果年龄超出范围，则应将其设置为字符串unset。打电话给unset($w)不会这样做。将其更改为：

$w = "unset";

Answer 2

你很接近。将变量、参数和参数命名为有意义的名称。

如果大于 130 或小于 0，则将$age值设置为未设置(仅 作为字符串(

可能意味着$this->age = 'unset';

<?php 
class User
{
    private $name;
    private $age;
    public $site;
    function __construct($name, $site, $age)
    {
        echo "User was successfully created!";
        $this->name = $name;
        $this->site = $site;
        if ($age > 130 || $age < 0) {
            $this->age = 'unset';
        } else {
            $this->age = $age;
        }
    }
    public function getFullInfo()
    {
        return "$this->name at the age of $this->age is a user of $this->site";
    }
}
$user = new User('Tony McTony', 'Tree House', 35);
echo $user->getFullInfo();
// Output: User was successfully created!Tony McTony at the age of 35 is a user of Tree House

<</div> div class="one_answers">

我认为在这里：

 return "$this->name at the age of $this->age is a user of $this->site";
// it should be : (i think)
return $this->name . "at the age of " . $this->age . " is a user of " . $this->site;

希望对你有帮助

Answer 3

您的问题很可能是取消设置$w变量。当他们说取消设置变量时，您应该对$this->age而不是$w这样做。

<?php
$this->age = $w;
if($w > 130 || $w < 0){
   $this->age = 'unset';
};
?>

如果这不能解决您的问题，那么对问题的更具描述性的解释会有所帮助。