创建的类有错误。帮助找到它，请）

Question

我正在学习 php，其中一项任务是创建某个类。我已经尝试了很多，但仍然有错误。任务：创建类User，将有$name和$age私人财产和公共$site财产，其价值应在类的构造函数，方法来设置getFullInfo()，将返回（不能打印！） $this->name at the age of $this->age is a user of $this->site。构造函数应该检查输入年龄，如果它大于130或小于0设置$age值unset（就像一个字符串）。成功创建实例后，构造函数应该打印User was successfully created!.我使用 PHP 5。我学习的站点不适用我的解决方案版本。希望，你会帮助我^^<?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

任务说如果年龄超出范围，您应该将其设置为字符串unset。打电话unset($w)不会那样做。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。

如果它大于 130 或小于 0，则将 $age 值设置为未设置（就像字符串一样）

大概意思 $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

Answer 3

我认为这里是：

 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;

希望能帮助到你