我正在學習 php，其中一項任務是創建某個類。我已經嘗試了很多，但仍然有錯誤。任務：創建類User，將有$name和$age私人財產和公共$site財產，其價值應在類的構造函數，方法來設置getFullInfo()，將返回（不能打?。。?$this->name at the age of $this->age is a user of $this->site。構造函數應該檢查輸入年齡，如果它大于130或小于0設置$age值unset（就像一個字符串）。成功創建實例后，構造函數應該打印User was successfully created!.我使用 PHP 5。我學習的站點不適用我的解決方案版本。希望，你會幫助我^^<?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"; } } ?>
3 回答
胡子哥哥
TA貢獻1825條經驗 獲得超6個贊
你很接近。將變量、參數和參數命名為有意義的名稱。
如果它大于 130 或小于 0，則將 $age 值設置為未設置（就像字符串一樣）
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 $user->getFullInfo();
// Output: User was successfully created!Tony McTony at the age of 35 is a user of Tree House
紫衣仙女
TA貢獻1839條經驗 獲得超15個贊
我認為這里是：
return "$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;
希望能幫助到你
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