在子PHP oop中導入父函數數據

Question

我試圖通過在子級中調用父函數數據來學習php oop的簡單方法。 但是我在某個地方犯了錯誤。

class name{
    var $firstname;
    var $lastname;
    var $name;
    public function name($firstname, $lastname){
        $this->firstname=$firstname;
        $this->lastname=$lastname;
        $this->name=$this->firstname." ".$this->lastname;
        return $this->name;
    }
}
class sentence extends name{
    var $name;
    var $letter;
    function sentence(){
        $this->letter="My name is ";
        echo $this->letter.$this->name;
    }
}
$name=new name(ABC, Xyz);
$letter=new sentence();

我創建了一個調用來獲取名稱輸入，並創建了另一個子類來編寫句子。 但是不能在孩子中叫名字。

Answer 1

快速修復，因為那可能就是您想要的。

<?php
class name {

private $firstname;
private $lastname;
private $name;
private $sentence;

public function __construct($firstname, $lastname){
    $this->firstname = $firstname;
    $this->lastname = $lastname;
    $this->name = $firstname . " " . $lastname;
    $this->sentence = "My name is : " . $this->name;

}

public function getName(){
    return $this->name;
}

public function getSentence(){
    return $this->sentence;
 }
}

$instance = new name("test","lastname");

echo $instance->getSentence();
?>

Answer 2

class name
{
    public $firstname;
    public $lastname;
    public $name;
    public function name($firstname, $lastname)
    {
        $this->firstname = $firstname;
        $this->lastname = $lastname;
        $this->name = $this->firstname.' '.$this->lastname;

        return $this->name;
    }
}
class sentence extends name
{
    public $name;
    public $letter;
    public function greeting()
    {
        $this->letter = 'My name is ';
        echo $this->letter.$this->name;
    }
}
//$name=new name(ABC, Xyz);
$letter = new sentence('ABC', 'Xyz');
$letter->greeting();

感謝@ManhNguyen

Answer 3

您是否要讀取父類中變量的狀態？

這不是OOP的工作方式。

您必須將所有變量組合到一個類中，或者將其作為參數傳遞給想要訪問數據的函數。

第二，您似乎並沒有使用__construct函數，而是將變量傳遞到也是NO-NO的new name()函數中。