我需要在PHP中有一个类构造函数调用父类的父类的(祖父母?)构造函数,而不调用父类构造函数。

// main class that everything inherits
class Grandpa 
{
    public function __construct()
    {

    }

}

class Papa extends Grandpa
{
    public function __construct()
    {
        // call Grandpa's constructor
        parent::__construct();
    }
}

class Kiddo extends Papa
{
    public function __construct()
    {
        // THIS IS WHERE I NEED TO CALL GRANDPA'S
        // CONSTRUCTOR AND NOT PAPA'S
    }
}

我知道这是一件很奇怪的事情,我正试图找到一种不难闻的方法,但尽管如此,我很好奇这是否可能。


当前回答

另一个不使用标志的选项可能适用于您的情况:

<?php
// main class that everything inherits
class Grandpa 
{
    public function __construct(){
        $this->GrandpaSetup();
    }

    public function GrandpaSetup(){
        $this->prop1 = 'foo';
        $this->prop2 = 'bar';
    }
}

class Papa extends Grandpa
{
    public function __construct()
    {
        // call Grandpa's constructor
        parent::__construct();
        $this->prop1 = 'foobar';
    }

}
class Kiddo extends Papa
{
    public function __construct()
    {
        $this->GrandpaSetup();
    }
}

$kid = new Kiddo();
echo "{$kid->prop1}\n{$kid->prop2}\n";

其他回答

从PHP 7你可以使用

家长:家长:__construct ();

丑陋的解决方法是将一个布尔参数传递给Papa,表明您不希望解析它的构造函数中包含的代码。即:

// main class that everything inherits
class Grandpa 
{
    public function __construct()
    {

    }

}

class Papa extends Grandpa
{
    public function __construct($bypass = false)
    {
        // only perform actions inside if not bypassing
        if (!$bypass) {

        }
        // call Grandpa's constructor
        parent::__construct();
    }
}

class Kiddo extends Papa
{
    public function __construct()
    {
        $bypassPapa = true;
        parent::__construct($bypassPapa);
    }
}

你必须使用外公::__construct(),没有其他的快捷方式。此外,这也破坏了Papa类的封装——当读取或处理Papa时,应该可以安全地假设__construct()方法将在构造过程中被调用,但Kiddo类不会这样做。

<?php

class grand_pa
{
    public function __construct()
    {
        echo "Hey I am Grand Pa <br>";
    }
}

class pa_pa extends grand_pa
{
    // no need for construct here unless you want to do something specifically within this class as init stuff
    // the construct for this class will be inherited from the parent.
}

class kiddo extends pa_pa
{
    public function __construct()
    {
        parent::__construct();
        echo "Hey I am a child <br>";
    }
}

new kiddo();
?>

当然,这期望您不需要在pa_pa的构造中做任何事情。运行该命令将输出:

嘿,我是爷爷 嘿,我是个孩子

    class Grandpa 
{
    public function __construct()
    {
        echo"Hello Kiddo";
    }    
}

class Papa extends Grandpa
{
    public function __construct()
    {            
    }
    public function CallGranddad()
    {
        parent::__construct();
    }
}

class Kiddo extends Papa
{
    public function __construct()
    {

    }
    public function needSomethingFromGrandDad
    {
       parent::CallGranddad();
    }
}