我有以下代码: 班级定义:
<?php
class Person{
var $name;
public $height;
protected $socialInsurance = "yes";
private $pinnNumber = 12345;
public function __construct($personsName){
$this->name = $personsName;
}
public function setName($newName){
$this->name = $newName;
}
public function getName(){
return $this->name;
}
public function sayIt(){
return $this->pinnNumber;
}
}
class Employee extends Person{
}
以及实例的部分:
<!DOCTYPE html>
<HTML>
<HEAD>
<META charset="UTF-8" />
<TITLE>Public, private and protected variables</TITLE>
</HEAD>
<BODY>
<?php
require_once("classes/person.php");
$Stefan = new Person("Stefan Mischook");
echo("Stefan's full name: " . $Stefan->getName() . ".<BR />");
echo("Tell me private stuff: " . $Stefan->sayIt() . "<BR />");
$Jake = new Employee("Jake Hull");
echo("Jake's full name: " . $Jake->getName() . ".<BR />");
echo("Tell me private stuff: " . $Jake->sayIt() . "<BR />");
?>
</BODY>
</HTML>
输出:
Stefan's full name: Stefan Mischook.
Tell me private stuff: 12345
Jake's full name: Jake Hull.
Tell me private stuff: 12345 // Here I was expecting an error
据我所知,私有变量只能从它自己的类中访问,受保护变量也可以从扩展类的类中访问。我有私有变量$pinnNumber
。所以我预计,如果我拨打$Jake->sayIt()
,我会收到错误消息。因为$Jake
是扩展class Employee
的{{1}}的成员。变量class Person
只能从$pinnNumber
访问,而不能从class Person
访问。
问题出在哪里?
答案 0 :(得分:4)
实际上,这不是它的工作原理。
由于您没有扩展sayIt()
方法,因此没有“可访问性问题”,如果您执行了类似的操作,则会有一个:
<?php
class Person{
var $name;
public $height;
protected $socialInsurance = "yes";
private $pinnNumber = 12345;
public function __construct($personsName){
$this->name = $personsName;
}
public function setName($newName){
$this->name = $newName;
}
public function getName(){
return $this->name;
}
public function sayIt(){
return $this->pinnNumber;
}
}
class Employee extends Person{
public function sayIt(){
return $this->pinnNumber;//not accessible from child class
}
}
答案 1 :(得分:-1)
受保护,公共和私人只是环境范围,在您的情况下,对于一个类。由于您的satIt()
功能为public
,因此您可以访问具有正确环境范围的功能,以访问任何private
或protected
个变量。
如果您尝试这样做:
$Jake->pinnNumber
在课外,你会得到错误。
您应该在方法和类中更多地研究Scopes,然后您可以转到匿名函数;)