无法查看从父类继承的属性 [重复]
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【中文标题】无法查看从父类继承的属性 [重复]【英文标题】:Unable to view property inherited from parent class [duplicate] 【发布时间】:2019-08-11 14:02:47 【问题描述】:我创建了一个带有以下变量的银行账户类,$Balance
变量是私有的,所以我使用 setter 和 getter 函数来访问 balance 的私有属性。
class BankAccount
public $HoldersName;
public $AccountNumber;
public $SortCode;
private $Balance = 1;
public $APR = 0;
public $Transactions = []; //array defined.
public function __construct($HoldersName, $AccountNumber, $SortCode, $Balance = 1) //constructor where i.e. HoldersName is the same as the variable $HoldersName.
//constructor automatically called when object is created.
echo "Holders Name: " . $this->HoldersName = $HoldersName . "</br>";
echo "Account Number: " . $this->AccountNumber = $AccountNumber . "</br>";
echo "Sort Code: " . $this->SortCode = $SortCode . "</br>";
$this->Balance = $Balance >= 1 ? $Balance : 1;
public function set_Balance($Balance) //SET balance.
return $this->Balance=$Balance;
public function get_Balance() //GET balance.
return $this->Balance; //Allows us to access the prviate property of $Balance.
SavingsAccount
类扩展了 BankAccount
,因此它继承了 BankAccount
类的所有内容。我创建了一个函数来计算利息,即余额 6800 * (0.8+0.25) * 1。
class SavingsAccount extends BankAccount
public $APR = 0.8;
public $APRPaymentPreference;
public $ExtraBonus = 0.25;
public function CalculateInterest()
$this->Balance = $this->Balance * ($this->APR + $this->ExtraBonus) * 1; //this is where i get the error now
public function FinalSavingsReturn()
这里我创建了一个SavingsAccount
类的实例,余额为6800,我尝试调用函数SavingsAccount::CalculateInterest()
却出现了这个错误:
注意未定义的属性:第 42 行的 SavingsAccount::$Balance
//define objects
$person2 = new SavingsAccount ('Peter Bond', 987654321, '11-11-11', 6800); //create instance of class SavingsAccount
echo "<b>Interest:</b>";
print_r ($person2->CalculateInterest());
【问题讨论】:
【参考方案1】:正如您所说,$Balance
属性是私有的。私有意味着它不能从定义类外部访问,甚至不能从继承该属性的子类访问。
<?php
class Foo
private $bar="baz";
public function __construct()echo $this->bar;
public function getBar()return $this->bar;
class Bar extends Foo
public function __construct()echo $this->bar;
new Foo;
new Bar;
如果我们运行这段代码,我们会得到错误:
baz
PHP Notice: Undefined property: Bar::$bar in php shell code on line 2
所以,你有两个选择。您可以改用您的吸气剂:
<?php
class Foo
private $bar="baz";
public function __construct()echo $this->bar;
public function getBar()return $this->bar;
class Bar extends Foo
public function __construct()echo $this->getBar();
new Foo;
new Bar;
或者,您可以将属性声明为protected
而不是private
:
<?php
class Foo
protected $bar="baz";
public function __construct()echo $this->bar;
public function getBar()return $this->bar;
class Bar extends Foo
public function __construct()echo $this->bar;
new Foo;
new Bar;
其中任何一个都给出了预期的输出:
baz
baz
有关属性和方法可见性的更多详细信息,请参阅documentation here。
【讨论】:
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