代码之家 › 专栏 › 技术社区 › Gufran Hasan

是否能够使用箭头运算符->通过对象访问静态属性?[副本]

oop php

-1

Gufran Hasan · 技术社区 · 7 年前

这个问题已经有了答案:

Reference â What does this symbol mean in PHP? 18个答案
PHP: What if I call a static method in non-static way 3个答案

正如我读到的 Static properties cannot be accessed through the object using the arrow operator ->. 静态属性可以使用resolution运算符通过类名访问。

在下面的示例中,我可以使用 arrow operator -> 是的。

class Foo {
   public static $name="I am php";
    public static function aStaticMethod() {
        // ...
        echo 'In Static method';
    }
}

Foo::aStaticMethod();//output: In Static method    
$obj = new Foo;   
$obj->aStaticMethod();//output: In Static method
$obj->name;

输出:

In Static methodIn Static method

但是当试图通过 operator -> 它给出以下错误:

PHP Notice:  Accessing static property Foo::$name as non static in /home/jdoodle.php on line 14
PHP Notice:  Undefined property: Foo::$name in /home/jdoodle.php on line 14

php真的支持oops吗?你说的是什么? Static properties cannot be accessed through the object using the arrow operator -> 是吗?

谢谢

2 回复 | 直到 7 年前

1

2

Gufran Hasan 7 年前

问题是由于对什么是财产的误解, a property is a variable defined in a class 是的。如果展开上面的示例并添加属性…

<?php
error_reporting(E_ALL);
ini_set('display_errors', 1);

class Foo {
    public static $a = "static property";

    public static function aStaticMethod() {
        // ...
        echo 'In Static method';
    }
}

Foo::aStaticMethod();//output: In Static method
echo Foo::$a;
$obj = new Foo;
$obj->aStaticMethod();//output: In Static method
echo $obj->a;

这试图引用 $a ,但输出是…

In Static methodstatic propertyIn Static methodPHP Notice:  Accessing static property Foo::$a as non static in /home/nigel/workspace2/Test/t2.php on line 18
PHP Stack trace:
PHP   1. {main}() /home/nigel/workspace2/Test/t2.php:0
PHP Notice:  Undefined property: Foo::$a in /home/nigel/workspace2/Test/t2.php on line 18
PHP Stack trace:
PHP   1. {main}() /home/nigel/workspace2/Test/t2.php:0

Notice: Accessing static property Foo::$a as non-static in /home/nigel/workspace2/Test/t2.php on line 18

Call Stack:
    0.0001     348024   1. {main}() /home/nigel/workspace2/Test/t2.php:0


Notice: Undefined property: Foo::$a in /home/nigel/workspace2/Test/t2.php on line 18

Call Stack:
    0.0001     348024   1. {main}() /home/nigel/workspace2/Test/t2.php:0

(请注意,我已将其设置为报告所有错误/警告等。)

2

0

dpattayath 7 年前

根据php文档:

“将类属性或方法声明为静态的使它们可以访问,而不需要类的实例化。声明为静态的属性不能与实例化的类对象一起访问(尽管静态方法可以)。”

Example from PHP doc