PHP-访问外部对象的属性和方法
我是php新手,我想知道为什么这不起作用:PHP-访问外部对象的属性和方法,php,Php,我是php新手,我想知道为什么这不起作用: class A { private $containigObject; public function __construct($containigObject){ $this->containigObject = $containigObject; } public function functionThatAccessOutsideObject(){ //...
class A {
private $containigObject;
public function __construct($containigObject){
$this->containigObject = $containigObject;
}
public function functionThatAccessOutsideObject(){
//...
$this->containgObject->b = 2;
//...
}
}
class B {
public $b = 1;
public function someFunction(){
//...
$a = new A($this);
$a->functionThatAccessOutsideObject();
//...
}
}
$b = new B();
$b->someFunction();
我试图在类A中使用类B中的属性和方法
class A {
private $**containigObject**;
public function __construct($containigObject){
$this->containigObject = $containigObject;
}
public function functionThatAccessOutsideObject(){
//...
$this->**containgObject**->b = 2;
//...
}
}您的代码确实有效!我只能在A的构造函数中更改$b,而不能更改它的方法。当我执行var_dump$b时,它会显示object(b)#1(1){[“b”]=>int(1)}您应该共享您的错误消息。如果上面的代码是从实际代码粘贴而来的,那么您的问题是拼写。您的私有var在declaration和use.OMG中的拼写不同。你说得对,我现在明白了。谢谢。