很抱歉这个奇怪的话题,但我不知道如何用其他方式表达它。
我正在尝试从调用类访问方法。就像这个例子一样:
class normalClass {
public function someMethod() {
[...]
//this method shall access the doSomething method from superClass
}
}
class superClass {
public function __construct() {
$inst = new normalClass;
$inst->someMethod();
}
public function doSomething() {
//this method shall be be accessed by domeMethod form normalClass
}
}
这两个类不通过继承相关,我不想将函数设置为静态。
有什么办法可以实现这一点吗?
感谢您的帮助!
您可以像这样传递对第一个对象的引用:
class normalClass {
protected $superObject;
public function __construct(superClass $obj) {
$this->superObject = $obj;
}
public function someMethod() {
//this method shall access the doSomething method from superClass
$this->superObject->doSomething();
}
}
class superClass {
public function __construct() {
//provide normalClass with a reference to ourself
$inst = new normalClass($this);
$inst->someMethod();
}
public function doSomething() {
//this method shall be be accessed by domeMethod form normalClass
}
}
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