假设我有3个对象:"Place"， "Person"， "Action"。
根据人所在的地点和年龄，人可以做出不同的动作。
例如:
$place->person->action->drive(); // OK if place is "parking" and "person" is 18+
$place->person->action->learn(); // OK if the place is "school" and person is less than 18.
如何从Action类中访问对象"Person"answers"Place"的数据?
类示例:
class Place {
public $person;
private $name;
function __construct($place, $person) {
$this->name = $place;
$this->person = $person;
}
}
class Person {
public $action;
private $name;
private $age;
function __construct($name, $age) {
$this->name = $name;
$this->age = $age;
$this->action = new Action();
}
}
class Action {
public function drive() {
// How can I access the person's Age ?
// How can I acess the place Name ?
}
public function learn() {
// ... Same problem.
}
}
我想当我创建动作对象时，我可以将"$this"从Person传输到Action。$this->action = new action ($this))，但是Place数据呢?
让Person成为Place的属性或者Action成为Person的属性是没有意义的。
我更倾向于为Person和Place的属性创建公共getter ，要么使它们成为Action的可注入属性，要么至少将它们作为参数传递给Action的方法，例如
class Place
{
private $name;
public function __construct($name)
{
$this->name = $name;
}
public function getName()
{
return $this->name;
}
}
class Person
{
private $name;
private $age;
public function __construct($name, $age)
{
$this->name = $name;
$this->age = $age;
}
public function getName()
{
return $this->name;
}
public function getAge()
{
return $this->age();
}
}
class Action
{
private $person;
private $place;
public function __constuct(Person $person, Place $place)
{
$this->person = $person;
$this->place = $place;
}
public function drive()
{
if ($this->person->getAge() < 18) {
throw new Exception('Too young to drive!');
}
if ($this->place->getName() != 'parking') {
throw new Exception("Not parking, can't drive!");
}
// start driving
}
}