我来自Java,在这里我们可以执行以下操作:
Action.java:
public interface Action {
public void performAction();
}
MainClass.java:
public class MainClass {
public static void main(String[] args) { //program entry point
Action action = new Action() {
public void performAction() {
// custom implementation of the performAction method
}
};
action.performAction(); //will execute the implemented method
}
}
如您所见,我不是在创建实现Action的类,而是在声明时直接实现接口.
这样的事情甚至可以用PHP实现吗?
我尝试过的
action.php:
<?php
interface Action {
public function performAction();
}
?>
myactions.php:
include "action.php";
$action = new Action() {
public function performAction() {
//do some stuff
}
};
我得到的是:
Parse error: syntax error, unexpected '{' in myactions.php on line 3
所以,我的问题是:PHP可能会发生这种情况吗?我该怎么办?
解决方法:
不行不行PHP不提供像Java这样的匿名类.但是,您可以尝试模拟所需的行为,但是结果将……最好混合在一起.
这是一些代码:
interface Action
{
public function performAction();
}
class MyClass
{
public function methodOne($object)
{
$object->performAction(); // can't call directly - fatal error
// work around
$closure = $object->performAction;
$closure();
}
public function methodTwo(Action $object)
{
$object->performAction();
}
}
$action = new stdClass();
$action->performAction = function() {
echo 'Hello';
};
$test = new MyClass();
$test->methodOne($action); // will work
$test->methodTwo($action); // fatal error - parameter fails type hinting
var_dump(method_exists($action, 'performAction')); // false
var_dump(is_callable(array($action, 'performAction'))); // false
希望能帮助到你!