Question

如何在PHP中调用作为参数传递的对象的方法？

这是thefirstclass.php

中定义的第一个类

class TheFirstClass {

    private $_city='Madrid';

    public function city() {
        return $_city
    }   
}

这是在secondcondclass.php中定义的第二个类

class TheSecondClass {

    public function myMethod($firstClassObject) {

        echo "City: " . $firstClassObject->city(); // <- Why This method doesn´t work?      

    }
}

最后，这是 include_once“class / thefirstclass.php”; include_once“class / thesecondclass.php”;

$firstClassObject = new TheFirstClass();
$secondClassObject = new TheSecondClass();

$secondClassObject->myMethod($firstClassObject);

Answer 1

问题不在于TheFirstClass::city内TheSecondClass::myMethod的调用，而是TheFirstClass::city返回局部变量（$_city）而不是实例变量（ $this->_city）。与C ++等语言不同，PHP实例变量必须始终通过对象访问，即使在方法中也是如此。

这是工作代码：

class TheFirstClass {
    private $_city = "a";

    public function city() {
        return $this->_city;
    }   
}

class TheSecondClass {
    public function myMethod($firstClassObject) {
        echo "City: " . $firstClassObject->city(); // <- Why This method doesn´t work?
    }
}

$test = new TheFirstClass();
$test2 = new TheSecondClass();
$test2->myMethod($test);

在PHP中将对象作为参数传递时如何访问对象方法？

1 个答案: