Question

我正在尝试使用以下PHP类：

<?php 

class Service {

    public $code, $description;

    public static $services = array(
        "A"         =>      "Shipping",
        "B"     =>      "Manufacturing",
        "C"         =>      "Legal",
        "D"         =>      "Accounts Receivable",
        "E"         =>      "Human Resources",
        "F"         =>      "Security",
        "G"         =>      "Executive",
        "H"         =>      "IT"
    );

    public function _construct( $c, $d) {
        $this->code = $c;
        $this->description = $d;
    }

    public static function getDescription( $c ){
        return $services[$c];
    }

    public static function generateServiceList() {

        $service_list[] = array();

        foreach ($services as $k => $v ){
            $service_list[] = new Service( $k, $v );
        }

        return $service_list;

    }

}

?>

......以下列方式：

<?php
    $services = Service::generateServiceList();
?>

...但收到以下错误：

Warning: Invalid argument supplied for foreach() in /classes/service.php on line 31

知道为什么吗？这是某种访问问题吗？感谢。

Answer 1

$services未定义。你的意思是self::$services吗？

Answer 2

$services变量在函数外声明。使用类时，您必须使用$this关键字来访问它：

...
foreach ($this->services as $k => $v ){
...

稍后修改：对于静态变量，请使用self::$services代替$this->services。

Answer 3

foreach ($services as $k => $v ){
            $service_list[] = new Service( $k, $v );
        }

到

foreach ($this->services as $k => $v ){
            $service_list[] = new Service( $k, $v );
        }

您想要引用Class的$ services而不是函数的本地$ services。

Answer 4

我现在无法测试它，但我的直觉是移动$service没有被实例化。在这种情况下，我倾向于使用init变量的方法。在这种情况下，您可以创建另一个方法

$services = array(
        "A"         =>      "Shipping",
        "B"     =>      "Manufacturing",
        "C"         =>      "Legal",
        "D"         =>      "Accounts Receivable",
        "E"         =>      "Human Resources",
        "F"         =>      "Security",
        "G"         =>      "Executive",
        "H"         =>      "IT"

实例化。

为什么这个php函数调用失败？

4 个答案: