为什么这个php函数调用失败？

Question

我尝试使用下面的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(); 
?> 

...但得到以下错误：

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

任何想法为什么？这是一种访问问题吗？谢谢。

Answer 1

$services未定义。你的意思是self::$services？

Answer 2

$services变量在函数外声明。当使用类，你必须使用$this关键字一样，访问它：

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

后来编辑：对于静态变量，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); 
     } 

要参考$类的服务，而不是函数的局部$服务。

Answer 4

我现在无法测试它，但我的直觉是移动$service未被实例化。我倾向于在这种情况下使用init变量的方法。在这种情况下，您可以创建另一种方法，使其生成

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

实例化。