爲什麼這個php函數調用失敗？

Question

我嘗試使用下面的PHP類：

<?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(); 
?> 

...但得到以下錯誤：

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

任何想法爲什麼？這是一種訪問問題嗎？謝謝。

Answer 1

$services未定義。你的意思是self::$services？

Answer 2

$services變量在函數外聲明。當使用類，你必須使用$this關鍵字一樣，訪問它：

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

後來編輯：對於靜態變量，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); 
     } 

要參考$類的服務，而不是函數的局部$服務。

Answer 4

我現在無法測試它，但我的直覺是移動$service未被實例化。我傾向於在這種情況下使用init變量的方法。在這種情況下，您可以創建另一種方法，使其生成

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

實例化。