Question

我有一个配置类，它是一个抽象类。我想设置它，以便自动检测网站所在的服务器，然后分配适当的常量。我在第10行遇到错误$this->hostName = $_SERVER['SERVER_NAME'];期待`T_FUNCTION。这样做的正确方法是什么？有更好的方法吗？这是我班级的第一部分

abstract class config{

 public $hostName;
 public $hostSlices;
 public $domain;
echo $_SERVER['SERVER_NAME'];
//strips out the "www" from the server name but only if it has the name on it . 
$this->hostName = $_SERVER['SERVER_NAME'];
$this->hostSlices = explode(".",$this->hostName);
if($this->hostSlices[0]=="www"){array_shift($this->hostSlices);}
$this->domain = join(".",$this->hostSlices);

//depending on which domain is used, different config setup is used. 
switch ($this->domain){
    case "localhost":*/

  const HOST = "localhost";//would http://localhost/ work as well. In that case, then this and the SITE_ROOT  could be the same variable and i would preferentially set them depending on the host that the site is on.
  const USER = "root";
  const PWD = "xxxxxxx";
  const NAME = "hurunuitconz";//database name

  //public $name = "hello from the config class";//you cannot access variables from an abstract class  you should define constants and then the can be used anywhere

 ###########Location of file groups########
  const SITE_ROOT = "http://localhost";
  const ADMIN_IMAGES = 'http://localhost/images/user_images/admin_images';

        break;



    case "charles.hughs.natcoll.net.nz":

  const HOST = "charles.hughs.natcoll.net.nz";//would http://localhost/ work as well. In that case, then this and the SITE_ROOT  could be the same variable and i would preferentially set them depending on the host that the site is on.
  const USER = "charles_andrew";
  const PWD = "xxxxxxx";
  const NAME = "charles_hurunuitconz";//database name
 ###########Location of file groups########
  const SITE_ROOT = "http://charles.hughs.natcoll.net.nz/_Assignments/Industry/www";//this is just confusing the way natcoll makes us do this. 
  const ADMIN_IMAGES = 'http://charles.hughs.natcoll.net.nz/_Assignments/Industry/www/images/user_images/admin_images';

        break;
    }

Answer 1

抽象类不应该允许您设置私有数据（只有一个继承的具体类）。

另外，请查看stackoverflow中的this link，以获取有关SERVER_NAME与HTTP_HOST的有趣讨论

Answer 2

您必须将所有代码包装在构造函数中，或者更好地包含一个名为init()或somthing的函数。然后，当您在子类中重写init时，您将调用parent::init()。你的意思是让这个类是静态的，而不是抽象的吗？

Answer 3

您构建代码的方式是不对的。 PHP确实允许许多疯狂的东西，但定义常量和输出代码作为抽象类的一部分在OOP方面没有任何意义。

您可能要做的是拥有一个帮助程序类，它根据本地服务器名称定义配置设置。为此，您有几个选择：

选项1）使用构造函数创建常规类。

class Config {
   public $x; 
   public $y;

   public function __construct() {
      switch (...) {
          $this->x = 2;
          $this->y = 3;
      }
   }
}

并像这样使用它：

$config = new Config();
echo "Variable y: " . $config->y;

选项2）抽象类中的静态方法。

abstract class Config {
   public $boolInitialized = false;

   public static function init() {
      if (self::$boolInitialized) {
          return;
      }

      self::$boolInitialized = true;

      switch (...) {
        self::$x = 1;
        self::$y = 2;
      }
   }

   public static function getX() {
      self::init();
      return self::$x;
   }
}

并像这样使用它：

echo Config::getX();

Answer 4

你不能在没有声明第一个方法的情况下输出类中的内容这将失败

abstract class fo
{
public $fo;
echo $fo;
}

**but this will work**

abstract class fo
{
public $fo;

public function __construct()
{
  $this->fo = $_SERVER['SERVER_NAME'];

}

public function sayFo()
{
  echo $this->fo;
}

}

Php抽象类站点配置

4 个答案: