Php 仅使用类的名称作为字符串访问类

我试图在一个函数中使用一个已经初始化的类，只将该类的字符串名传递给该函数

例如：

class art{  
    var $moduleInfo = "Hello World";  
}

$art = new Art;

getModuleInfo("art");

 function getModuleInfo($moduleName){
   //I know I need something to happen right here.  I don't want to reinitialize the class since it has already been done.  I would just like to make it accessible within this function.

   echo $moduleName->moduleInfo;
}

---

谢谢你的帮助

var$moduleInfo

使$moduleInfo成为一个（php4样式，公共）实例属性，即类

art

的每个实例都有自己的（单独的）成员$moduleInfo。因此，该类的名称对您没有多大好处。您必须指定/传递要引用的实例。
也许您正在寻找静态属性，请参见

---

将对象本身作为参数传递到函数中

class art{  
    var $moduleInfo = "Hello World";  
}

function getModuleInfo($moduleName){
   //I know I need something to happen right here.  I don't want to reinitialize the class since it has already been done.  I would just like to make it accessible within this function.

   echo $moduleName->moduleInfo;
}

$art = new Art;

getModuleInfo($art);

class art{  
    var $moduleInfo = "Hello World";  
}

function getModuleInfo($moduleName){
   //I know I need something to happen right here.  I don't want to reinitialize the class since it has already been done.  I would just like to make it accessible within this function.

   echo $moduleName->moduleInfo;
}

$art = new Art;

getModuleInfo($art);