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获取类构造函数(Get class constructor)

 如何获得一个类构造函数名称而不实例化类？  
 例：  
$class = 'someClass';
$constructor = somehow get constructor;

$args = array();
$object = call_user_func_array(array($class,$constructor),$args);
 
 我需要的是通过将不确定数量的变量传递给它的构造函数来创建一个对象。 

how to get a class constructor function name without instantiating the class? 
example: 
$class = 'someClass';
$constructor = somehow get constructor;

$args = array();
$object = call_user_func_array(array($class,$constructor),$args);
 
what I need is to create a object by passing a undetermined number of variables into it's constructor.

原文：https://stackoverflow.com/questions/3378276

更新时间：2024-01-14 14:01

最满意答案

 尝试调整transform-origin：  
document.body.style.transformOrigin = 'top left';
document.body.style.transform = 'scale(' + scaleFactor + ')';
 
 您可能还需要调整身体的宽度和高度以匹配缩放比例。  
var scaleFactor = 1.5;
document.body.style.transformOrigin = 'top left';
document.body.style.transform = 'scale(' + scaleFactor + ')';
document.body.style.width = 100 * scaleFactor + "%";
document.body.style.height = 100 * scaleFactor + "%";
 
 有关变换的注释。 变换是虚构的，不会改变物理尺寸，包括x，y，宽度和高度。 所以你必须手动管理这些物理尺寸，以匹配你的“变换”尺寸，以保持滚动条的快乐。 

Try adjusting the transform-origin: 
document.body.style.transformOrigin = 'top left';
document.body.style.transform = 'scale(' + scaleFactor + ')';
 
You may also need to adjust the width and height of the body to match the scaling. 
var scaleFactor = 1.5;
document.body.style.transformOrigin = 'top left';
document.body.style.transform = 'scale(' + scaleFactor + ')';
document.body.style.width = 100 * scaleFactor + "%";
document.body.style.height = 100 * scaleFactor + "%";
 
A note concerning transforms. Transforms are imaginary and don't alter physical dimensions including x, y, width and height. So you'll have to manage these physical dimensions manually to match your "transform'd" dimensions in order to keep the scroll bars happy.

获取类构造函数(Get class constructor)

最满意答案

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