如何在PHP中编写一个简单的对象

时间:2011-11-30 13:58:36

标签: php variables object

这是一个非常简单的问题,但我似乎找不到合适的答案。

假设我在动作脚本3中写了这样一个对象:

var myCar = new Object();
myCar.engine = "Nice Engine";
myCar.numberOfDoors = 4;
myCar.howFast= 150;

我如何在PHP中编写这样的东西?

2 个答案:

答案 0 :(得分:22)

$myCar = new stdClass;
$myCar->engine = 'Nice Engine';
$myCar->numberOfDoors = 4;
$myCar->howFast = 150;

请查看documentation for objects以进行更深入的讨论。

答案 1 :(得分:9)

您可以使用类,例如:

class Car {

public $engine;
public $numberOfDoors;
public $howFast;

}
$myCar = new Car();
$myCar->engine = 'Nice Engine';
$myCar->numberOfDoors = 4;
$myCar->howFast = 150;

或者如果您只需要将此对象用于属性存储,则可以使用关联数组,例如:

 $myCar['engine'] = "Nice engine";
 $myCar['numberOfDoors'] = 4;
 $myCar['howFast'] = 150;