PHP 实例化类的一点摘记_php实例
下面是调用一个模型(Module)的函数。这个函数的基本功能是指定一个模型(抽象化为类)的名称,然后它会在模型目录下面寻找这个类的脚本实例化以后返回。这样的做法有一点好处就是载入和实例化是自动的,你可以获得最大的灵活性。下面请看下面的代码,它并不长而且不复杂:
function &load_class($class_name, $param = null, $instantiate = true)
{
static $objects = array();
$class_name = ucfirst(strtolower($class_name));
if (isset($objects[$class_name])) {
return $objects[$class_name];
}
$class_file = DIR_MODELS . "/{$class_name}.inc.php";
if (file_exists($class_file)) {
require_once $class_file;
if (!class_exists($class_name)) {
return false;
} else {
$objects[$class_name] =& new $class_name($param);
return $objects[$class_name];
}
} else {
if ($instantiate) {
$objects[$class_name] = null;
}
return null;
}
}函数只有三个参数,分别是 $class_name 、$param 以及 $instaniate ,其中 $param 是构造函数的参数,$instaniate 是可选的。请注意函数中的 $objects 数组是一个静态变量,也就是当调用完这个函数的时候数组并不会释放,下次调用此函数时这个数组的数据是会保存的。这样做的好处就是可以将大部分的类实例了以后,如需要重复调用则直接返回这个类的实例就可以了,避免了重复调用,提高了性能。代码如下:
static $objects = array();
if (isset($objects[$class_name])) {
return $objects[$class_name];
}其它继续的代码就是检测是否有这个类名称的文件,如果有载入这个文件并寻找指定名称的类,如找到了这个类以后就实例化。这要求脚本中类的名称必须和脚本的文件名是一致的。我想这也有利于以后的代码管理。
$instaniate 参数这个时候就发挥了功效,这个参数会告诉函数如果未找到则在 $objects 下面做一个标记位(null)避免函数又重复的寻找文件名并重复载入和寻找。
$class_file = DIR_MODELS . "/{$class_name}.inc.php";
if (file_exists($class_file)) {
require_once $class_file;
if (!class_exists($class_name)) {
return false;
} else {
$objects[$class_name] =& new $class_name($param);
return $objects[$class_name];
}
} else {
if ($instantiate) {
$objects[$class_name] = null;
}
return null;
}其中语句:
$objects[$class_name] =& new $class_name($param);可以多次的推敲一下。$class_name 在函数中是一个字符串变量。关键字 new 可以动态的实例化指定字符串的类(如果存在的话)。有关此调用方法可以参见 PHP 手册和这里。
此函数的不足之处就是如何去考虑传递不同个数的参数给每个不同的类的构造函数。或许可以使用 call_user_func_array 等函数实现,但是这样的做法非常的不 Grace。在这里需要推敲一下。其实 file_exists 等文件存在的测试可以交给 __autoload 函数处理,不过由于其他的函数比如 interface_exists 等也会调用 __autolaod 函数,出于兼容性的考虑,所以只在函数内做一个简单的测试。
PHP5 相对 PHP4 而言更加的面向对象。我想是时候更新我们我们的编码思想了。有关 PHP5 的类和对象,这里有一个非常好的教程。

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