问题描述
所以我在主类中有这种struct函数
So i have this kind of struct function in main class
function __construct(){
$this->conf = $GLOBALS['conf'];
$this->dbi = new dbinfo;
$this->modOpt = new modOptions;
$this->lang = new language;
/** Connect DB extended Class **/
parent::__construct($GLOBALS['connect']);
}
我在其中定义类,但是这些类进入到库文件中,该文件包含在文件的开头,除了一个外,当请求后出现时,该文件将包含在该文件中:
where i define classes, but this classes is into library file which is included at the start of file except one, which is included when post request will appear like this:
if (isset($_POST['delGroup']) && isset($_SESSION['content_viewer']) && $_SESSION['content_viewer']['code'] >= 1){
include_once(realpath(dirname(__FILE__) . '/../..')."/mod/dbinfo/proc.php");
}
所以我想将检查添加到我的dbinfo类的构造函数中
so i want add check into my construct function for dbinfo class like this
function __construct(){
$this->conf = $GLOBALS['conf'];
if (isset(new dbinfo))
$this->dbi = new dbinfo;
$this->modOpt = new modOptions;
$this->lang = new language;
/** Connect DB extended Class **/
parent::__construct($GLOBALS['connect']);
}
,但是此方法与 if isset
一起使用时不起作用,请向我展示如何检查类是否存在于文件中的正确方法.谢谢
but this method with if isset
does not works, please show me correct way how to check if class exists into file. thanks
推荐答案
尝试使用 class_exists()
http://php.net/manual/en/function.class-exists.php
在您的情况下,寻找 dbinfo
类可以做到这一点:
In your case, looking for dbinfo
class do this:
if(class_exists('dbinfo')){
//do something
如果您的类具有名称空间,请包含ful名称空间的类名.
If your class has a namespace, include the ful namespaced classname.
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