我刚刚创建了一个类来控制我的php应用程序,我有一个大问题(我花了2天来思考和搜索它,但找不到任何解决方案)。我的类包含一个名为 register()
的方法,该方法将脚本加载到页面中。我的班级是:
class Apps
{
protected $_remember; // remember something
public function register($appName)
{
include "$appName.php"; //include this php script into other pages
}
public function set($value)
{
$this->_remember = $value; // try to save something
}
public function watch()
{
return $this->_remember; // return what I saved
}
}
并在time.php
文件中
$time = 'haha';
$apps->set($time);
作为我问题的标题,当我纯粹将time.php
包含在main.php
中时,我可以使用$apps->set($time)
($apps
已在main.php
中定义)。像这样main.php
:
$apps = new Apps();// create Apps object
include "time.php";
echo $apps->watch(); // **this successfully outputs 'haha'**
但是当我从 Apps 类调用方法 register()
以包含 time.php
时,我得到了错误 undefined variable $apps
并为time.php
call set method from none object
(听起来它不接受$apps
time.php
对我来说)。我main.php
是:
$apps = new Apps();// create Apps object
$apps->register('time'); // this simply include time.php into page and it has
//included but time.php doesn't accept $apps from main.php
echo $apps->watch(); // **this outputs errors as I said**
对了,我不擅长写作。所以如果你什么都不懂,就问我。我感谢任何答复。:D
如果您希望第二个代码片段正常工作,请将time.php
的内容替换为:
$time = 'haha';
$this->set($time); // instead of $apps->set($time);
由于此代码由 Apps
类的实例方法include
,因此它将有权访问实例本身,$this
.