最近,我完成了一个web项目,其中php版本为5.3,然后我将其上传到一个服务器中,其中php版本为5.2.*。Class name must be a valid object or a string in /home/user_folder/public_html/_includes/class.myclass.php= get_called_class(); //
$object = new $className; //
/ObjectManager/ObjectManager.php:70]
#4 Magento\Framework\ObjectManager\ObjectManager->get() called at\Data->__construct() called at [vendor/magento/framework/ObjectManager/Factory/AbstractFactory.php:121/Data.php<
\Factory\Compiled->create() called at [vendor/magento/framework/ObjectManager/Factory/Compiled.php:150\Compiled->get() called at [vendor/magento/framework/ObjectManager/Factory/Compiled.php:79]
#9 Magento
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一个叫web1.php,另一个叫web2.php。我有一个菜单在两个页面的顶部,它通过链接调用两个页面。我必须能够传递并显示按下的“提交”按钮(在web1.php中有3个按钮,一个名为"Hi“、"Hello”、"Hola"),然后在我的标题上按下web1.php的链接,当我返回到“您以前按了hi”的页面时我可以让它显示在web2.php上,但不知道如何再次显示后,按下链接。echo "called using the HI";
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