Php 严格的标准错误,无法解决
我试过了,但没能解决,检查过了,在严格的标准上得到了错误 严格的标准:只有变量应该通过引用传递 我做错了什么-如何纠正Php 严格的标准错误,无法解决,php,Php,我试过了,但没能解决,检查过了,在严格的标准上得到了错误 严格的标准:只有变量应该通过引用传递 我做错了什么-如何纠正 $this->assignRef('prodDet' , $prodDet); $this->assignRef('CatName' , $modelNewcar->getCatName($id)); $this->assignRef('nav' , $nav); $this->assignRef('CatList' , $modelNe
$this->assignRef('prodDet' , $prodDet);
$this->assignRef('CatName' , $modelNewcar->getCatName($id));
$this->assignRef('nav' , $nav);
$this->assignRef('CatList' , $modelNewcar->loadMainCat($brand,$Carmodel,$minprice,$maxprice,$fuel_type));
$this->assignRef('CatName' , $modelNewcar->getCatName);
parent::display($tpl);
下面是原始代码
function display($tpl = null)
{
$mainframe =JFactory::getApplication();
$option = JRequest::getCmd('option');
$db =JFactory::getDBO();
$user = JFactory::getUser();
// Push a model into the view
$model = $this->getModel();
$modelNewcar = $this->getModel( 'product' );
$id = JRequest::getVar('id','','default','int');
$vid = JRequest::getVar('vid','','default','int');
$prodDet = $modelNewcar->loadProduct($id,$vid);
$this->assignRef('prodDet' , $prodDet);
$this->assignRef('CatName' , $modelNewcar->getCatName($id));
parent::display($tpl);
}
function display($tpl = null)
{
$db =JFactory::getDBO();
$user = JFactory::getUser();
// Push a model into the view
$model =$this->getModel();
$modelNewcar =$this->getModel( 'category' );
$brand = JRequest::getVar('brand','','default','int');
$Carmodel = JRequest::getVar('model','','default','int');
$minprice = JRequest::getVar('minprice','','default','int');
$maxprice = JRequest::getVar('maxprice','','default','int');
$fuel_type = JRequest::getVar('fuel_type','','default','');
$this->assignRef('nav' , $nav);
$this->assignRef('CatList' , $modelNewcar->loadMainCat($brand,$Carmodel,$minprice,$maxprice,$fuel_type));
$this->assignRef('CatName' , $modelNewcar->getCatName);
parent::display($tpl);
}
这正是错误所说的
$this->assignRef()
通过引用分配变量
例如:
$this->assignRef('CatName' , $modelNewcar->getCatName($id));
在这里,您尝试将函数getCatName()
传递到assignRef()
。解决方案是首先将其分配给一个变量
$catName = $modelNewcar->getCatName($id);
$this->assignRef('CatName' , $catName);
文档:这正是错误所说的
$this->assignRef()
通过引用分配变量
例如:
$this->assignRef('CatName' , $modelNewcar->getCatName($id));
在这里,您尝试将函数getCatName()
传递到assignRef()
。解决方案是首先将其分配给一个变量
$catName = $modelNewcar->getCatName($id);
$this->assignRef('CatName' , $catName);
文件: