严格的标准错误,无法解决

时间:2016-07-28 14:51:59

标签: php

我尝试但无法解决,检查并且在严格标准

上收到错误
  

严格的标准:只有变量才能通过引用传递给......

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'  , $modelNewcar->loadMainCat($brand,$Carmodel,$minprice,$maxprice,$fuel_type));
    $this->assignRef('CatName'  , $modelNewcar->getCatName);
parent::display($tpl);

以下是原始代码

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);
}

1 个答案:

答案 0 :(得分:2)

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()传递给assignRef()。解决方案是首先将其分配给变量。

$catName = $modelNewcar->getCatName($id);
$this->assignRef('CatName'  , $catName);

文档:http://php.net/manual/en/language.references.pass.php