PHP的未定义变量和试图获取非对象属性的通知

我正在学习通过在Joomla项目上执行php
如何改进此代码和解决PHP公告-任何建议-解决方案-非常感谢！

注意：未定义的变量：在第140行的* / home / mygames / public_html / components / com_toys / models / category.php中
(这是$sql行)*

   function loadSubCat($id,$Carmodel,$minprice,$maxprice){
   $mainframe =& JFactory::getApplication();
   $option = JRequest::getCmd('option');
   $database =& JFactory::getDBO();
   global $Itemid;  
   if($Carmodel!="")
   $cond=" and prod_id='$Carmodel' ";
   $sql = "Select * from #__toycar_products Where prod_cat_id='".$id."' $cond and prod_status='1' and prod_id in (select v_prod_id from #__toycar_variants)  Order By prod_sorder";

注意：试图在第200行的/home/truecar7/public_html/components/com_toys/models/category.php中获取非对象的属性

第200行是return $row-> id;

   function getItemIdByName($Name){
   $mainframe =& JFactory::getApplication();
   $option = JRequest::getCmd('option');
   $database =& JFactory::getDBO();
   $sql = "Select id  from #__menu Where name = '".$Name."'";
   $database->setQuery($sql);
   $row = $database->loadObject();
   return $row->id;
}

编辑

你好罗德& Elin,它起作用了,但是像这样,否则它在return $row行上显示了未定义的变量通知.

function getItemIdByName($Name){
$db = JFactory::getDBO();
$query = $db->getQuery(true);
$query->select('*')
 ->from('#__menu')      
 ->where('id = ' . $db->quote($Name));      
$db->setQuery($query);
$rows = $db->loadObjectList();

foreach ($rows as $row){
    $row = $row->msg;
     }    
$row='';
return $row;
}

解决方法:

尝试使用以下内容.我对您的功能进行了一些更改,并使用Joomla 2.5编码标准进行数据库查询.

$Name = "XXXXXXXXX";  //define the name variable

function getItemIdByName($Name){
    $db = JFactory::getDBO();
    $query = $db->getQuery(true);
    $query->select('*')
     ->from('#__menu')      
     ->where('id = ' . $db->quote($Name));      
    $db->setQuery($query);
    $rows = $db->loadObjectList();

    foreach ($rows as $row){
        $row = $row->msg;
    }
    return $row;
}

echo getItemIdByName($Name); //echo the result of the function

标签：joomla2-5,joomla,php
来源： https://codeday.me/bug/20191030/1968383.html