php未定义变量的注意事项&；正在尝试获取非对象的属性 - php notices of Undefined Variable & Trying to get property of non-object

php notices of Undefined Variable & Trying to get property of non-object

Am learning&通过Joomla项目执行php如何改进此代码&解决PHP通知-任何建议-解决方案-非常感谢！！

注意：未定义的变量：第140行*/home/mygames/public_html/contensions/com_toys/models/category.php中的cond（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：正在尝试获取第200行上/home/truecar7/public_html/contensions/com_toys/models/category.php中非对象的属性

第200行返回$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Lodder&Elin，它是有效的，但像这样，否则它在返回$row行上显示行的未定义变量通知。

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;
}

尝试使用以下方法。我对您的函数进行了一些更改，并在数据库查询中使用了Joomla2.5编码标准。

$Name = "XXXXXXXXX";  //define the name variable
function getItemIdByName($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

对于您的Undefined Notice，您必须像这样修改您的代码

$cond = '';
if($Carmodel!="") {
   $cond = " and prod_id='$Carmodel' ";
}

对于Trying to get property of non-object通知：我认为$row是空的，这就是抛出通知的原因。检查$row

var_dump($row);

问题：

$database->loadObject(); // This line