Question

我在我的app.pagination中使用zend paginator工作得很好但我得到这个错误试图获取非对象的属性

$user = new Zend_Session_Namespace('user');
$user_id =$user->user_id;
 $DB = Zend_Db_Table_Abstract::getDefaultAdapter();
 $select = $DB->select()
     ->from(array('p' => 'phone_service'))
     ->join(array('u' => 'user_preferences'), 'u.phone_service_id = p.phone_service_id')
     ->where('u.user_preferences_name = ?', 'is_user_package_active')
     ->where('p.user_id = ?', $user_id);

     $adapter = new Zend_Paginator_Adapter_DbSelect($select);
     $paginator = new Zend_Paginator($adapter);
     $this->view->paginator=$paginator;

在我看来我正在这样做

 foreach($this->paginator as $record){
 <td><?php echo $record->phone_service_name;?></td>
  <td><?php echo $record->phone_service_type;?></td>
  <td ><?php echo $record->phone_service_Duration;?></td>}

给出以下错误

Notice: Trying to get property of non-object

当我在我的观点中这样做时

var_dump($this->paginator);
 [_tableCols:protected] => Array
                    (
                    )
there are no data of columns

任何提示PLZ ???

Answer 1

看起来您的查询返回一个空记录集。要调试它，请尝试：

$select = $DB->select()
     ->from(array('p' => 'phone_service'))
     ->join(array('u' => 'user_preferences'), 'u.phone_service_id = p.phone_service_id')
     ->where('u.user_preferences_name = ?', 'is_user_package_active')
     ->where('p.user_id = ?', $user_id);
echo $select;
die();

这应该输出发送到mysql的确切查询，你可以将它复制粘贴到phpmyadmin / heidisql或者用于管理mysql的任何内容，看看它是否返回你期望的内容。从那里你可以尝试使查询工作，然后将其移植回你的PHP代码。

Answer 2

make it sure column names in DB and in your code are same.mean spelling are same
or $record['phone_service_name']
 once i got this error and the problem was i miss spell column name hope it helps

试图在Zend Paginator中获取非对象的属性

2 个答案: