1

我在我的 app.pagination 中使用了 zend 分页器，但我得到了这个错误 Trying to get property of non-object in

$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>}

这是给出以下错误

Notice: Trying to get property of non-object

在我看来，当我这样做时

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

任何提示PLZ ???

4

2 回答 2

1

看起来您的查询正在返回一个空记录集。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，您可以将其复制粘贴到 phpmyadmin/heidisql 或您用来管理 mysql 的任何内容中，看看它是否返回您期望的结果。从那里您可以尝试使查询工作，然后将其移植回您的 php 代码。

于 2012-02-07T15:05:42.093 回答
0 
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
于 2012-02-27T11:22:21.473 回答