查询Zend框架
问题描述:
这是用户控制的我的索引行动:查询Zend框架
$users = new Application_Model_DbTable_Users();
$this->view->users = $users->fetchAll(
$users->select('userid,username')
->order('userid ASC')
->limit(10, 0));
这是我的观点:
<?php
echo "<pre>";
echo print_r($this->users);
?>
在输出我想用户表的结果的JSON,但数组在未来的观点是 是
Zend_Db_Table_Rowset Object
(
[_data:protected] => Array
(
[0] => Array
(
[userid] => 1
[username] => rahul
[firstname] => rahul1
[lastname] => Khan2
[password] => ��2jr�``�(E]_�=^
[email] => [email protected]
[avatar] => 4cfe07efd2e1c.jpg
[updatedon] => 2011-01-23 18:45:49
[createdon] => 0000-00-00 00:00:00
[featuredgibs] =>
[defaultgib] =>
)
完全
但我想只有JSON:
{
"userid":"1",
"username": "rahul"
}
答
你必须定义您在从部分所需的列,试试这个:
$users = new Application_Model_DbTable_Users();
$select = $users->select()
->from('table', array('userid', 'username')) // important
->order('userid ASC')
->limit(10, 0);
$this->view->users = Zend_Json::encode($users->fetchAll($select)); // output json
$用户 - >使用fetchall($选择)是返回一个Zend_Db_Table_Rowset对象,我将如何访问数据,因为它受保护? – XMen 2011-01-24 06:26:45