我正在使用CakePHP v2.2.1稳定版。我有UsersController
行动add()
。我正在尝试通过ajax(从主页到/users/add
)发送用户信息并保存数据。我的代码是这样的:
// /app/View/Pages/home.ctp
<?php
$data = array('User' => array('username' => 'vegeta_super_sayajin',
'password' => 'over9000!', 'email' => 'vegeta@supersayajin.com',
'profile_pic' => '/home/pics/scouter.jpg', 'firstname' => 'Vegeta',
'lastname' => 'Vegeta', 'level_id' => '9001'));
?>
<script type="text/javascript">
var data = <?php echo json_encode($data); ?> //convert $data into json format
$.ajax({url: '/users/add', data: "data="+data, type: 'post'});
</script>
如何在UsersController
中收到此数据,以便我可以处理并保存?
目前,我正在尝试:
// /app/Controller/UsersController.php
function add() {
if($this->request->is('post') {
//returns "Error: [object Object] in logfile
$this->log($this->request->data);
} else {
$this->Session->setFlash(__("The user could not be saved :("));
}
$this->autoRender = false;
}
$this->log($this->request->data)
会在Error: [object Object]
文件中返回/app/tmp/logs/error.log
,并且此用户信息在$this->request->params
的任何索引中都不存在。到目前为止,我所有的谷歌搜索都只返回了复杂的cakephp v1.3技术。怎么在cakephp v2.2.1中完成?
答案 0 :(得分:1)
您可以尝试以下代码。它会对你有用。
<?php
$data = array(
'User' => array(
'username' => 'vegeta_super_sayajin',
'password' => 'over9000!',
'email' => 'vegeta@supersayajin.com',
'profile_pic' => '/home/pics/scouter.jpg',
'firstname' => 'Vegeta',
'lastname' => 'Vegeta',
'level_id' => '9001')
);
?>
<script type="text/javascript">
var data = [<?php echo json_encode($data); ?>] //convert $data into json format
$.ajax({
url: 'checks/add',
data: "data="+JSON.stringify(data),
type: 'post'});
</script>
在你的控制器代码中:
// /app/Controller/UsersController.php
function add() {
if($this->request->is('post') {
$this->log(json_encode($this->request->data, true)); //returns "Error: [object Object] in logfile
} else {
$this->Session->setFlash(__("The user could not be saved :("));
}
$this->autoRender = false;
}
这是json_decode documentation。第二个参数true
会将对象转换为数组。