Yii的异步JSONP请求 [英] Yii Async jsonp request
问题描述
我是新来的Yii的PHP框架如此忍受我。
I am new to the Yii PHP Framework so bear with me.
我需要做一个跨域JSONP请求（从非警予的应用程序），以创造纪录的Yii的应用服务数据库。一旦创建，它应该返回通过getVisit应用/ JSON内容
I need to make a cross-domain JSONP request(from a non-yii app) to create a record in the Yii apps DB. Upon creating it should return Application/json content via getVisit
该控制器：
public function actionGetVisit($id)
{
header('Content-type: application/json');
$visit = Visit::model()->findByPK((int)$id);
echo CJSON::encode($visit);
Yii::app()->end();
}
/**
* Creates a new model.
* If creation is successful, the browser will be redirected to the 'view' page.
*/
public function actionCreate()
{
$model=new Visit;
// Uncomment the following line if AJAX validation is needed
// $this->performAjaxValidation($model);
if(isset($_POST['Visit']))
{
$model->attributes=$_POST['Visit'];
if($model->save())
$this->redirect(array('getVisit','id'=>$model->id));
}
$this->render('create',array(
'model'=>$model,
));
}
表单：
<form id="visit-form" action="http://host/index.php?r=visit/create" method="post">
<p class="note">Fields with <span class="required">*</span> are required.</p>
<div class="row">
<label for="Visit_rvc_id" class="required">Rvc <span class="required">*</span></label> <input name="Visit[rvc_id]" id="Visit_rvc_id" type="text" value="1"> </div>
<div class="row">
<label for="Visit_zone" class="required">Zone <span class="required">*</span></label> <input name="Visit[zone]" id="Visit_zone" type="text" value="1"> </div>
<div class="row">
<label for="Visit_table" class="required">Table <span class="required">*</span></label> <input name="Visit[table]" id="Visit_table" type="text" value="1"> </div>
<div class="row">
<label for="Visit_seat" class="required">Seat <span class="required">*</span></label> <input name="Visit[seat]" id="Visit_seat" type="text" value="1"> </div>
<div class="row">
<label for="Visit_user_id" class="required">User <span class="required">*</span></label> <input name="Visit[user_id]" id="Visit_user_id" type="text" value="1"> </div>
<div class="row">
<label for="Visit_guest_name" class="required">Guest Name <span class="required">*</span></label> <input size="60" maxlength="256" name="Visit[guest_name]" id="Visit_guest_name" type="text" value="1"> </div>
<div class="row">
<label for="Visit_created" class="required">Created <span class="required">*</span></label> <input name="Visit[created]" id="Visit_created" type="text" value="1"> </div>
<div class="row buttons">
<input type="submit" name="yt0" value="Create"> </div>
</form>
在JS：
$('#visit-form').submit(function(event)
{
alert('submit');
event.preventDefault();
var $form = $(this);
$.ajax({
url: $(this).attr('action'),
dataType: 'jsonp',
type: 'POST',
data : $form.serialize()+'&ajax='+$form.attr('id'),
success: function(data, textStatus, XMLHttpRequest)
{
alert('success');
if (data != null && typeof data == 'object'){
$.each(data, function(key, value){
$('#error').append(value);
});
}
},
error: function(XMLHttpRequest, textStatus, errorThrown)
{
alert(errorThrown);
alert('error');
}
});
return false;
});
在提交： 它看起来像它不是示数或打成功。回应说：
Upon Submission: It looks like its not erroring or hitting success. The response says:
GET http://host/index.php?r=visit/create&callback=jQuery15102636089683510363_1317230765087&Visit%5Brvc_id%5D=1&Visit%5Bzone%5D=1&Visit%5Btable%5D=1&Visit%5Bseat%5D=1&Visit%5Buser_id%5D=1&Visit%5Bguest_name%5D=1&Visit%5Bcreated%5D=1&_=1317230785272 The URL can’t be shown
响应设置为输出文本/
The response is set to output text/
有谁知道这个错误意味着什么？表单提交完全没有JS。但我只是不能似乎得到Ajax请求的工作。我将它设置为JSONP所以跨域的问题会消失。但我不知道，如果Yii的后端可以处理将作为JSONP数据。任何帮助AP preciated！
Does anyone know what this error means? The form submits perfectly without the js. But I just cant seem to get the ajax request working. I set it to 'jsonp' so cross-domain issues would go away. But I'm not sure if the Yii backend can handle the data sent as jsonp. Any help is appreciated!
推荐答案
这不是一个真正的Yii的问题，更多的是JSONP问题;这里就是你的GetVisit功能应该是这样的：
It's not really a Yii question, more of a JSONP issue; here's what your GetVisit function should look like:
public function actionGetVisit($id)
{
header('Content-type: application/json');
$visit = Visit::model()->findByPK((int)$id);
$json = CJSON::encode($visit);
echo $_GET['callback'] . ' (' . $json . ');';
Yii::app()->end();
}
jQuery的重视全局临时函数时，脚本JSONP请求期间被插入时调用的窗口对象。 jQuery的替换？与调用内联函数生成的函数名（即jsonp1232617941775）。您传递给函数的窗口对象。
jQuery attaches a global temporary function to the window object that is called when the script is inserted during a JSONP request. jQuery replaces the ? with a generated function name (ie jsonp1232617941775) that calls the inline function. You are passing that function to the window object.
希望帮助，道歉，如果它不能正确解释，因为我工作的一个项目。
Hope that helps, apologize if it's not correctly explained as I am working on a project.
这篇关于Yii的异步JSONP请求的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
