Yii 异步 jsonp 请求
[英] Yii Async jsonp request
本文介绍了Yii 异步 jsonp 请求的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！
问题描述
我是 Yii PHP 框架的新手，请耐心等待.
我需要发出跨域 JSONP 请求(来自非 yii 应用程序)以在 Yii 应用程序数据库中创建记录.创建后应通过 getVisit
返回 Application/json 内容
控制器:
公共函数 actionGetVisit($id){header('内容类型:应用程序/json');$visit = Visit::model()->findByPK((int)$id);echo CJSON::encode($visit);Yii::app()->end();}/*** 创建一个新模型.* 如果创建成功，浏览器将被重定向到查看"页面.*/公共函数 actionCreate(){$model=新访问；//如果需要 AJAX 验证，取消注释以下行//$this->performAjaxValidation($model);if(isset($_POST['访问'])){$model->attributes=$_POST['访问'];if($model->save())$this->redirect(array('getVisit','id'=>$model->id));}$this->render('create',array('模型'=>$模型，));}
表格:
<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 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 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 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 class="row"><label for="Visit_guest_name" class="required">客人姓名<span class="required">*</span></label><input size="60" maxlength="256" name="Visit[guest_name]" id="Visit_guest_name" type="text" value="1">
<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 class="行按钮"><input type="submit" name="yt0" value="Create">
</表单>
JS:
$('#visit-form').submit(function(event){警报('提交')；event.preventDefault();var $form = $(this);$.ajax({网址:$(this).attr('action'),数据类型:'jsonp'，类型:'POST'，数据:$form.serialize()+'&ajax='+$form.attr('id'),成功:功能(数据，文本状态，XMLHttpRequest){alert('成功');if (data != null && typeof data == 'object'){$.each(data, function(key, value){$('#error').append(value);});}},错误:函数(XMLHttpRequest，textStatus，errorThrown){警报(错误抛出)；警报('错误')；}});返回假；});
提交后:看起来它没有出错或成功.回复说:
GET http://host/index.php?r=visit/create&callback=jQuery15102636089683510363_1317230765087&Visit%5Brvc_id%5D=1&Visit%5Bzone%5D=1&Visit%;Visit%5Bseat%5D=1&Visit%5Buser_id%5D=1&Visit%5Bguest_name%5D=1&Visit%5Bcreated%5D=1&_=1317230785272 无法显示网址
响应设置为输出文本/
有谁知道这个错误是什么意思?表单在没有js的情况下完美提交.但我似乎无法让 ajax 请求工作.我将它设置为jsonp"，这样跨域问题就会消失.但是我不确定 Yii 后端是否可以处理作为 jsonp 发送的数据.任何帮助表示赞赏！
解决方案
这不是一个 Yii 的问题，更像是一个 JSONP 问题；您的 GetVisit 函数应如下所示:
公共函数 actionGetVisit($id){header('内容类型:应用程序/json');$visit = Visit::model()->findByPK((int)$id);$json = CJSON::encode($visit);回声 $_GET['回调'] .' (' . $json . ');';Yii::app()->end();}
jQuery 将全局临时函数附加到 window 对象，该函数在 JSONP 请求期间插入脚本时调用.jQuery 替换了 ?使用生成的函数名称(即 jsonp1232617941775)调用内联函数.您正在将该函数传递给 window 对象.
希望对您有所帮助，如果我在做一个项目时没有正确解释，请道歉.
I am new to the Yii PHP Framework so bear with me.
I need to make a cross-domain JSONP request(from a non-yii app) to create a record in the Yii apps DB. Upon creating it should return Application/json content via getVisit
The Controllers:
public function actionGetVisit($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,
));
}
The Form:
<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>
The 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){
$('#error').append(value);
});
}
},
error: function(XMLHttpRequest, textStatus, errorThrown)
{
alert(errorThrown);
alert('error');
}
});
return false;
});
Upon Submission:
It looks like its not erroring or hitting success. The response says:
GET http://host/index.php?r=visit/create&callback=jQuery15102636089683510363_1317230765087&Visit%5Brvc_id%5D=1&Visit%5Bzone%5D=1&Visit%5Btable%5D=1&Visit%5Bseat%5D=1&Visit%5Buser_id%5D=1&Visit%5Bguest_name%5D=1&Visit%5Bcreated%5D=1&_=1317230785272 The URL can’t be shown
The response is set to output text/
Does anyone know what this error means? The form submits perfectly without the js. But I just cant seem to get the ajax request working. I set it to 'jsonp' so cross-domain issues would go away. But I'm not sure if the Yii backend can handle the data sent as jsonp. Any help is appreciated!
解决方案
It's not really a Yii question, more of a JSONP issue; here's what your GetVisit function should look like:
public function actionGetVisit($id)
{
header('Content-type: application/json');
$visit = Visit::model()->findByPK((int)$id);
$json = CJSON::encode($visit);
echo $_GET['callback'] . ' (' . $json . ');';
Yii::app()->end();
}
jQuery 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.
Hope that helps, apologize if it's not correctly explained as I am working on a project.
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