Php 使用Ajax将表单数据发布到ZF2控制器
编辑-Php 使用Ajax将表单数据发布到ZF2控制器,php,ajax,controller,xmlhttprequest,zend-framework2,Php,Ajax,Controller,Xmlhttprequest,Zend Framework2,编辑- 我已经在下面发布了答案。 问题是，当按下提交按钮时，我不理解ZF2如何/在何处发布表单数据。因此，如果（$this->getRequest（）->isPost（））{ 在下面的ajax调用之后，它告诉我没有发布任何数据。 当我执行上面的isPost（）if语句时，当我点击提交按钮，告诉我数据已经发布，然后告诉我表单数据 这是有效的 这里是ajax调用- <script> $.ajax({
我已经在下面发布了答案。 问题是，当按下提交按钮时，我不理解ZF2如何/在何处发布表单数据。因此，如果（$this->getRequest（）->isPost（））{
在下面的ajax调用之后，它告诉我没有发布任何数据。
当我执行上面的
isPost（） <script>
$.ajax({
url: urlform,
type: 'POST',
dataType: 'json',
contentType: "application/json; charset=utf-8",
async: true,
data: ($("#newThoughtForm").serialize() + '&submit=go'),
success: function () {
console.log('SUBMIT WORKS');
setTimeout(function () { <?php echo $this->invokeIndexAction()->test(); ?> ;
}, 1000);
},
//This keeps getting executed because there is no response, as the controller action is not run on a Post()
error: function () {
console.log('There is error while submit');
setTimeout(function () { <?php echo $this->invokeIndexAction()->test(); ?> ;
}, 1000);
}
//I assume the data won't get pushed to the server if there is no response,
//but I can't figure out how to give a response in ZF2 since the controller is not
//run when the Post() is made.
});
Request URL:http://test/newAlbum.html
Request Method:POST
Status Code:200 OK
Request Headersview source
Accept:*/*
Accept-Encoding:gzip,deflate,sdch
Accept-Language:en-US,en;q=0.8
Connection:keep-alive
Content-Length:46
Content-Type:application/x-www-form-urlencoded; charset=UTF-8
Cookie:PHPSESSID=h46r1fmj35d1vu11nua3r49he4
Host:test
Origin:http://test
Referer:http://test/newAlbum.html
User-Agent:Mozilla/5.0 (Windows NT 6.2; WOW64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/30.0.1599.101 Safari/537.36
X-Requested-With:XMLHttpRequest
Form Dataview sourceview URL encoded
album[albumText]:hello world
submit:go
Response Headersview source
Connection:Keep-Alive
Content-Length:4139
Content-Type:text/html
Date:Sun, 20 Oct 2013 16:52:15 GMT
Keep-Alive:timeout=5, max=99
Server:Apache/2.4.4 (Win64) PHP/5.4.12
X-Powered-By:PHP/5.4.12
对ajax调用的请求-
<script>
$.ajax({
url: urlform,
type: 'POST',
dataType: 'json',
contentType: "application/json; charset=utf-8",
async: true,
data: ($("#newThoughtForm").serialize() + '&submit=go'),
success: function () {
console.log('SUBMIT WORKS');
setTimeout(function () { <?php echo $this->invokeIndexAction()->test(); ?> ;
}, 1000);
},
//This keeps getting executed because there is no response, as the controller action is not run on a Post()
error: function () {
console.log('There is error while submit');
setTimeout(function () { <?php echo $this->invokeIndexAction()->test(); ?> ;
}, 1000);
}
//I assume the data won't get pushed to the server if there is no response,
//but I can't figure out how to give a response in ZF2 since the controller is not
//run when the Post() is made.
});
Request URL:http://test/newAlbum.html
Request Method:POST
Status Code:200 OK
Request Headersview source
Accept:*/*
Accept-Encoding:gzip,deflate,sdch
Accept-Language:en-US,en;q=0.8
Connection:keep-alive
Content-Length:46
Content-Type:application/x-www-form-urlencoded; charset=UTF-8
Cookie:PHPSESSID=h46r1fmj35d1vu11nua3r49he4
Host:test
Origin:http://test
Referer:http://test/newAlbum.html
User-Agent:Mozilla/5.0 (Windows NT 6.2; WOW64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/30.0.1599.101 Safari/537.36
X-Requested-With:XMLHttpRequest
Form Dataview sourceview URL encoded
album[albumText]:hello world
submit:go
Response Headersview source
Connection:Keep-Alive
Content-Length:4139
Content-Type:text/html
Date:Sun, 20 Oct 2013 16:52:15 GMT
Keep-Alive:timeout=5, max=99
Server:Apache/2.4.4 (Win64) PHP/5.4.12
X-Powered-By:PHP/5.4.12
Request URL:http://test/newAlbum.html
Request Method:POST
Status Code:200 OK
Request Headersview source
Accept:text/html,application/xhtml+xml,application/xml;q=0.9,image/webp,*/*;q=0.8
Accept-Encoding:gzip,deflate,sdch
Accept-Language:en-US,en;q=0.8
Connection:keep-alive
Content-Length:46
Content-Type:application/x-www-form-urlencoded
Cookie:PHPSESSID=h46r1fmj35d1vu11nua3r49he4
Host:test
Origin:http://test
Referer:http://test/newAlbum.html
User-Agent:Mozilla/5.0 (Windows NT 6.2; WOW64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/30.0.1599.101 Safari/537.36
Form Dataview sourceview URL encoded
album[albumText]:hello world
submit:go
Response Headersview source
Connection:Keep-Alive
Content-Length:4139
Content-Type:text/html
Date:Sun, 20 Oct 2013 16:52:14 GMT
Keep-Alive:timeout=5, max=100
Server:Apache/2.4.4 (Win64) PHP/5.4.12
X-Powered-By:PHP/5.4.12
以下是控制器上的indexAction（），以确保完整性-
public function indexAction()
{
echo 'console.log("Index Action is Called");';
$form = new \AlbumModule\Form\newAlbumForm();
if ($this->getRequest()->isPost()){
echo 'console.log("Data posted");';
$form->setData($this->getRequest()->getPost());
if ($form->isValid()){
echo 'console.log("Form Valid");';
//todo
$this->forward()->dispatch('newAlbum', array('action' => 'submitAlbum'));
return new ViewModel(
array(
'form' => $form
)
);
} else {
echo 'console.log("Form Invalid");';
return new ViewModel(
array(
'form' => $form
)
);
}
} else {
echo 'console.log("No data posted")';
return new ViewModel(
array(
'form' => $form
)
);
}
}
isPost（）我已经在下面发布了答案。通常，当您从ajax发送数据时，不需要再次呈现模板，而ViewModel就是这样做的 尝试将json策略添加到module.config.php中
'view_manager' => array(
//other configuration
'strategies' => array(
'ViewJsonStrategy',
),
),
return [
'view_manager'=>[
'Strategies'=> 'ViewJsonStrategy',
]
]
public function ajaxAction()
{
$request = $this->getRequest();
if ($request->isXmlHttpRequest()){ // If it's ajax call
$data = $request->getPost('data'));
...
}
return new JsonModel($formData);
}
多亏了SzymonM我才弄明白 基本上，您似乎必须以json类型发布到操作，这意味着您发布到的任何控制器/操作都必须为jQueryAjax调用返回“成功”响应，才能返回成功。
因此，尝试发布到索引操作会产生问题，因为您可能试图通过许多if语句返回viewModel对象或json响应。
最好的选择是将json请求发布到另一个操作，该操作将管理请求和响应 Ajax操作-
public function ajaxAction()
{
$form = new \AlbumModule\Form\newAlbumForm();
$request = $this->getRequest();
$response = $this->getResponse();
if ($request->isPost()) {
//$hello is a test variable used for checking if the form is valid
//by checking the response
$hello = 1;
$form->setData($request->getPost());
if ($form->isValid()){
$hello = 4020;
};
}
$messages = array();
if (!empty($messages)){
$response->setContent(\Zend\Json\Json::encode($messages));
} else {
$response->setContent(\Zend\Json\Json::encode(array('success'=>1,'hello'=>$hello)));
}
return $response;
}
Ajax调用-
var urlform=''；
$.ajax({
url:urlform，
键入：“POST”，
数据类型：“json”，
async:true，
数据：$（“#newAlbumForm”）.serialize（），
成功：功能（数据）{
控制台日志（数据）；
}
错误：函数（数据）{
控制台日志（数据）；
}
});
'view_manager' => array(
//other configuration
'strategies' => array(
'ViewJsonStrategy',
),
),
return [
'view_manager'=>[
'Strategies'=> 'ViewJsonStrategy',
]
]
class IndexController extends AbstarctActionController{
protected $accptCretiria = [
'Zend\View\Model\ViewModel'=>'text/html',
'Zend\View\Model\JsonModel'=>'application/json, text/json'
];
public function indexAction(){
//here if is ajax call it returns jsonView ,and if is normal call it return ViewModel
$viewModel = $this->acceptableViewModelSelector($this->acceptCretiria);
return $viewModel
}
}
Zend Framework 2有一个名为WasabiLib的模块。它几乎拥有所有功能，可以以非常方便的方式管理ajax请求和响应。请看一个简单的示例，它提供了一个简单的表单：
//inside the phtml
<form id="simpleForm" class="ajax_element" action="simpleFormExample" method="POST"
data-ajax-loader="myLoader">
<input type="text" name="written_text">
<input type="submit" value="try it">
<i id="myLoader" class="fa fa-spinner fa-pulse fa-lg" style="display: none;"></i>
</form>
//Server-side code
public function simpleFormExampleAction(){
$postArray = $this->getRequest()->getPost();
$input = $postArray['written_text'];
$response = new Response(new InnerHtml("#element_simple_form","Server Response: ".$input));
return $this->getResponse()->setContent($response);
}
//在phtml内部
//服务器端代码
公共函数simpleFormExampleAction（）{
$postArray=$this->getRequest（）->getPost（）；
$input=$postArray['writed_text']；
$response=newresponse（新的InnerHtml（“#元素_简单_形式”，“服务器响应：”.$input））；
返回$this->getResponse（）->setContent（$response）；
}
$this->getRequest”；$data=$request->getPost（）->toArray（）；