How to correctly return html template from ajax request with Symfony
I was wondering how to return correctly an HTML template (SEO friendly) from an ajax call.
In my app, I use 2 differents ways to return response:
For simple template:
public function ajaxCallAction() {
//.....
$response = array(
"code" => 202,
"success" => true,
"simpleData" => $simpleData
);
return new JsonResponse($response);
}
And in the JS I do something like:
$("div#target").click(function(event) {
$.ajax({
type: "POST",
success: function(response) {
if(response.code === 202 && response.success) {
$("div#box").append(response.simpleData);
}
}
});
});
For complexe template (more than 20 lines and various var):
public function ajaxCallAction() {
//...
$listOfObjects = $repo->findAll();
$viewsDatas = [
'listOfObjects' => $listOfObjects,
//....other vars
];
return $this->render('myAppBundle:template:complexTemplate.html.twig', $viewsDatas);
//in complexTemplate.html.twig, I loop on listOfObjects for example...
}
And for this kind of call, the JS looks like:
$("div#target").click(function(event) {
$.ajax({
type: "POST",
success: function(response) {
$("div#box").append(response);
}
});
});
All those methods are working, but with the second one, we dont have status code (does it matter?) and I know that returning directly formated html template can be heavy (according to for example this topic Why is it a bad practice to return generated HTML instead of JSON? Or is it?).
How are you guys doing ajax calls? What are the best practices here?
In my apps, I typically use something like that:
$response = array(
"code" => 200,
"response" => $this->render('yourTemplate.html.twig')->getContent() );
Hope this help!
Edit
To return your response, you should use JsonResponse as explained in docs: http://symfony.com/doc/current/components/http_foundation.html#creating-a-json-response
Simply use that:
return new JsonResponse($response);