Laravel 5.1 save something in database with ajax call - php

Question

I am building an application where users can upload projects. I am implementing a system where users can 'Like/Unlike' other projects. I am trying to use an AJAX call to save likes. Users are able to like projects on the detail page of a project (/projects/{id})
I have a table users, projects and likes. My plan is to save the likes in the likes table obviously so a record looks like this: id, user_id, project_id. In the future I can do a COUNT query and find out how many likes each project has, etc.
For some reason when I click on the like button my application gets stuck and keeps loading until it crashes.
My files:
Ajax call
$(document).ready(function(){
console.log("js loaded");
$('#likeform').on('submit', function(e) {
console.log("like submitted");
e.preventDefault();
$.ajax({
url: $(this).parent().attr('action'),
type: "post",
data: {'_token': token, 'user_id': $('input[name=user_id]').val(), 'project_id': $('input[name=project_id]').val()},
success: function(data){
alert(data);
}
})
});
});
My Form
{!! Form::open(array('url'=>'projects/'.$project->id.'/like','method'=>'POST', 'id'=>'likeform')) !!}
<input type="hidden" id="token" value="{{ csrf_token() }}">
{!! Form::Submit('Like', array('class'=>'send-btn')) !!}
{!! Form::close() !!}
My Routes:
Route::get('/', 'LikeController#index');
Route::post('projects/{id}/like', 'LikeController#like');
My like function in the LikeController:
public function like()
{
if(Request::ajax()) {
$data = Input::all();
print_r($data);die;
}
}

Answer 1

The
gets stuck and keeps loading until it crashes
part may be related to the fact that the ajax post expects some kind of response, but you are not responding at all in your controller. Try something like
public function like()
{
if(Request::ajax()) {
$data = Input::all();
return Response::json([
'error' => false,
'insertedData' => $data
], 200);
}
}
Remenber to use Response in your controller. Now you are returning something to the ajax call and you should be able to access $data
success: function(data){
alert(data.insertedData);
}
Are you experiencing the same behaviour? If not you can move on and perform the actual db insert.