How to use AJAX with Laravel?

Question

I have a select that when I change its value it should reorder a list, e.g.: by id, name, etc. without Laravel I would use a function onchange and pass the sorting pattern to a PHP page that would print out this sorted list for me...

But how do I do that with Laravel? I’ll have to generate one view? I tried but I couldn’t...

I was trying here I think I’m missing the time to generate the views..

I have a layout view that has a @yeld ('content'), in case I would need to update only this content, however, when I do:

$this->layout->content = View::make('usuarios.index', $variaveis);

it duplicates my layout within div content

//Follow the controller

   if(Request::ajax()){
       $usuarios = Usuario::orderBy(Input::get('order'),'ASC')->get();
       $variaveis = array('usuarios' => $usuarios);
       return View::make('usuarios.index', $variaveis);;
   }else{
       $usuarios = Usuario::orderBy('id','ASC')->get();
       $variaveis = array('usuarios' => $usuarios);
       $this->layout->content = View::make('usuarios.index', $variaveis);
   }

But he’s never entering Request::ajax() and I can’t figure out why

Answer 1

You don’t have a problem with Laravel. You have to do the same thing, take the events with javascript and assemble the result the same way you did before Laravel.

In the Laravel you will take your function that makes the select and do something more or less like this

public function getConsulta()
{
   //tua consulta normal...  

   if (Request::ajax()) 
   {
      return Response::json($dados);
   }
}

In the javascript you should take this json and treat to update the table

As you are returning all your view and rendering all of it, you can do so in your layout

@if(! Request::ajax())

   // teu código normal quando não é ajax

@Else

   @yield('content') // isso vai fazer renderizar só a tua view

@endif

It is necessary to do this because you are giving a extends in your view

in your controller you will until you can change leaving it so

   $usuarios = Usuario::orderBy(isset(Input::get('order'))?Input::get('order'):'id','ASC')->get();
   $variaveis = array('usuarios' => $usuarios);
   return View::make('usuarios.index', $variaveis);;

It will work for your two situations, when it is or not ajax.

Answer 2

It’s just gonna get into Request::ajax() when the request is made via Ajax! :-)

But, attention: this is checked through the HTTP header HTTP_X_REQUESTED_WITH. That one header is automatically sent in Ajax requests made by jQuery and other Javascript libraries, but if you are not using you need to add it so that Laravel recognizes that the call is Ajax:

XHR.setRequestHeader("X-Requested-With", "XmlHttpRequest");

---

Another alternative is to use different functions in the Controller: one for Ajax another for normal requests, and leave out the if (Request::ajax())

Answer 3

I don’t understand of lavarel, work with the zend and I imagine that it works basically the same way, in ajax, you’re going to have to make some kind of request to have that return, ie in your controller will be necessary to create a action just to meet this request, and a view to assemble the content...if I were doing in zend, I would have to give one disable layout so that it uses only the specific view content, probably in lavarel this also has to be done...

Answer 4

It’s basically the same thing, just swap the directories and insert the token in the form submission, follow an example of what I implemented here:

$.ajax({
    headers: {
        'X-CSRF-Token': $('input[name="_token"]').val()
    },
    type: 'GET',
    url: "{{ URL::to(teste/ajax) }}",
    data: 'nome='+$('input[name="nome"]').val(),
    enctype: 'multipart/form-data',
    success: function(data){
        $(location).attr('href', "{{ URL::to(Request::path()) }}");
    },
    error: function(){
        alert('Erro no Ajax !');
    }
});