Get variable in jquery method and use it in view (laravel 5.3)

Question

I am using jquery to get data from controller from controller i am returning a $user from controller method but don't know how to use it in view.

Here is my Code..

    <li>
<a href="#suppression" data-toggle="modal" onclick="getForm('{{ $data->id }}')"> Get Modal</a>
</li>

and my jquery method is

function getForm(id)
{
    $.ajax({
        url: '/student-forms/get-form/'+id,
        type: "GET",
    });
}

Controller method is

public function getForm($id)
{

    $user = DB::table('users')->select('name','type','options')->whereIn('id' , $id)->get();
    return $user;

}

I want to get above variable $user here in my model

<div id="suppression" class="modal" role="dialog" aria-hidden="true">
<div class="modal-dialog" style="width: 500px;">
    <div class="modal-content">
        <div class="modal-header">
            <button type="button" class="close" data-dismiss="modal" aria-hidden="true">&times;</button>
        </div>
        <div class="modal-body">

        </div>
    </div>
</div>

Please help how can i achieve using above jquery method Thanks

You will have to use jquery to append that information to the dom in the success method of your ajax request. — Joe
– Joe, Commented May 10, 2017 at 11:23

madalinivascu · Accepted Answer · 2017-05-10 12:14:04Z

1

Change your ajax to append the data to the page when the ajax is executed successfully

function getForm(id)
{
    $.ajax({
        url: '/student-forms/get-form/'+id,
        type: "GET",
        success:function(data) {
          $('modal-body').append(data.name);
        }
    });
}

Remove the onclick event , add a attribute to the link to store the id

<a href="#suppression" data-toggle="modal" data-id="{{ $data->id }}"> Get Modal</a>

trigger the ajax when the modal is shown

$('#suppression').on('show.bs.modal', function (e) {
  var id = $(e.relatedTarget).attr('data-id');
  getForm(id);
})

Controller should return json

public function getForm($id)
{

    $user = DB::table('users')->select('name','type','options')->whereIn('id' , $id)->get();
    return response()->json($user);

}

edited May 10, 2017 at 12:14

answered May 10, 2017 at 11:24

madalinivascu

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10 Comments

Koray Küpe Over a year ago

Laravel DB Facade already returns Collection. You don't need to encode it.

madalinivascu Over a year ago

is your modal opening?

madalinivascu Over a year ago

then the show.bs.modal isn't triggered, are you sure the modal opens the default bootstrap way?

madalinivascu Over a year ago

try the shown.bs.modal event and drop a console.log in there to see if the event triggers

Ramzan Mahmood Over a year ago

what is data.name in jquery method above in your answer @madalinivascu

|

Joe · Accepted Answer · 2017-05-10 11:26:12Z

0

As you're using jQuery you'll need to also use it to edit the dom.

$.ajax({
    url: '/student-forms/get-form/'+id,
    type: "GET",
    success: function(response){
        $('#elementToChange').text(response.thingfromyourusermodel)
    }
});

answered May 10, 2017 at 11:26

Joe

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2 Comments

Ramzan Mahmood Over a year ago

dear please explain what will be thingfromyourusermodel

Joe Over a year ago

@bluemoon response will be the response from your controller ajax call, so it going on your code it will contain the name, type and options attributes from your user model.

Koray Küpe · Accepted Answer · 2017-05-10 11:26:24Z

0

If you use jQuery you can use this code:

function getForm(id)
{
    $.ajax({
        url: '/student-forms/get-form/'+id,
        type: "GET",
    }).done(function(msg) {
        $(".modal-body").val(msg);
    });
}

It will add returned value in your modal-body div. Since Laravel returns collection you will need format your user data.

e.g

.val('name:' + msg.name);

By the way don't forget to add crsf token field to your forms.

answered May 10, 2017 at 11:26

Koray Küpe

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