How to show data passed from Controller to View using jQuery in Laravel

Question

I am trying to show data which is passed from controller to blade and manipulate something with it in my view page. But I can see the my data passed correctly in my console but i am not getting it in my view file. I am using Laravel 4

Controller:

public function addRow()
{
    if(Request::ajax()){
        $row = Input::all();        
    }
    return View::make('add-stock')
        ->with('rows', $row);
}

View :

@if( !empty($rows) )
    @foreach($rows as $row)
        {{ $row[0]  }}
    @endforeach
@endif

route:

Route::post('add-stock/row','StockController@addRow');

jQuery :

<script>
    $(document).ready(function(){

        $('.rowForm').click(function(e){
            e.preventDefault();

            var rowVal = $('input[name=row]').val();

            //ajax post
            $.post('add-stock/row', {row:rowVal}, function(data){
                console.log(data);
            })
        })          


    })
</script>

Sohel0415 · Accepted Answer · 2015-04-06 13:43:03Z

I would do like this. Your ajax function........

public function addRow(){  

    $row = Input::all();

    return Response::json($row);
 }

And then your JQuery............

<script>
$(document).ready(function(){

    $('.rowForm').click(function(e){
        e.preventDefault();

        var rowVal = $('input[name=row]').val();

        //ajax post
        $.get('/add-stock/row?rowVal =' +rowVal , function(data){
            console.log(data);
             $('#rowVal ').empty();
        $.each(data,function(index,subcatObj)
        {
            $('#rowVal ').append('<option value="'+subcatObj.value+'">'+subcatObj.value+'</option>');
        })
        })
    })          


})

here #rowVal id of the div or select attribute

And atlast ...........the route file

Route::get('add-stock/row','StockController@addRow');

By the way you didn't specify from where JQuery will be called

mininoz · Accepted Answer · 2015-04-05 15:03:01Z

0

You can return the response with JSON content type. Luckily, Laravel provides a tool for that. You could do something right this

if(Request::ajax()){
        return Response::json(Input::all()); 
}

for more information: http://laravel.com/docs/4.2/responses#special-responses

edited Apr 5, 2015 at 15:03

answered Apr 4, 2015 at 9:02

mininoz

5,9584 gold badges27 silver badges24 bronze badges

5 Comments

Wahidul Alam Over a year ago

how can i return the value to my desired view ??

mininoz Over a year ago

Do you get anything at "console.log(data);"?

Wahidul Alam Over a year ago

no it shows -- {"error":{"type":"Symfony\\Component\\Debug\\Exception\\FatalErrorException","message":"Call to undefined function response()","file":"C:\\xampp\\htdocs\\easybiz\\app\\controllers\\StockController.php","line":93}}

mininoz Over a year ago

Are you using Laraval 4 or 5?

mininoz Over a year ago

laravel.com/docs/4.2/responses#special-responses . Using the code in the link instead.

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