Show a form at on click and show it with value of a input box

Question

I want to show the value of a form on clicking of a checkbox and after selecting a checkbox I want that one of our input box of my form which is disabled show the name of checkbox in that field.I also want to get the id of that checkbox. I am working in laravel .

 <table class="table table-bordered" style="width: 80%;margin:5% 10%;background:slateblue">
        <tr>
            <th>#</th>
            <th>item</th>
            <th>item name</th>
            <th>Added_at</th>

        <tr>
        <?php $i=1 ?>
        @foreach($items as $item)
         <tr>
             <td><input type="checkbox" id="bought" name="{{$item->item_name}}" class="checkbox1"></td>
             <td>{{$item->id}}</td>
             <td>{{$item->item_name}}</td>
             <td>{{$item->date}}
            </tr>
        @endforeach
    </table>


<form class="shopped" method="post" action="{{url('post-data')}}" hidden><br>
    {!! csrf_field()!!}

    <h4 align="center">Add some information to mark as shopped</h4>
    <label for ="item">Items</label>
    <input type="text" class ="item"  disabled><br>
    <label for ="price">Price</label>
    <input type="text" name="price"><br>
    <label for ="store_name">Store</label>
    <input type="text" name="store_name"><br>
    <input type="submit" value="Add as purchased" class="btn btn-info" style="margin:20px 0 20px 60px"><br>
</form>

My Jquery Code....

 $('.checkbox1').on('change',function(e){
        e.preventDefault();

        var a = this.name;
            $('.item').attr('val',a);
            $('.shopped').show();

        });

Answer 1

Try this:

$('.checkbox1').on('change',function(e){
    e.preventDefault();
    var a = this.name;
    $('.item').val(a);
    $('.shopped').show();
 });