Value values ​​in php array do not change. How to fix?

Question

MySQL

There is this kind:

<?php if (!empty($products)):?>
                 <?php foreach ($products as $product):?>
                    <div class="col-xl-4">

                        <div class="dishes-item">
                            <img src="/img/products/no-foto.png" alt="3">
                            <h4><?= $product->name?></h4>
                            <p><?= $product->description?></p>
                            <div class="dishes-item-details">
                                <span class="weight"><span class="weight-size"><?= $product->weight?></span> гр</span> <span class="sum"><?= $product->price?> руб.</span>
                            </div>
                            <div class="dishes-item-counter">
                                Кол/во <input type="text" value="1" id="qty" data-id="<?= $product->id?>" class="dishes-item-counter-field" />
                                <a class="add_to_cart"  data-id="<?= $product->id?>"   href="<?= \yii\helpers\Url::to(['cart/add', 'id' => $product->id])?>"><button  class="btn btn-choose">Добавить</button></a>
                            </div>
                        </div>
                    </div>
                    <?php endforeach;?>
                    <?php else:?>
                    <h2>Здесь товаров пока нет</h2>
                   <?php endif;?>

Script for it:

$('.add_to_cart').on('click',  function (e) {
            e.preventDefault();
            var id = $(this).data('id'), qty = $('#qty').val();
            console.log(qty)
              $.ajax({
               url: '/cart/add',
               data: {id: id, qty: qty},
               type: 'GET',
               success: function(res){
               	if(!res) alert(ошибка);
               	 
                  showCart(res);
               },
             error: function(){
                   alert('Ошибка');
              }
           });
        });

I need the get method to get the value value in the input field.
Behind the array, this field accepts all incoming values, in the array it returns the value 1, regardless of the entered data.

Answer 1

A lot of id is a mistake, only the first input is considered and not always correct.
Rewrite to a selection like $(this).closest('.dishes-item').find('input').val()
If it doesn't help, you need Back's code.

Answer 2

If I understand everything correctly, then you in the back:
It is necessary to throw in the value="1"incoming data instead of 1