Unable to save table column value from dropdown in Laravel - php

Question

I am using Laravel 7 and PHP 7.4.
I' m trying to make a dependent dropdown into my form using ajax. I'm sending the column_id and column in ajax response to my dropdown. My dropdown is working absolutely fine but there is mess with my either database query or ajax response. My dropdown shows the values of column but when I try to save into database, it only saves the "Id" of column not the value.
I just wanted to send the column_id and column both to my dropdown but only save the value into database.
Blade
<form action="{{route('form_for_private_sellers')}}>
<div class="form-group">
<select name="make" id="make" data-dependent="model">
<option value="">Select Make</option>
<option value="254">Abarth</option>
</select>
</div>
<div class="form-group">
<select name="model" id="model">
<option selected value="">Select Model*</option>
</select>
</div>
</form>
Javascript
$("#make").change(function()
{
var id=$(this).val();
var make_id = id;
var dependent = $(this).data('dependent');
var _token = $('input[name="_token"]').val();
$.ajax({
url:"{{ route('fetchModel') }}",
method:"POST",
data:{make_id:make_id, _token:_token,dependent:dependent},
success:function(result)
{
$('#'+dependent).html(result);
}
});
});
Controller
public function fetchModel(Request $request)
{
$get_make_id = $request->get('make_id');
$dependent = $request->get('dependent');
$fetch_model = DB::table('auto_databases_one')
->select('model','model_id')
->distinct()
->where('make_id', '=', $get_make_id)
->where(function ($query) {
$query->orWhere('is_active', '=', '1');
})
->orderBy('model', 'ASC')
->get();
$show_model = '<option value="">Select ' . ucfirst(dependent) . '</option>';
foreach ($fetch_model as $row) {
$show_model .= '<option value="' . $row->model_id . '">' . $row->model . '</option>';
}
echo $show_model;
}
public function form_for_private_sellers(Request $request)
{
$store_seller = new Sellers();
$store_seller->make = $request['make'];
$store_seller->model = $request['model'];
}

Answer 1

I assume there is a button with id="save".
$("#save").click(function(e)
{
e.preventDefault();
const make = $("#make option:selected").val();
const model = $("#model option:selected").val();
const _token = $('input[name="_token"]').val();
$.ajax({
url:"{{route('form_for_private_sellers')}}",
method:"POST",
data:{make, _token, model},
success:function(result)
{
console.log('all done');
}
});
});
public function form_for_private_sellers(Request $request)
{
$store_seller = new Sellers();
//you may need to adjust this part as I do not know your properties called.
$store_seller->make = $request->make;
$store_seller->model = $request->model;
$store_seller->save();
}