How I return data from database separately (integers) using ajax

Question

I want to return data from database using AJAX here the AJAX code for retrieving the data but it can't give it saparataly.

  $(document).ready(function(){
    $('#bathroom-select').change(function(){
        var bathroom_option1 =$(this).val();
        console.log(bathroom_option1);
        $.ajax({
            type:'POST',
            data:({bathroom_option1}),
            success:function(data){
                price1=parseInt(data);
                console.log(price1);
                var rows;
                $.each(data, function (key, name) {    //this will not work 
                    console.log(item[i]);
                    });
                  }
             });
        });
    });

here the database image where the data is stored.

anybody plz explain me i m new on the stackoverflow so if there is any mistake then sorry. And thankyou for replying me the answers.

this is the server site processing using php

 $con=mysqli_connect("localhost","root","","price");
   if (isset($_POST['bathroom_option1'])) {
    $query=mysqli_query($con,"select * from bathroom where number_of_bathrooms ='$_POST[bathroom_option1]'");
   while ($row = mysqli_fetch_array($query)) {
    echo json_encode($row['price'],JSON_NUMERIC_CHECK);
    echo json_encode($row['hours'],JSON_NUMERIC_CHECK);
    echo json_encode($row['minutes'],JSON_NUMERIC_CHECK);
  }
  }

Answer 1

This probably because item is not defined try this

$.each(data, function () {    
   console.log($(this));
});

Answer 2

Ok so try this maybe :

1/ Your ajax call :

$(document).ready(function(){
    $('#bathroom-select').change(function(){
        var bathroom_option1 =$(this).val();
        console.log(bathroom_option1);
        $.ajax({
            type:'POST', // you send data with POST method
            data:{ "bo1" : bathroom_option1}, //the data you send
            dataType : "JSON", // what you will get as anwser data type
            url : yourphp.php, // your .php where you send your data
            success:function(response){
                var data = response.data; // your anwser array with one row = one item with price, hours and minutes

                // See why in my PHP, but here you can get the message and the type you send with "response.message" and "response.type", can be usefull if you want to do something different according to what happen in your .php

                $.each(data, function (key, name) {   
                    console.log(this); // one row 
                });
              }
         });
    });
});

2/ Your .php :

$result = array();    

$con=mysqli_connect("localhost","root","","price");
if (isset($_POST['bo1'])) { // you get your data with post method here, with the key you used in your call, here 'bo1'
    $query=mysqli_query($con,"select * from bathroom where number_of_bathrooms ='$_POST['bo1']'");
    while ($row = mysqli_fetch_array($query)) {
       // You add each row in your $result['data'] array
       $result['data'][] = array(
           "price" => $row['price'],
           "hours" => $row['hours'],
           "minutes" => $row['minutes']
       );
      } // end while
  } // end if

  $result['message'] = "All is ok !";
  $result['type'] = "success";

  echo json_encode($result);

 // You can send back what you want as message or type for example, if you have some test to do in your php and it fails send back $result['type'] = "warning" for example

Is this what you are looking for?