Ajax PHP update using 2 query

Question

im trying to update the reserve table column additional where ill get the value from table price multiply by quantity from post but my problem is i can't update it heres my code

for the form php

  <form method="post">

              <div class="col-md-12">
                <label>ID</label>
                <input type="text" class="form-control" id="id" name="id">
              </div>

              <div class="col-md-12" >
                <label>Charge</label>
                <select id="name" name="name" class="form-control">
                  <?php
                  while ($reserve=mysqli_fetch_array($charge)) { ?>
                    <option value=" <?php echo $reserve['name']?>">
                      <?php echo $reserve['name']; ?>
                    </option><?php } ?>
                </select>
              </div>

              <div class="col-md-12">
                <br>
                <label>Quantity</label>
                <input type="number" class="form-control" id="quantity" name="quantity">
              </div>

              <div class="col-md-12">
                <br>
                <button class="Huge ui teal button" id="charge" name="charge">Add Charge</button>
              </div>

            </form>

for the ajax

  $(document).ready(function(e){
  $('#charge').click(function(){
  var id = $('#id').val();
  var name = $('#name').val();
  var quantity = $('#quantity').val();

         $.ajax({
           type   : 'POST',
           data   :{
              id:id,
              name:name,
              quantity:quantity,
                  },
            url     :"charge.php",
            success : function(result){
              if(result)
              {
                 $('#error').html("<span class='text-success' >Success Man</span>");
             }else{
               $('#error').html("<span class='text-danger'>Check mo information Man</span>");
           }
          }
         })
    });
  });

and for the charge.php

<?php

$connect = mysqli_connect("localhost", "root", "", "tobedetermined");

   $id = $_POST['id'];
   $name = $_POST['name'];
   $quantity = $_POST['quantity'];

   $sql = "SELECT price FROM charge where name ='$name'";
   $result = $connect->query($sql);
   $additional = $result * $quantity;

 mysqli_query($connect,"update reserve set additional= '$additional' where id = $id");


 mysqli_close($connect);


?>

please don't mind the Select tag i just used that to put all the value of column name frm charge table

here `$result = $connect->query($sql);` $result is a result object, not an integer. You need to fetch the values out of that with [fetch_array()](http://php.net/manual/en/mysqli-result.fetch-array.php) for example — Jeff, Sep 12 '17 at 10:06

score 0 · Answer 1 · answered Sep 12 '17 at 10:15

If you want to use Ajax, you could not use form.

Just remove form tag

<div class="col-md-12">
                <label>ID</label>
                <input type="text" class="form-control" id="id" name="id">
              </div>

              <div class="col-md-12" >
                <label>Charge</label>
                <select id="name" name="name" class="form-control">
                  <?php
                  while ($reserve=mysqli_fetch_array($charge)) { ?>
                    <option value=" <?php echo $reserve['name']?>">
                      <?php echo $reserve['name']; ?>
                    </option><?php } ?>
                </select>
              </div>

              <div class="col-md-12">
                <br>
                <label>Quantity</label>
                <input type="number" class="form-control" id="quantity" name="quantity">
              </div>

              <div class="col-md-12">
                <br>
                <button class="Huge ui teal button" id="charge" name="charge">Add Charge</button>
              </div>

Ajax PHP update using 2 query

1 Answers1