Ajax gives error and status 200?

Question

PHP

Good afternoon. Help me to understand.
There is a form

<form  id="feedback">
  <div class="form-group">
    <label>Name</label>
    <input type="text" class="form-control" id="name">
  </div>
  <div class="form-group">
    <label>email</label>
    <input type="email" class="form-control" id="email" >
  </div>
  <div class="form-group">
    <label>theme</label>
    <input type="text" class="form-control" id="theme" >
  </div>
  <div class="form-check">
    <label>
      massage
    </label>
    <textarea id="message" class="form-control"></textarea>
  </div>

  <button type="submit" class="btn btn-primary">Submit</button>
</form>

there is an ajax request

$(document).ready(function(){
    
    $('button').on('click', function(e){
      e.preventDefault();
      
        sendMessage();
  
    });

    function sendMessage()
    {
      var user_name = $('#name').val(),
        user_email = $('#email').val(),
        user_theme = $('#theme').val(),
        user_message = $('#message').val();
      $.ajax({
        url: 'action.php',
        type: 'POST',
        dataType: 'JSON',
        data: {
          name: user_name,
          email: user_email,
          theme: user_theme,
          message: user_message
        },
        success: function(data){
            alert('done');
        },
        error: function(request, status, error){
          var statusCode = request.status; // вот он код ответа
          console.log(statusCode);
        }
      })
    }
  })

and there is an action.php file that handles ajax and should write to the database

$name = $_POST['name'];
$email = $_POST['email'];
$theme = $_POST['theme'];
$message = $_POST['message'];

$db_host = "*****"; 
$db_user = "*****"; 
$db_password = "******"; 
$db_table = "feedback";     


$db = mysql_connect($db_host,$db_user,$db_password) OR DIE("Не могу создать соединение ");
// Выборка базы
mysql_select_db("*****_db",$db);
 
mysql_query ("INSERT INTO feedback (name, email, theme, message,) VALUES ('$name', '$email', '$theme', '$message')");

during execution, ajax fulfills error: and displays status 200 in the console and nothing is written to the database

Answer 1

Let's start with how the form handler should look like:
1. The event should be hung on the form, not on the button - $('#feedback').on('submit', ...)
2. Leave the preventDefault line
3. Have set the name attribute of each input
4. In the data parameter, write $(this).serializeArray() - all inputs with a name will be serialized into an array and sent to the server, where they will be visible via $_POST
Next, via var_dump in php, see the values ​​of the variables, the cheekbone will display error if something goes wrong.

Answer 2

dataType: 'JSON',
What do you get in response from php? Make an answer of this type and we will be successful for you
and to all this recommendations from grisha2217