T
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Thomas Laybe2022-02-04 11:36:52
AJAX
Thomas Laybe, 2022-02-04 11:36:52

How to display title ajax?

Good afternoon. I send data from the form to the database, I don’t know how to display the title of the form immediately on the page via ajax.

<div class="display-flex" style="display: flex;">
    
    <div class="from" style="margin-right: 200px;">
      <input type="text" name="title" class="title"> <br>
      <textarea name="content" class="content" id="" cols="30" rows="10"></textarea> <br>
      <button type="button" class="button">Submit</button> <br>
    </div>

  </div>


<script>
    
    $(document).ready(function(){
      $('button').on('click', function(){
        var titleValue = $('input.title').val();
        var contentValue = $('textarea.content').val();

        $.ajax({
          method: "POST",
          url: "some.php",
          data: { title: titleValue, content: contentValue }
        })
          .done(function(  ) {
          	  
          });
        $('input.title').val('');
        $('textarea.content').val('');

      })
    });

  </script>


<?php 

require 'db.php';

$data = [
  "title" => $_POST['title'],
  "content" => $_POST['content']
];

$sql = 'INSERT INTO posts (title, content) VALUES (:title, :content)';
$stmt = $connect->prepare($sql);
$result = $stmt->execute($data);
var_dump($result);

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1 answer(s)
A
Alexander, 2022-02-04
@thomasLaybe

You need to check the result of the write in PHP and, if successful, return a confirmation in response to the post request.
Then, in you process the response. For example, like this:.done(function() {...});

.done(function(rez) {
  if (rez.isDone) {
    $('input.title').val(titleValue);
  }
});

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