Loois

comment.sql

CREATE TABLE `comment` (
  `user_id` int(8) NOT NULL,
  `photo_id` int(8) NOT NULL, 
  `comment` text NOT NULL,
  `date` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP,
  PRIMARY KEY (`user_id``photo_id`)
);

ajax_comment.php

<?php
// code will run if request through ajax
if (isset( $_SERVER['HTTP_X_REQUESTED_WITH'] )):
  include('../config.php');
  // connecting to db
  dbConnect();
  
  if (!empty($_SESSION['user_id']) AND !empty($_POST['photo_id']) AND !empty($_POST['comment'])) {
    // preventing sql injection
    $user_id = $_SESSION['user'];
    $photo_id = $_POST['photo_id'];
    $comment = $_POST['comment'];

    // insert new comment into comment table
    $query = "INSERT INTO comment (user_id, photo_id, comment) VALUES('$user_id', '$photo_id', '$comment')");  
  }
?>
<!-- sending response with new comment and html markup-->
<div class="comment-item">
  <div class="comment-avatar">
        <a href="<?php echo $baseurl . "/" . $photo_username ?>"><img src="./core/getimg.php?profiloimg=<?php echo $photo_userid ?>" class="home-foto-profilofoto" /></a>  </div>
  <div class="comment-post">
    <h3><?php echo $photo_username . "/" . $user_id ?> <span>ha commentato:</span></h3>
    <p><?php echo $comment?></p>
  </div>
</div>

<?php
  // close connection
  dbConnect(0);
endif?> 

jQuery.js

<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.10.2/jquery.min.js"></script>
<script>
$(document).ready(function(){
  var form = $('form');
  var submit = $('#submit');

  form.on('submit', function(e) {
    // prevent default action
    e.preventDefault();
    // send ajax request
    $.ajax({
      url: 'ajax_comment.php',
      type: 'POST',
      cache: false,
      data: form.serialize(), //form serizlize data
      beforeSend: function(){
        // change submit button value text and disabled it
        submit.val('Submitting...').attr('disabled', 'disabled');
      },
      success: function(data){
        // Append with fadeIn see http://stackoverflow.com/a/978731
        var item = $(data).hide().fadeIn(800);
        $('.comment-block').append(item);

        // reset form and button
        form.trigger('reset');
        submit.val('Submit Comment').removeAttr('disabled');
      },
      error: function(e){
        alert(e);
      }
    });
  });
});
</script> 

home.php (Where there are my users posts)

		
		<!--  Commenti -->
		
	<form id="form" method="post">
    <!-- need to supply post id with hidden fild -->
    <input type="hidden" name="comment" value="1">
        <label>
      <a href="<?php echo $baseurl . "/" . $photo_username ?>"><img 	width="25"
	height="25"src="./core/getimg.php?profiloimg=<?php echo $photo_userid ?>" class="home-foto-profilofoto" /></a>
		</label>
		<label>
        <a href="<?php echo $baseurl . "/" . $photo_username; ?>"><?php echo $photo_username; ?></a>

		</label>
	<label>
      <span>Commenta</span>
      <textarea name="comment" id="comment" cols="0" rows="0" placeholder="Scrivi un commento.." required></textarea>
    </label>
    <input type="submit" id="submit1" value="Submit Comment">
  </form>
  <?php 
include ("ajax_comment.php");
include ("jquery.js");
  ?>
      <!-- Fine commenti -->

Someone would give me a hand in creating a comment form for my users posts, unfortunately I can not.

from contact : leccialoois@me.com

my website: http://www.socialedge.altervista.org/