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copied!<p>The technical problem is this line:</p> <pre><code>$.get("./scripts/likecounter.php", {_choice : choice, _site : site, _id : id}... </code></pre> <p>as site and id are undefined you'll be getting a javascript error <code>ReferenceError: site is not defined</code></p> <p>Try:</p> <pre><code>$.get("./scripts/likecounter.php", {_choice : choice, _site : "<?php echo $site; ?>", _id : "<?php echo $imgid; ?>"}... </code></pre> <p>You don't need the hidden inputs, also <code>onClick</code> should be <code>onclick</code>.</p> <p>Other quick suggestions:</p> <p>Make sure $imgid is actually an integer:</p> <pre><code>$imgid = intval($_GET['_id']); if($imgid == 0) exit('Invalid id'); </code></pre> <p>The following doesn't need to be separate:</p> <pre><code>$likes = $mysqli->query("SELECT like FROM $table WHERE id = $imgid"); $dislikes = $mysqli->query("SELECT dislike FROM $table WHERE id = $imgid"); $result = $mysqli->query("SELECT like_count, dislike_count FROM $table WHERE id = $imgid"); </code></pre> <p><code>$mysqli->query</code> will only return a mysqli_result, meaning you can't just echo it out, you'll have to do a fetch, and because you know you are only getting a single result you can just use <code>list($likes, $dislikes) = $result->fetch_array(MYSQLI_NUM);</code></p> <p>Have a read of the <a href="http://php.net/manual/en/mysqli-result.fetch-array.php" rel="nofollow">docs</a> to help you understand what the heck I'm going on about.</p> <p>As you mentioned people can just keep hitting like/dislike all they want, you'll need to limit this with an IP log of some sort. For example setup a new database table with the following fields; ip, table, imgid. Then when someone "likes" log their IP:</p> <pre><code>$sql = "INSERT INTO xxx (ip, table_name, imgid) VALUES(\"".$_SERVER['REMOTE_ADDR']."\", \"$table\", $imgid)"; $mysqli->query($sql); </code></pre> <p>Then check if they have a record before adding a new like:</p> <pre><code>$sql = "SELECT ip FROM xxx WHERE ip = \"".$_SERVER['REMOTE_ADDR']."\" AND table_name = "$table" AND imgid = $imgid"; $result = $mysqli->query($sql); if($result->num_rows == 0) { ...if($input ==... } </code></pre>

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