Implementing A Like/dislike Counter With Php And Jquery

I have a comics website, and I'd like to keep track of how many people click like and dislike for a single comic id... I'd like to store the value in my database, and also report i

Solution 1:

The technical problem is this line:

$.get("./scripts/likecounter.php", {_choice : choice, _site : site, _id : id}...

as site and id are undefined you'll be getting a javascript error ReferenceError: site is not defined

Try:

$.get("./scripts/likecounter.php", {_choice : choice, _site : "<?php echo $site; ?>", _id : "<?php echo $imgid; ?>"}...

You don't need the hidden inputs, also onClick should be onclick.

Other quick suggestions:

Make sure $imgid is actually an integer:

$imgid = intval($_GET['_id']);
if($imgid == 0) exit('Invalid id');

The following doesn't need to be separate:

$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");

$mysqli->query will only return a mysqli_result, meaning you can't just echo it out, you'll have to do a fetch, and because you know you are only getting a single result you can just use list($likes, $dislikes) = $result->fetch_array(MYSQLI_NUM);

Have a read of the docs to help you understand what the heck I'm going on about.

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:

$sql = "INSERT INTO xxx (ip, table_name, imgid) VALUES(\"".$_SERVER['REMOTE_ADDR']."\", \"$table\", $imgid)";
$mysqli->query($sql);

Then check if they have a record before adding a new like:

$sql = "SELECT ip FROM xxx WHERE ip = \"".$_SERVER['REMOTE_ADDR']."\" AND table_name = "$table" AND imgid = $imgid";
$result = $mysqli->query($sql);
if($result->num_rows == 0) { ...if($input ==... }

Solution 2:

I think

$site = $_GET['site'];`

Should be

$site = $_GET['_site'];` 

You are also not sending the id- $imgid = $_GET['id'];

Please look into PDO or MySQLi as your code is unsafe (look up SQL injection).

Solution 3:

I believe you will need to change the column name "like" to something else as that word is reserved in SQL here is a list of reserved words

just change it to "likes" or something like that.

another minor thing is that you can get both values in a single query

Select likes,dislikes from $table where id = $ImgID