Question

I’m making a facemash analog and I came across with an issue. When I vote for second image, the winner is the first one anyway. Where’s the problem? Web-site: http://facemash123.shimansky.ru

<form METHOD=POST ACTION="rate.php">
<table width="auto" align="center">
    <tr align="center" width="auto" valign="top">
        <td><img src="images/<?=$images[0]->filename?>" /></td>
        <td><img src="images/<?=$images[1]->filename?>" /></td>
    </tr>
    <tr>
        <td><input type="submit" name="winner" value="Vote"></td>
        <td><input type="submit" name="winner" value="Vote"><td>

        <input type="hidden" name="first" value="<?=$images[0]->image_id?>">
        <input type="hidden" name="second" value="<?=$images[1]->image_id?>">
    </tr>
    <tr>
        <td><center>Wins: <?=$images[0]->wins?>, Fails: <?=$images[0]->losses?></center></td>
        <td><center>Wins: <?=$images[1]->wins?>, Fails: <?=$images[1]->losses?></center></td>
    </tr>
</table>
</form>

rate.php http://jsfiddle.net/Rg7vf/
index.php http://jsfiddle.net/ad3PM/

THANKS in advance!

Answer 1

Your two submit buttons submit the same value, and there’s no way to differentiate between the two images.

Maybe change the code to:

    <td><input type="submit" name="winner" value="Vote for Img #1"></td>
    <td><input type="submit" name="winner" value="Vote for Img #2"><td>

and then in the vote handling code:

switch($_POST['winner']) {
    case "Vote for Img #1":
        $vote_id = $_POST['first'];
        break;
    case "Vote for Img #2":
        $vote_id = $_POST['second'];
        break;
    default:
        die("VOTER FRAUD! CALL THE FEC!");
}