PHP投票，每天每IP一票：需要代码帮助

我在一个非常基本的网站上工作，它允许人们每24小时对每个IP地址的一个项目（游戏）进行一次投票。投票存储在名为Votes的表中，其中的列包含投票ID、游戏ID、votedate和IP地址

用户通过…vote.php？id=#（其中#是gid）进入该游戏的投票页面。下面是my current vote.php的简化版本，它不起作用：

<?php
$gid = $_GET["id"];
$ip = $_SERVER['REMOTE_ADDR'];
require_once ('config.php');
$q = "SELECT * FROM votes WHERE (gid=$gid)";
$r = @mysqli_query($dbc, $q);

if ($r) {
while ($row = mysqli_fetch_array($r)) {

    if ($row['ip'] != $ip) { // If no IP exists, go ahead and vote.

        $q2 = "INSERT INTO votes (gid, votedate, ip) VALUES ('$gid', NOW(), '$ip' )";
        $r2 = @mysqli_query($dbc, $q2);
        if ($r2) {
                echo 'Thank you for voting';
        } else {
            echo 'There was a problem with your vote'
            }

    } elseif ($ip == $row['ip']) {
    // If IP address exists, check date of last time they voted for this game

        // Define the two dates
        $votedate = $row['votedate'];
        $now = date("Y-m-d H:i:s");
        // Calculate hours between the dates
        $diff = round((strtotime($now) - strtotime($votedate)) / (60 * 60));
        // Create variable for when user can return.
        $return = '24' - $diff;

        if ($diff > '24') { // Allow to vote once every 24 hours
            $q3 = "INSERT INTO votes (gid, votedate, ip) VALUES ('$gid', NOW(), '$ip' )";
            $r3 = @mysqli_query($dbc, $q3);

            // Display outcome of insert query:
            if ($r3) {
                echo 'Thank you for voting, please come back in 24 hours to vote again!';
            } else {
                echo 'There was a problem with your vote:' . mysqli_error($dbc) . ' in reference to query ' . $q3;
            }
        } else {
            echo 'Sorry, you\'ve already voted in past 24 hours, please come back again in ' . $return . ' hours.';
        }

    } 

} // End of while $r loop

} else {

echo 'Sorry, your query did not work.';

} // End of overall if $r ran

mysqli_close($dbc);

exit();
?>

---

对于使用相同类型的数学运算，请小心：

尝试使用floatval（）

什么东西不能“工作”呢？

$diff = floatval(round((strtotime($now) - strtotime($votedate)) / (60 * 60)));

$return = 24.0 - $diff;

if ($diff > 24.0) 
{
    /* Your code */
}