Question

**大家好，我需要一个评分系统，显示php的平均评分。实际上我完成了评级过程（保存和更新过程）。我只需要在php中显示平均评级（使用php-Ajax评级系统）当我从数据库中检索数据时，我得到了这样的错误

$add_coun= "SELECT sum(rating) sum, count(id) count from comments WHERE item_id = $itemID AND status=1";

        $result = mysql_query($add_coun,$con);
        if(!$result)
        {
        echo "query was not successfully";  
        }

        $result = mysql_fetch_object($result);

        $sum = $result->sum;
        $count = $result->count;
        $rating = $sum / $count;

        echo $rating;
?>

我收到类似这样的错误警告：mysql_fetch_object（）：提供的参数不是第19行的C：\ wamp \ www \ final work_apr51 \ final work_apr51 \ calculation.php中的有效MySQL结果资源

警告：在第23行的C：\ wamp \ www \ final work_apr51 \ final work_apr51 \ calculation.php中除以零 **

Answer 1

也许这可以帮到你？

    $link = mysqli_connect("localhost","mysqlusername","mysqlpassword","dbname");
    $rating = mysql_real_escape_string($_GET['id']);
    $q = mysqli_query($link,"SELECT * FROM ratings WHERE id='{$rating}'"); //Get our ratings by the page that has rated

    //Die if id dont exist!
    if(mysqli_num_rows($q) == 0) die("Wrong page id!");


    //Select good & bad ratings
    $good = mysqli_query($link,"SELECT * FROM ratings WHERE id='{$rating}' AND value ='yes'");
    $bad = mysqli_query($link,"SELECT * FROM ratings WHERE id='{$rating}' AND value ='no'");

    //Count good & bad ratings
    $gcnt = mysqli_num_rows($good);
    $bcnt = mysqli_num_rows($bad);

    //Calculate
    $totalVotes = $gcnt + $bcnt;

    if($totalVotes == 0){
      echo $totalVotes." votes";
    }
    if($totalVotes > 0){
      echo "<font color='green'>".$totalVotes." votes</font>";
    }
    if($totalVotes < 0){
      echo "<font color='red'>".$totalVotes." votes</font>";
    }

ajax-php星级评分系统（只显示mysql数据库的平均日期）

1 个答案: