我似乎不能让这个sql查询以我想要的方式工作

我有一个表，里面有一些图书的评级;

这些是我为测试这个查询而创建的一些虚拟记录;

 ID    BookID   RatersID     Rating    Date
 1     2        3            5         (date)
 2     2        4            4         (date)

SELECT `RatersID`,
COUNT(*) AS `raters`, `Rating`, COUNT(*) AS `Ratings`
FROM
`BOOKS_ratings`
GROUP BY
`BookID`

当我运行查询时，我期望

BookID    Raters     Ratings
2         2          9

结果:

RatersID    raters    Rating    Ratings
3           2         5         2

我不明白为什么会发生这种情况?///////////////////以上内容已得到解答

我已经得到了查询工作，但当试图在php中接收信息的数字是重复的

E。g Raters = 2 PHP显示22评分= 9

    $getratingq = mysqli_query($con,"SELECT `RatersID`, COUNT(*) AS `Raters`, sum(Rating) AS `Ratings` FROM `BOOKS_ratings` WHERE `BookID` ='$bookid' GROUP BY `BookID` LIMIT 1") or die("Get ratings query error");
if($getrating = mysqli_fetch_array($getratingq))
{
   echo $ratings = $getrating['Ratings'];
   $raters= $getrating['Raters'];
   $rating =  $ratings/$raters;
   $stars = floor("$rating");
}