Php 我能';我似乎无法让这个sql查询按我想要的方式工作
我有一张桌子,上面有一些书籍的评级 我创建了一些虚拟记录来测试这个查询Php 我能';我似乎无法让这个sql查询按我想要的方式工作,php,mysql,sql,mysqli,phpmyadmin,Php,Mysql,Sql,Mysqli,Phpmyadmin,我有一张桌子,上面有一些书籍的评级 我创建了一些虚拟记录来测试这个查询 ID BookID RatersID Rating Date 1 2 3 5 (date) 2 2 4 4 (date) SELECT `RatersID`, COUNT(*) AS `raters`, `Rating`, COUNT(*) AS `Ratings` FRO
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`
当我运行我期望的查询时
BookID Raters Ratings
2 2 9
我得到的是:
RatersID raters Rating Ratings
3 2 5 2
我不明白为什么会这样?
///////////////////以上问题已得到答复
我已经让查询工作,但当试图在php中接收信息的数字是重复的
例如,评分员=2 PHP显示22
评级=9 PHP显示99
$getratingq = mysqli_query($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");
}
您需要使用sum()
来获得评级的总和
SELECT
`BookID`,
COUNT(*) AS `raters`
sum(Rating) as Ratings
FROM
`BOOKS_ratings`
GROUP BY
`BookID`
谢谢你,我意识到了我的错误,我使用了计数来计算有多少人被评分,而不是我应该使用总数