Question

此查询出了什么问题：

SELECT co.*, mod.COUNT(*) as moduleCount, vid.COUNT(*) as vidCount 
 FROM courses as co, modules as mod, videos as vid 
 WHERE mod.course_id=co.id AND vid.course_id=co.id ORDER BY co.id DESC

换句话说，我怎么能这样做，从'课程'返回的每一条记录，都有一个名为'modCount'的附加列，它显示了该course_id的modules表中的记录数，另一个名为'vidCount'，它对视频表执行相同的操作。

错误：

错误号码：1064

您的SQL语法有错误;   检查对应的手册   您的MySQL服务器版本   正确的语法在“”附近使用   moduleCount，vid.COUNT（）作为vidCount   从作为co的课程，'在第1行

Answer 1

使用子选择，你可以这样做：

SELECT co.*, 
    (SELECT COUNT(*) FROM modules mod WHERE mod.course_id=co.id) AS moduleCount, 
    (SELECT COUNT(*) FROM videos vid WHERE vid.course_id=co.id) AS vidCount
FROM courses AS co
ORDER BY co.id DESC

但要小心，因为当课程有很多行时，这是一个昂贵的查询。

修改如果您的表格非常大，则以下查询应该执行得更好（有利于阅读和理解更复杂）。

SELECT co.*, COALESCE(mod.moduleCount,0) AS moduleCount, COALESCE(vid.vidCount,0) AS vidCount FROM courses AS co LEFT JOIN ( SELECT COUNT(*) AS moduleCount, course_id AS courseId FROM modules GROUP BY course_id ) AS mod ON mod.courseId = co.id LEFT JOIN ( SELECT COUNT(*) AS vidCount, course_id AS courseId FROM videos GROUP BY course_id ) AS vid ON vid.courseId = co.id ORDER BY co.id DESC

Answer 2

我有更好的解决方案和轻松

SELECT COUNT(*),(SELECT COUNT(*) FROM table2) FROM table1

Answer 3

SELECT co.*,
       (
       SELECT  COUNT(*)
       FROM    modules mod
       WHERE   mod.course_id = co.id
       ) AS modCount,
       (
       SELECT  COUNT(*)
       FROM    videos vid
       WHERE   vid.course_id = co.id
       ) AS vidCount
FROM   courses co
ORDER BY
        co.id DESC

Answer 4

SELECT co.*, m.ModCnt as moduleCount, v.VidCnt as vidCount 
FROM courses co
INNER JOIN (
        select count(*) AS ModCnt, co.id AS CoID
        from modules 
        group by co) m
    ON m.CoID = co.id
INNER JOIN (
        select count(*) AS VidCnt, co.id AS CoID
        from videos
        group by co) v
    ON v.CoID = co.id   
INNER JOIN videos vid 
    ON vid.course_id = co.id 
ORDER BY co.id DESC

Answer 5

拍这个。我用一些非mysql代码完成了这项工作：

function getAllWithStats($info='*',$order='',$id=0)
{
    $courses=$this->getAll($info,$order,$id);

    foreach ($courses as $k=>$v)
    {
        $courses[$k]['modCount']=$this->getModuleCount($v['id']);
        $courses[$k]['vidCount']=$this->getVideoCount($v['id']);
    }

    return $courses;
}

多个表上的Mysql COUNT（*）

5 个答案: