使用多个左连接查询 - 点列值不正确

Question

我有以下数据库结构,我正在尝试运行一个查询,该查询将显示教室,有多少学生是教室的一部分,教室分配了多少奖励,以及分配给单个教授的积分数量教室(基于classroom_id专栏).

在最底层使用查询我正在尝试收集教室分配的'totalPoints' - 基于在classroom_redeemed_codes表中计算points列并将其作为单个整数返回.

由于某种原因,totalPoints的值不正确 - 我做错了但不确定是什么......

- 更新 - 这是sqlfiddle: - http://sqlfiddle.com/#!2/a9f45

我的结构:

CREATE TABLE `organisation_classrooms` (
  `classroom_id` int(11) NOT NULL AUTO_INCREMENT,
  `title` varchar(255) NOT NULL,
  `active` tinyint(1) NOT NULL,
  `organisation_id` int(11) NOT NULL,
  `period` int(1) DEFAULT '0',
  `classroom_bg` int(2) DEFAULT '3',
  `sortby` varchar(6) NOT NULL DEFAULT 'points',
  `sound` int(1) DEFAULT '0',
  PRIMARY KEY (`classroom_id`)
);

CREATE TABLE organisation_classrooms_myusers (
  `classroom_id` int(11) NOT NULL,
  `user_id` bigint(11) unsigned NOT NULL,
);

CREATE TABLE `classroom_redeemed_codes` (
  `redeemed_code_id` int(11) NOT NULL AUTO_INCREMENT,
  `myuser_id` bigint(11) unsigned NOT NULL DEFAULT '0',
  `ssuser_id` bigint(11) NOT NULL DEFAULT '0',
  `classroom_id` int(11) NOT NULL,
  `order_product_id` int(11) NOT NULL DEFAULT '0',
  `order_product_images_id` int(11) NOT NULL DEFAULT '0',
  `date_redeemed` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP,
  `points` int(11) NOT NULL,
  `type` int(1) NOT NULL DEFAULT '0',
  `notified` int(1) NOT NULL DEFAULT '0',
  `inactive` tinyint(3) NOT NULL,
   PRIMARY KEY (`redeemed_code_id`),
);

SELECT
  t.classroom_id,
  title,
  COALESCE (
    COUNT(DISTINCT r.redeemed_code_id),
      0
   ) AS totalRewards,
  COALESCE (
    COUNT(DISTINCT ocm.user_id),
    0
   ) AS totalStudents,
  COALESCE (sum(r.points), 0) AS totalPoints
  FROM
  `organisation_classrooms` `t`
   LEFT OUTER JOIN classroom_redeemed_codes r ON (
   r.classroom_id = t.classroom_id
   AND r.inactive = 0
   AND (
    r.date_redeemed >= 1393286400
    OR r.date_redeemed = 0
   )
   )
   LEFT OUTER JOIN organisation_classrooms_myusers ocm ON (
   ocm.classroom_id = t.classroom_id
   )
   WHERE
    t.organisation_id =37383
   GROUP BY title
   ORDER BY t.classroom_id ASC
   LIMIT 10

- 编辑 -

OOPS!我有时讨厌SQL ...我犯了一个大错误,我试图计算在classroom_redeemed_codes而不是organisation_classrooms_myuser表中的学生人数.我真的很抱歉我应该早点把它拿走？!

classroom_id | totalUniqueStudents
     16             1
     17             2
     46             1
     51             1
     52             1

在classroom_redeemed_codes表中有7行,但由于teacher_id 46有两行,尽管具有相同的myuser_id(这是学生ID),这应该显示为一个唯一的学生.

这有意义吗？基本上试图根据myuser_id列获取classroom_redeemed_codes表中唯一学生的数量.

例如,一个教室id 46可以在classroom_redeemed_codes表中有100行,但如果它们是相同的myuser_id,则应该显示totalUniqueStudents计为1而不是100.

如果不清楚,请告诉我......

- 更新 - 我有以下查询似乎工作借用了一个似乎工作的用户...(我的头痛)我会再次接受答案.很抱歉这个混乱 - 我想我只是在想这个

select crc.classroom_id,
    COUNT(DISTINCT crc.myuser_id) AS users,
    COUNT( DISTINCT crc.redeemed_code_id ) AS classRewards,
    SUM( crc.points ) as classPoints, t.title
  from classroom_redeemed_codes crc
       JOIN organisation_classrooms t
         ON crc.classroom_id = t.classroom_id 
        AND t.organisation_id = 37383
        where crc.inactive = 0
        AND ( crc.date_redeemed >= 1393286400
        OR crc.date_redeemed = 0 )
        group by crc.classroom_id

Answer 1

我首先运行每个特定类的点数的预查询聚合,然后使用左连接到它.我在结果集中获得的行数多于预期的样本,但没有MySQL直接测试/确认.但是,这是一个查询的SQLFiddle 通过使用点的总和进行查询,并在应用users表时具有笛卡尔结果,它可能是重复点的基础.通过预先查询兑换代码本身,您只需获取该值,然后加入用户.

SELECT
      t.classroom_id,
      title,
      COALESCE ( r.classRewards, 0 ) AS totalRewards,
      COALESCE ( r.classPoints, 0) AS totalPoints,
      COALESCE ( r.uniqStudents, 0 ) as totalUniqRedeemStudents,
      COALESCE ( COUNT(DISTINCT ocm.user_id), 0 ) AS totalStudents
   FROM
      organisation_classrooms t
         LEFT JOIN ( select crc.classroom_id,
                            COUNT( DISTINCT crc.redeemed_code_id ) AS classRewards,
                            COUNT( DISTINCT crc.myuser_id ) as uniqStudents,
                            SUM( crc.points ) as classPoints
                        from classroom_redeemed_codes crc
                           JOIN organisation_classrooms t
                              ON crc.classroom_id = t.classroom_id 
                              AND t.organisation_id = 37383
                        where crc.inactive = 0
                          AND ( crc.date_redeemed >= 1393286400
                           OR crc.date_redeemed = 0 )
                        group by crc.classroom_id ) r
            ON t.classroom_id = r.classroom_id

         LEFT OUTER JOIN organisation_classrooms_myusers ocm 
            ON t.classroom_id = ocm.classroom_id
   WHERE
      t.organisation_id = 37383
   GROUP BY 
      title
   ORDER BY 
      t.classroom_id ASC
   LIMIT 10

Answer 2

你需要sum(r.points)和左外连接中的子查询,见下文

 SELECT
 t.classroom_id,
 title,
 COALESCE (
   COUNT(DISTINCT r.redeemed_code_id),
     0
  ) AS totalRewards,
 COALESCE(sum(r.points),0) AS totalPoints
,COALESCE(sum(T1.cnt),0) as totalStudents
 FROM
  `organisation_classrooms` `t`
left outer join (select classroom_id, count(user_id) cnt
                  from organisation_classrooms_myusers 
 group by classroom_id) T1 on (T1.classroom_id=t.classroom_id)
  LEFT OUTER JOIN classroom_redeemed_codes r ON (
  r.classroom_id = t.classroom_id
  AND r.inactive = 0
  AND (
   r.date_redeemed >= 1393286400
   OR r.date_redeemed = 0
  )
  )
  WHERE
   t.organisation_id =37383
  GROUP BY title
  ORDER BY t.classroom_id ASC
  LIMIT 10