SQL 计数多个字段

Question

这是上一个问题的后续问题：Complicated COUNT query in MySQL。没有一个答案在所有条件下都有效，而且我也很难找到解决方案。我将向第一个提供完全正确答案的人奖励 75 分奖励（我将在奖励可用时立即奖励，作为参考，我之前已经这样做过：Improving Python/django view code）。

我想获取用户拥有的视频积分计数并且不允许重复（即，对于每个视频，用户可以在其中获得 0 或 1 次积分。我想找到三个计数：用户的视频数量已上传（简单）--Uploads；用户未上传的视频中计入的视频数--Credited_by_others；以及用户已计入的视频总数--Total_credits。

我有三张桌子：

CREATE TABLE `userprofile_userprofile` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `full_name` varchar(100) NOT NULL,
   ...
 )

CREATE TABLE `videos_video` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `title` int(11) NOT NULL,
  `uploaded_by_id` int(11) NOT NULL,
  ...
  KEY `userprofile_video_e43a31e7` (`uploaded_by_id`),
  CONSTRAINT `uploaded_by_id_refs_id_492ba9396be0968c` FOREIGN KEY (`uploaded_by_id`) REFERENCES `userprofile_userprofile` (`id`)
)

注意uploaded_by_id与userprofile.id相同

CREATE TABLE `videos_videocredit` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `video_id` int(11) NOT NULL,
  `profile_id` int(11) DEFAULT NULL,
  `position` int(11) NOT NULL
  ...
  KEY `videos_videocredit_fa26288c` (`video_id`),
  KEY `videos_videocredit_141c6eec` (`profile_id`),
  CONSTRAINT `profile_id_refs_id_31fc4a6405dffd9f` FOREIGN KEY (`profile_id`) REFERENCES `userprofile_userprofile` (`id`),
  CONSTRAINT `video_id_refs_id_4dcff2eeed362a80` FOREIGN KEY (`video_id`) REFERENCES `videos_video` (`id`)
)

下面是一步一步来说明：

1) 创建 2 个用户：

insert into userprofile_userprofile (id, full_name) values (1, 'John Smith');
insert into userprofile_userprofile (id, full_name) values (2, 'Jane Doe');

2) 用户上传视频。他还没有将任何人——包括他自己——归功于其中。

insert into videos_video (id, title, uploaded_by_id) values (1, 'Hamlet', 1);

结果应该如下：

**User**     **Uploads**  **Credited_by_others**  **Total_credits**
John Smith       1                0                      1
Jane Doe         0                0                      0

3) 上传视频的用户现在将自己的功劳归于视频。请注意，这不应该改变任何事情，因为用户已经收到了上传电影的信用并且我不允许重复信用：

insert into videos_videocredit (id, video_id, profile_id, position) values (1, 1, 1, 'director')

结果现在应该如下：

**User**     **Uploads**  **Credited_by_others**  **Total_credits**
John Smith       1                0                      1
Jane Doe         0                0                      0

4) 用户现在在同一个视频中又两次将自己的功劳归于自己（即，他在视频中有多个“位置”）。此外，他在该视频中三度称赞 Jane Doe：

insert into videos_videocredit (id, video_id, profile_id, position) values (2, 1, 1, 'writer')
insert into videos_videocredit (id, video_id, profile_id, position) values (3, 1, 1, 'producer')
insert into videos_videocredit (id, video_id, profile_id, position) values (4, 1, 2, 'director')
insert into videos_videocredit (id, video_id, profile_id, position) values (5, 1, 2, 'editor')
insert into videos_videocredit (id, video_id, profile_id, position) values (6, 1, 2, 'decorator')

结果现在应该如下：

**User**     **Uploads**  **Credited_by_others**  **Total_credits**
John Smith       1                0                      1
Jane Doe         0                1                      1

5) Jane Doe 现在上传视频。她没有归功于自己，但在视频中两次归功于 John Smith：

insert into videos_video (id, title, uploaded_by_id) values (2, 'Othello', 2)
insert into videos_videocredit (id, video_id, profile_id, position) values (7, 2, 1, 'writer')
insert into videos_videocredit (id, video_id, profile_id, position) values (8, 2, 1, 'producer')

结果现在应该如下：

**User**     **Uploads**  **Credited_by_others**  **Total_credits**
John Smith       1                1                      2
Jane Doe         1                1                      2

所以，我想为每个用户找到这三个字段——Uploads、Credited_by_others 和 Total_credits。数据永远不应为 Null，而应在字段没有计数时为 0。谢谢。

Answer 1

我使用连接重写了查询，因此服务器更容易优化。

前两个视图用于简化查询

CREATE VIEW IF NOT EXISTS vperson_videos AS
    SELECT
        v.uploaded_by_id AS id,
        COUNT(*) AS uploads
    FROM vvideo v
    GROUP BY v.uploaded_by_id;

上面的视图只是统计用户上传的视频数量。

CREATE VIEW vperson_credits AS
    SELECT
        c.profile_id AS id,
        COUNT(DISTINCT c.video_id) AS credits
    FROM vcredit c
    INNER JOIN vvideo cv ON cv.id = c.video_id
    WHERE cv.uploaded_by_id <> c.profile_id
    GROUP BY c.profile_id;

上面的视图计算了归功于用户的（不同的）视频的数量，但忽略了用户自己上传的视频。

然后是查询本身：

SELECT
    p.id,
    p.full_name,
    IFNULL(pv.uploads,0) AS uploads,
    IFNULL(pc.credits,0) AS credits,
    IFNULL(pv.uploads,0) + IFNULL(pc.credits,0) AS total_credits
FROM vperson p
LEFT OUTER JOIN vperson_videos pv ON pv.id = p.id
LEFT OUTER JOIN vperson_credits pc ON pc.id = p.id;

我使用LEFT OUTER JOIN 来包含那些没有上传任何视频或没有被记录在任何视频中的用户。 IFNULL() 是必要的，因为我会得到 NULL 而不是 0。

最终结果是：

+----+------------+---------+---------+---------------+
| id | full_name  | uploads | credits | total_credits |
+----+------------+---------+---------+---------------+
|  1 | John Smith |       1 |       1 |             2 | 
|  2 | Jane Doe   |       1 |       1 |             2 | 
+----+------------+---------+---------+---------------+

Answer 2

首先，我认为您的问题描述有几个错误。

• 在第 5 步中，您在视频 2 中描述了 Jane 两次将 John 归功于 John。我认为您只是在 values 子句中的某些列排序错误。应该是：

  insert into videos_videocredit (id, video_id, profile_id, position) values (7, 2, 1, 'writer');
  insert into videos_videocredit (id, video_id, profile_id, position) values (8, 2, 1, 'producer');
  

• 您的结果应显示 John 在 2 个视频中获得荣誉，而 Jane 在 1 个视频中获得荣誉。

  +------------+---------+--------------------+---------------+
  | full_name  | Uploads | Credited_by_others | Total_credits |
  +------------+---------+--------------------+---------------+
  | John Smith |       1 |                  1 |             2 | 
  | Jane Doe   |       1 |                  1 |             1 | 
  +------------+---------+--------------------+---------------+
  

我在 MySQL 5.1.57 上测试了以下查询，它给出了上述结果。

SELECT
  u.full_name,
  COUNT(DISTINCT myvideos.id) AS Uploads,
  COUNT(DISTINCT byothers.id) AS Credited_by_others,
  COUNT(DISTINCT credited.id) AS Total_credits
FROM userprofile_userprofile AS u
LEFT OUTER JOIN videos_video AS myvideos ON myvideos.uploaded_by_id = u.id
LEFT OUTER JOIN (
  videos_videocredit AS c USE INDEX (videocredit_profileid_videoid)
  INNER JOIN videos_video AS credited
    ON c.video_id = credited.id
) ON c.profile_id = u.id
LEFT OUTER JOIN videos_video AS byothers USE INDEX (video_up_id)
  ON c.video_id = byothers.id
  AND byothers.uploaded_by_id <> u.id
GROUP BY u.id

我创建了几个额外的索引并给出了使用它们的查询提示。

CREATE INDEX video_up_id ON videos_video (id,uploaded_by_id);

CREATE INDEX videocredit_profileid_videoid ON videos_videocredit (profile_id,video_id);

这样可以保证所有的表（除了userprofile）都是用Using index模式访问的，也就是说只读取索引B树就可以满足查询，不需要读取表数据.这是解释报告：

*************************** 1. row ***************************
           id: 1
  select_type: SIMPLE
        table: u
         type: index
possible_keys: NULL
          key: PRIMARY
      key_len: 4
          ref: NULL
         rows: 2
        Extra: 
*************************** 2. row ***************************
           id: 1
  select_type: SIMPLE
        table: myvideos
         type: ref
possible_keys: userprofile_video_e43a31e7
          key: userprofile_video_e43a31e7
      key_len: 4
          ref: test.u.id
         rows: 1
        Extra: Using index
*************************** 3. row ***************************
           id: 1
  select_type: SIMPLE
        table: c
         type: ref
possible_keys: videocredit_profileid_videoid
          key: videocredit_profileid_videoid
      key_len: 5
          ref: test.u.id
         rows: 1
        Extra: Using index
*************************** 4. row ***************************
           id: 1
  select_type: SIMPLE
        table: credited
         type: eq_ref
possible_keys: PRIMARY,video_up_id
          key: PRIMARY
      key_len: 4
          ref: test.c.video_id
         rows: 1
        Extra: Using index
*************************** 5. row ***************************
           id: 1
  select_type: SIMPLE
        table: byothers
         type: ref
possible_keys: video_up_id
          key: video_up_id
      key_len: 4
          ref: test.c.video_id
         rows: 1
        Extra: Using index
5 rows in set (0.00 sec)

在针对少量行进行测试时，优化可以提供可变报告。因此，在针对真实数据集进行测试时，我们可能会看到不同的结果，然后可能没有必要给出USE INDEX 提示。

---

但是，尽管有上述解决方案，我还是希望在单独的查询中完成每项任务。在一个查询中完成所有事情对于开发和测试来说是复杂的，并且对于 RDBMS 来说执行起来通常很昂贵。如果您需要添加另一个计数，它将更加复杂。

SELECT
  u.full_name,
  COUNT(DISTINCT myvideos.id) AS Uploads
FROM userprofile_userprofile AS u
LEFT OUTER JOIN videos_video AS myvideos ON myvideos.uploaded_by_id = u.id
GROUP BY u.id;

SELECT
  u.full_name,
  COUNT(DISTINCT byothers.id) AS Credited_by_others
FROM userprofile_userprofile AS u
LEFT OUTER JOIN videos_videocredit AS c
  USE INDEX (videocredit_profileid_videoid)
  ON c.profile_id = u.id
LEFT OUTER JOIN videos_video AS byothers
  USE INDEX (video_up_id)
  ON c.video_id = byothers.id AND byothers.uploaded_by_id <> u.id
GROUP BY u.id;

SELECT
  u.full_name,
  COUNT(DISTINCT credited.id) AS Total_credits
FROM userprofile_userprofile AS u
LEFT OUTER JOIN (
  videos_videocredit AS c
  USE INDEX (videocredit_profileid_videoid)
  INNER JOIN videos_video AS credited
    ON c.video_id = credited.id
) ON c.profile_id = u.id
GROUP BY u.id;

Answer 3

总信用只是上传信用与国外信用的总和。由于上传信用很容易，这里只是外国信用。屏住呼吸进行两次子查询。

SELECT profile_id, COUNT(video_id) AS foreign_credit
       FROM (SELECT DISTINCT profile_id, video_id FROM videos_videocredit
             WHERE (profile_id, video_id) NOT IN (SELECT uploaded_by_id, id FROM videos_video)) AS crsq
GROUP BY profile_id;

---

这在视图中变得更加明显。我们制作了一个视图，只选择(profile_id, video_id) 的人在他们自己没有上传的视频中的功劳。我们将视图称为vfcredits。

CREATE VIEW vfcredits AS
  SELECT DISTINCT profile_id, video_id FROM videos_credit
  WHERE (profile_id, video_id) NOT IN (SELECT uploaded_by_id, id FROM videos_video);

现在我们可以愉快地将其粘贴到汇总外国信用的主查询中：

SELECT profile_id, COUNT(video_id) AS foreign_credit
FROM vfcredits
GROUP BY profile_id;

---

现在让我们把它们放在一起。我们再提出两种观点，一种是计算自己的学分，一种是计算国外的学分：

CREATE VIEW vowncount AS
  SELECT uploaded_by_id AS profile_id, COUNT(*) AS own_credits
  FROM videos_video
  GROUP BY uploaded_by_id;

CREATE VIEW vforeigncount AS
  SELECT profile_id, COUNT(video_id) AS foreign_credits
  FROM vfcredits
  GROUP BY profile_id;

最后，完整的选择：

SELECT name,
       own_credits,
       foreign_credits,
       own_credits + foreign_credits AS total_credits
FROM userprofile_userprofile
JOIN vowncount ON(userprofile_userprofile.id = vowncount.profile_id)
JOIN vforeigncount ON(userprofile_userprofile.id = vforeigncount.profile_id);