MySQL 中复杂的 COUNT 查询

我正在尝试查找特定用户拥有的视频积分数量。

以下是相关的三个表：

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

CREATE TABLE `userprofile_videoinfo` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `title` varchar(256) NOT NULL,
  `uploaded_by_id` int(11) NOT NULL,
  ...
  KEY `userprofile_videoinfo_e43a31e7` (`uploaded_by_id`),
  CONSTRAINT `uploaded_by_id_refs_id_492ba9396be0968c` FOREIGN KEY (`uploaded_by_id`) REFERENCES `userprofile_userprofile` (`id`)
)

CREATE TABLE `userprofile_videocredit` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `video_id` int(11) NOT NULL,
  `profile_id` int(11) DEFAULT NULL,
  KEY `userprofile_videocredit_fa26288c` (`video_id`),
  KEY `userprofile_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 `userprofile_videoinfo` (`id`)
)

The videoinfo表是当用户上传视频时，他将获得“uploaded_by”列表。这videocredit表是给定电影的所有学分。它完全独立于上传电影（即，用户可以上传视频而无需注明自己的姓名，并且可以在用户未上传的视频中注明姓名）。

在尝试查找用户已记入的视频 COUNT 时，我想查找：

# videos a user has uploaded + # of non duplicate-video credits uploaded by others

举例来说：如果用户上传 5 个视频，名为：

VideoME1, VideoME2, VideoME3, VideoME4, and VideoME5 
(total = 5 videos [`videoinfo.uploaded_by_id`])

并拥有以下视频制作人员：

VideoME1 (4 credits - director, writer, editor, choreographer)
VideoME2 (1 credit)
VideoOTHER1 (2 credits - writer, editor)
VideoOTHER2 (1 credit - writer)
(total = 8 video credits [`videocredit.profile_id`])

COUNT 应为 5（上传的视频）+ 2（其他人上传的非重复视频片尾）= 7。如果用户没有视频片尾，则应 = 0（即，LEFT OUTER JOIN).

我已经能够计算出每个上传/积分的计数，但无法弄清楚如何将两者结合起来并消除重复项。我需要什么 SQL 来执行此操作？谢谢。

顺便说一句，这是我目前为每个（个人）COUNT 拥有的：

mysql> SELECT full_name, v.video_id, COUNT(DISTINCT v.video_id) as cnt
    -> FROM userprofile_userprofile u LEFT OUTER JOIN userprofile_videocredit v
    -> ON u.id = V.profile_id
    -> GROUP BY full_name
    -> ORDER BY cnt DESC;

mysql> SELECT full_name, v.id, COUNT(v.uploaded_by_id) as cnt
    -> FROM userprofile_userprofile u LEFT OUTER JOIN userprofile_videoinfo v
    -> ON u.id = v.uploaded_by_id
    -> GROUP BY full_name
    -> ORDER BY cnt DESC;

X-Zero 建议在数据中添加“上传者信用”，这是保持查询简单的最佳方法。如果这不是一个选项，请在 userprofile_videoinfo 和 userprofile_videocredit 之间进行内部联接，以便轻松消除重复项：

SELECT u.id, u.full_name, COUNT(DISTINCT v.video_id) as credit_count
FROM userprofile_userprofile u
LEFT JOIN (SELECT vi.video_id, vi.uploaded_by_id, vc.profile_id as credited_to_id
    FROM userprofile_videoinfo vi
    JOIN userprofile_videocredit vc ON vi.id = vc.video_id
    ) v ON u.id = v.uploaded_by_id OR u.id = v.credited_to_id
GROUP BY u.id, u.full_name
ORDER BY credit_count DESC

子查询对于创建视图可能很有用。