MySQL中复杂的COUNT查询

本文介绍了MySQL中复杂的COUNT查询的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我正在尝试查找特定用户拥有的视频点数。

I am trying to find the number of video credits a particular user has.

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`) )

videoinfo 表是用户上传视频时，获取 uploaded_by列表。 videocredit 表是给定电影的所有片数。它完全独立于上载影片（即，用户可以在不信誉自己的情况下上传视频，而用户可以在他尚未上传的视频中获得信誉）。

The videoinfo table is when a user uploads a video, he will get an "uploaded_by" listing. The videocredit table are all the credits for a given film. It is entirely independent of uploading the film (i.e., a video can be uploaded by a user without crediting himself, and a user can be credited in a video he has not uploaded).

在尝试查找用户已记入的COUNT个视频时，我想查找：

In trying to find the COUNT of videos a user has been credited in, I want to find:

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

例如：用户上传了5个视频，分别为：

By way of example: if a user uploads 5 videos called:

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

并且具有以下视频片数：

And has the following video credits:

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应该是5（视频上传）+ 2（其他人上传的非重复的视频积分）=7。如果用户没有视频积分，则应= 0（即 LEFT OUTER JOIN ）。

The COUNT should be 5 (videos uploaded) + 2 (non-duplicate video credits uploaded by others) = 7. If a user has no video credits, it should = 0 (i.e., LEFT OUTER JOIN).

我已经能够计算出每个上载/贷项的COUNTS，但无法弄清楚如何合并两者并消除重复项。我需要执行什么SQL？谢谢。

I've been able to figure out the COUNTS for each of the uploads/credits, but can't figure out how to combine the two and get rid of duplicates. What SQL do I need to do this? Thank you.

顺便说一句，这是我目前每个（个人）COUNT所拥有的：

By the way, this is what I currently have for each (individual) COUNT:

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;

推荐答案

X-Zero建议添加上载额度数据是保持查询简单的最佳方法。如果不是这样，请在userprofile_videoinfo和userprofile_videocredit之间进行内部联接，以轻松消除重复项：

X-Zero's suggestion of adding an "uploader credit" to the data is the best way to keep the query simple. If that's not an option, do an inner join between userprofile_videoinfo and userprofile_videocredit to make it easy to eliminate duplicates:

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

子查询对于作为视图创建可能有用。

The subquery may be useful to create as a view.