我有一些 mysql 表，我想从中提取一些信息，这些表是:

• Videos - 代表带有分数的视频。
• 标签 - 包含标签的全局列表。
• VideoTags - 创建视频和标签之间的关联。

除了视频资源，我还有图片资源:

• 图片 - 代表带有分数的图片。
• PictureTopic - 在图片和主题之间建立关联。

还有一个用于记录视频和图片所有权的用户表

• 用户 - 可以拥有视频和图片

我想做的是为每个标签/主题找到得分最高的视频或图片。有很多具有相同标签/主题的视频和图片，但我的结果集将具有 与标签/主题相同的行数。最终目标是为每个唯一标签(标签是一个以散列为前缀的主题)列出最佳视频或图片(按点)。

使用上一个问题的解决方案 (http://stackoverflow.com/questions/12778329/mysql-data-extraction-from-3-tables-joins-and-max) 我能够获得每个标签最高分的所有视频。

SELECT SUBSTR(Tags.content,2) as topic_id, Videos.id as resource_id, 'video' as resource_type, Videos.owner_id as resource_owner_id, Videos.points FROM Videos JOIN (
  SELECT   VideoTags.tag_id, MAX(points) points
  FROM     Videos JOIN VideoTags ON Videos.id = VideoTags.video_id
  GROUP BY VideoTags.tag_id
) t USING (points) JOIN Tags ON t.tag_id = Tags.id and Tags.content LIKE "#%"

我也可以(有点)用这个表达式得到每个主题最高分的图片:

SELECT   PictureTopic.topic_id, Pictures.id as resource_id, 'picture' as resource_type, Pictures.owner_id as resource_owner_id, MAX(points) points
FROM     Pictures JOIN PictureTopic ON Pictures.id = PictureTopic.picture_id
GROUP BY PictureTopic.topic_id

我想要的是为每个标签/主题获取最高分的图片或视频，并处理以下边缘情况:

• 如果给定主题有多个图片或视频(即它们具有相同的高分)，则遵从资源所有者的分数，如果它们也具有相同的分数(不太可能)，则两个资源都可以在结果集中(除非资源归同一用户所有，在这种情况下，结果集中应该只有一个结果)。
• 如果视频或图片的点数小于 20，则将该资源从结果集中排除。

作为一名经常使用 Grails 的软件开发人员，我喜欢依赖对象关系映射，因此我的 sql 技能很差。到目前为止，我能做的最好的事情就是将两个选择的结果放在一起:

SELECT SUBSTR(Tags.content,2) as topic_id, Videos.id as resource_id, 'video' as resource_type, Videos.owner_id as resource_owner_id, Videos.points FROM Videos JOIN (
  SELECT   VideoTags.tag_id, MAX(points) points
  FROM     Videos JOIN VideoTags ON Videos.id = VideoTags.video_id
  GROUP BY VideoTags.tag_id
) t USING (points) JOIN Tags ON t.tag_id = Tags.id and Tags.content LIKE "#%"
UNION
SELECT   PictureTopic.topic_id, Pictures.id as resource_id, 'picture' as resource_type, Pictures.owner_id as resource_owner_id, MAX(points) points
FROM     Pictures JOIN PictureTopic ON Pictures.id = PictureTopic.picture_id
GROUP BY PictureTopic.topic_id

但不幸的是，这甚至没有获得预期的高分图片。从 sqlfiddle ( http://sqlfiddle.com/#!2/6650d/1 ) 可以看出

这个查询的输出是:

TOPIC_ID    RESOURCE_ID         RESOURCE_TYPE   RESOURCE_OWNER_ID   POINTS
topic-1     owner-x-video-a     video           owner-x             20
topic-2     owner-y-video-m     video           owner-y             44
topic-1     owner-j-pic-1       picture         owner-j             50
topic-3     owner-k-pic-2       picture         owner-k             22

但我也希望这一行:

TOPIC_ID    RESOURCE_ID         RESOURCE_TYPE   RESOURCE_OWNER_ID   POINTS
topic-3     owner-l-pic-3       picture         owner-l             22

在相同高分和分数阈值的边缘情况之后，我希望看到:

TOPIC_ID    RESOURCE_ID         RESOURCE_TYPE   RESOURCE_OWNER_ID   POINTS
topic-1     owner-j-pic-1       picture         owner-j             50
topic-2     owner-y-video-m     video           owner-y             44
topic-3     owner-l-pic-3       picture         owner-l             22

这是供引用的架构和示例数据:

CREATE TABLE `Users` (
  `id`       VARCHAR(24) NOT NULL DEFAULT '',
  `points`   DOUBLE      NOT NULL DEFAULT 0,
  PRIMARY KEY (id)
) Engine=InnoDB;

DROP TABLE IF EXISTS `Videos`;
CREATE TABLE `Videos` (
  `id` varchar(24) NOT NULL default '',
  `owner_id` varchar(24) NOT NULL default '',
  `points` DOUBLE NOT NULL default 0
);

DROP TABLE IF EXISTS `Tags`;
CREATE TABLE `Tags` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `content` varchar(32) NOT NULL default ''
PRIMARY KEY (id)
);

DROP TABLE IF EXISTS `VideoTags`;
CREATE TABLE `VideoTags` (
  `video_id` varchar(24) NOT NULL default '',
  `tag_id` int(11) NOT NULL
);

DROP TABLE IF EXISTS `Pictures`;
CREATE TABLE `Pictures` (
  `id` varchar(24) NOT NULL default '',
  `owner_id` varchar(24) NOT NULL default '',
  `points` DOUBLE NOT NULL default 0
);

DROP TABLE IF EXISTS `PictureTopic`;
CREATE TABLE `PictureTopic` (
  `picture_id` varchar(24) NOT NULL,
  `topic_id` varchar(31) NOT NULL
);

INSERT INTO Users (id, points) VALUES ('owner-x', 0);
INSERT INTO Users (id, points) VALUES ('owner-y', 0);
INSERT INTO Users (id, points) VALUES ('owner-j', 0);
INSERT INTO Users (id, points) VALUES ('owner-k', 5);
INSERT INTO Users (id, points) VALUES ('owner-l', 14);

INSERT INTO Videos (id,owner_id,points) VALUES
  ('owner-x-video-a','owner-x', 20),
  ('owner-x-video-b','owner-x', 15),
  ('owner-y-video-k','owner-y', 12),
  ('owner-y-video-l','owner-y', 17),
  ('owner-y-video-m','owner-y', 44);

INSERT INTO Tags (id, content) VALUES
  (111, '#topic-1'),
  (222, '#topic-2');

INSERT INTO VideoTags (video_id,tag_id) VALUES
  ('owner-x-video-a',111),
  ('owner-x-video-b',111),
  ('owner-y-video-k',111),
  ('owner-y-video-l',222),
  ('owner-y-video-m',222);

INSERT INTO Pictures (id, owner_id, points) VALUES ('owner-j-pic-1','owner-j', 50);
INSERT INTO Pictures (id, owner_id, points) VALUES ('owner-k-pic-2','owner-k', 22);
INSERT INTO Pictures (id, owner_id, points) VALUES ('owner-l-pic-3','owner-l', 22);

INSERT INTO PictureTopic (picture_id, topic_id) VALUES ('owner-j-pic-1','topic-1');
INSERT INTO PictureTopic (picture_id, topic_id) VALUES ('owner-k-pic-2','topic-3');
INSERT INTO PictureTopic (picture_id, topic_id) VALUES ('owner-l-pic-3','topic-3');

关于如何最好地提取此信息的任何指示？干杯:)

最佳答案

SELECT TOPIC_ID, RESOURCE_ID, RESOURCE_TYPE, RESOURCE_OWNER_ID, POINTS
FROM (( SELECT pt.topic_id AS TOPIC_ID,
            p.id AS RESOURCE_ID,
            'picture' AS RESOURCE_TYPE,
            p.owner_id AS RESOURCE_OWNER_ID,
            p.points AS POINTS,
            u.points AS user_points
        FROM Pictures AS p
        INNER JOIN PictureTopic AS pt
        ON p.id = pt.picture_id
        INNER JOIN Users AS u
        ON p.owner_id = u.id)
        UNION ALL
    (   SELECT SUBSTR(t.content, 1), v.id, 'video', v.owner_id, v.points, u.points
        FROM Videos AS v
        INNER JOIN VideoTags AS vt
        ON v.id = vt.video_id
        INNER JOIN Tags AS t
        ON vt.tag_id = t.id
        INNER JOIN Users AS u2
        ON v.owner_id = u2.id)
        ORDER BY POINTS DESC, user_points DESC) AS h
GROUP BY TOPIC_ID
ORDER BY TOPIC_ID ASC

此查询使用了 INNER JOIN , subqueries , UNION , GROUP BY和非官方的 MySQL 假设 GROUP BY 将返回基于 ORDER BY POINTS DESC

关于MySQL计算多个列/表的最大值，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/12793915/