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MySQL的计算跨越多个列/表

2012-10-09 93 views
3

我有一些MySQL表，我想从提取一些信息最大值，该表是：MySQL的计算跨越多个列/表

  • 影片 - 代表了一个得分的视频。
  • 标签 - 包含全局标签列表。
  • VideoTags - 创建视频和标签之间的关联。

而且除了视频资源，我也有图片资源：

  • 照片 - 代表了一个得分的画面。
  • PictureTopic - 创建图片和主题之间的关联。

而对于影片的所有权和图片

  • 用户一个用户表 - 可以自己的视频和图片

我想要做的就是找到视频或图片最高每个标签/主题的点数。有许多视频和图片具有相同的标记/主题，但我的结果集的行数与标记/主题的行数相同。最终目标是为每个唯一标签（标签是以哈希为前缀的主题）提供最佳视频或图片列表（按点数）。

从上一个问题的解决方案（http://stackoverflow.com/questions/12778329/mysql-data-extraction-from-3-tables-joins-and-max） 我能够得到所有每个标签的点数最高的视频。

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）获得最高分的图片与此表达式每个主题：

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

我想要的是得到的图片或视频的每个最高点标签/话题，下面的边缘情况处理：

  • 如果有一个以上的图片或视频的特定主题（即它们具有相同的高分），然后推迟到资源所有者的点，如果他们都也有相同的点（不太可能），那么两个资源都可以在结果集中（除非资源由同一用户拥有，在这种情况下，结果集中只应该有一个结果）。
  • 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。

作为一个软件开发者，他使用Grails很多我喜欢依赖于对象关系映射，因此我的SQL技能是跛脚的。我到目前为止做的最好的是将两个共同选择的结果：

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

但不幸的是这还没有得到高分图片预期。由于可以在sqlfiddle可以看出（http://sqlfiddle.com/#!2/6650d/1）

从这个查询的输出是：

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

但我希望此行太：

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

和平等高的边缘情况后分数和分数阈值我想看看：

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

这里是模式和样本数据以供参考：

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');

有关如何最好地提取此信息的任何指针？干杯:)

来源

2012-10-09 Tetsuo

+0

你可以请你清楚你的情况我无法理解 –

回答

2 
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

查询此查询需要使用INNER JOIN，subqueries，UNION，GROUP BY和非官方MySQL的假设，即GROUP BY将返回基于ORDER BY POINTS DESC第1行

来源

2012-10-09 07:36:13

+0

这是伟大的感谢@ robin-castlin，关于我如何推迟当用户A的视频和用户B的图片都具有相同的点时，向用户A和B的点发现高分？正如用户所有者-k'和'所有者-l'的样本数据中的情况那样，即它们都具有22点的图片，因此属于'所有者-1'的图片应该是最高得分， l'Users表中的点数多于'owner-k'点数。 – Tetsuo

+0

只需在'Users'表中用'2 SELECTs'连接'JOIN'并用'user_points DESC'更新'ORDER BY'。以上查询已更新。 –

+0

谢谢罗宾！您的支票在邮件中;） – Tetsuo

0

下面是视频

select 
    t.content as `TOPIC-ID`, 
    vt.video_id as `RESOURCE-ID`, 
    'video' as `RESOURCE-TYPE`, 
    vt.owner_id as `RESOURCE-OWNER-ID`, 
    vt.MaxPoints 

from tags as t 
inner join 
    (SELECT 
     vt.tag_id, 
     vt.video_id, 
     MAX(v.points) as MaxPoints, 
     v.id, 
     v.owner_id 
    FROM videotags as vt 
    left join videos as v on v.id = vt.video_id 
    group by vt.tag_id 

    ) as vt on vt.tag_id = t.id 

union all 
SELECT 
    topic_id as `TOPIC-ID`, 
    picture_id as `RESOURCE-ID`, 
    'picture' as `RESOURCE-TYPE`, 
    p.owner_id as `RESOURCE-OWNER-ID`, 
    p.points as MaxPoints 
FROM picturetopic 
LEFT JOIN (SELECT id , owner_id , points FROM pictures) as p on p.id = picturetopic.picture_id

来源

2012-10-09 06:35:04

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