mysql sql子查询还是多重嵌套连接?

2w3rbyxf  于 2023-02-28  发布在  Mysql
关注(0)|答案(2)|浏览(120)

我的目标是得到一个结果,其中它将输出所有练习条目,以及一个相关视频,然后是所有关联标签的连接字符串(其中标签是一个表,标签链接是标签与练习的关联方式)。
理想的最终结果:

My Exercise Name, Somepath/video.mp4, TagName1|TagName2|TagName3
Another Exercise Name, Somepath/video.mp4, TagName2|TagName5|TagName6
and so on...

我有这样的结构:

CREATE TABLE `exercises` (
  `id` int(11) unsigned NOT NULL AUTO_INCREMENT,
  `name` varchar(255) CHARACTER SET latin1 DEFAULT NULL,
  `video_id` int(11) DEFAULT NULL,
  PRIMARY KEY (`id`)
)

CREATE TABLE `videos` (
  `id` int(11) unsigned NOT NULL AUTO_INCREMENT,
  `filename` varchar(255) CHARACTER SET latin1 DEFAULT NULL,
  PRIMARY KEY (`id`)
)

CREATE TABLE `tags` (
  `id` int(11) unsigned NOT NULL AUTO_INCREMENT,
  `name` varchar(255) CHARACTER SET latin1 DEFAULT NULL,
  PRIMARY KEY (`id`)
)

CREATE TABLE `tag_linkage` (
  `id` int(11) unsigned NOT NULL AUTO_INCREMENT,
  `exercise_id` int(11) DEFAULT NULL,
  `tag_id` int(11) DEFAULT NULL,
  PRIMARY KEY (`id`)
)

这是我目前的疑问:

SELECT 
exercises.name AS Exercise, 
videos.filename AS VideoName, 
tags.name as Tag,
FROM exercises
  INNER JOIN tag_linkage AS tl ON  exercises.id = tl.exercise_id
  INNER JOIN tags ON tags.id = tl.tag_id 
    LEFT JOIN videos
    ON exercises.video_id = videos.id    
    ORDER BY exercises.id 
    LIMIT 20000 
    OFFSET 0;

标记的连接是我的难点。tag_linkage表有exercise_id和tag_id。我需要能够使用exercises.id作为tag_linkage.exercise_id的连接,但tag.name从另一个表返回www.example.com,但我不知道如何处理它。上面的当前查询将Tag作为结果中的一列,但每个标记都是一个新行:

My Exercise Name, Somepath/video.mp4, TagName1
My Exercise Name, Somepath/video.mp4, TagName2
My Exercise Name, Somepath/video.mp4, TagName3
Another Exercise Name, Somepath/video.mp4, TagName1
Another Exercise Name, Somepath/video.mp4, TagName3
Another Exercise Name, Somepath/video.mp4, TagName5
and so on...

我想删除重复的行并将标记连接到1列中。
编辑:想通了。

SELECT 
exercises.name as Exercise, 
videos.filename AS VideoName, 
GROUP_CONCAT(DISTINCT(tags.name)  separator ', ') Tags,
FROM exercises
LEFT JOIN videos
ON exercises.video_id = videos.id
JOIN tag_linkage ON exercises.id = tag_linkage.exercise_id
JOIN tags ON tags.id = tag_linkage.tag_id
GROUP BY exercises.name
LIMIT 1000
OFFSET 0;
643ylb08

643ylb081#

您需要单独连接标签

SELECT 
exercises.name AS Exercise, 
videos.filename AS VideoName, 
files.name as FileName, 
GROUP_CONCAT(tags.name SEPARATOR '|') AS Tags
FROM exercises
  INNER JOIN tag_linkage AS tl ON  exercises.id = tl.exercise_id
  INNER JOIN Tags  ON Tags.id = tl.tag_id 
    LEFT JOIN videos
    ON exercises.video_id = videos.id
    LEFT JOIN files
    ON exercises.file_id = files.id
    

    ORDER BY exercises.id 
    LIMIT 100 
    OFFSET 0;
3pmvbmvn

3pmvbmvn2#

我想明白了:

SELECT 
exercises.name as Exercise, 
videos.filename AS VideoName, 
GROUP_CONCAT(DISTINCT(tags.name)  separator ', ') Tags,
FROM exercises
LEFT JOIN videos
ON exercises.video_id = videos.id
JOIN tag_linkage ON exercises.id = tag_linkage.exercise_id
JOIN tags ON tags.id = tag_linkage.tag_id
GROUP BY exercises.name
LIMIT 1000
OFFSET 0;

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