选择具有匹配标签的所有项目

Question

我正在努力寻找最有效的方法来处理这个问题，但我必须先告诉你，我把事情搞砸了。环顾四周，发现没有任何相关的东西，所以就这样了。

如何选择与所需项目具有相似标签的所有项目？

以此表为例：
_{（用于重新创建表的 SQL 代码如下）}

project 1 -> tagA | tagB | tagC
project 2 -> tagA | tagB
project 3 -> tagA
project 4 -> tagC

选择项目 1 应返回所有项目。
选择项目 4 应该只返回项目项目 1

到目前为止，我的查询非常依赖于左连接，并且肯定有更好的方法来执行此操作：

SELECT all_tags.project_id, all_tags.tag_id, final.title, tag.tag
FROM projects AS p
LEFT JOIN projects_to_tags AS pt ON p.num = pt.project_id
LEFT JOIN projects_to_tags AS all_tags ON pt.tag_id = all_tags.tag_id
LEFT JOIN projects AS final ON all_tags.project_id = final.num
LEFT JOIN tags AS tag ON all_tags.tag_id = tag.tag_id
WHERE p.num = 4
GROUP BY final.num

---

谢谢大家的输入。我想与大家分享 100k 个项目数据库、100k 个标签数据库以及 100k 个projects_to_tags 关系的所有查询的平均结果。所有查询均更改为询问project_1。

甜蜜而简短：

0.0160 sec - OMG Ponies - Using JOINS  
0.0208 sec - jdelard  
0.2581 sec - OMG Ponies - Using EXISTS  
0.2777 sec - OMG Ponies - Using IN  
0.5295 sec - Emtucifor - updated query  
0.5088 sec - Emtucifor - first query  

非常感谢大家。将相应地更新我的所有查询。

这里是所有查询和相应的 MySQL EXPLAIN 以及时间的

===============================================================================================================================================
Emtucifor - updated query
===============================================================================================================================================
Showing rows 0 - 1 (2 total, Query took 0.5295 sec)
SELECT * 
FROM projects AS L
WHERE L.num !=1-- instead of <> PT2.project_id inside

AND EXISTS (

SELECT 1 
FROM projects_to_tags PT
INNER JOIN projects_to_tags PT2 ON PT.tag_id = PT2.tag_id
WHERE L.num = PT.project_id
AND PT2.project_id =1
)
LIMIT 0 , 30

id  select_type table   type    possible_keys   key key_len ref rows    Extra
1   PRIMARY L   ALL PRIMARY NULL    NULL    NULL    100000  Using where
2   DEPENDENT SUBQUERY  PT2 ref project_id  project_id  4   const   1   Using index
2   DEPENDENT SUBQUERY  PT  ref project_id  project_id  8   test.L.num,test.PT2.tag_id  12000   Using index




===============================================================================================================================================
Emtucifor - first query
===============================================================================================================================================
Showing rows 0 - 1 (2 total, Query took 0.5088 sec)
SELECT * 
FROM projects AS L
WHERE 
EXISTS (

SELECT 1 
FROM projects_to_tags PT
INNER JOIN projects_to_tags PT2 ON PT.tag_id = PT2.tag_id
WHERE L.num = PT.project_id
AND PT2.project_id =1
AND PT2.project_id <> L.num
)
LIMIT 0 , 30

id  select_type table   type    possible_keys   key key_len ref rows    Extra
1   PRIMARY L   ALL NULL    NULL    NULL    NULL    100000  Using where
2   DEPENDENT SUBQUERY  PT2 ref project_id  project_id  4   const   1   Using index
2   DEPENDENT SUBQUERY  PT  ref project_id  project_id  8   test.L.num,test.PT2.tag_id  12000   Using where; Using index




===============================================================================================================================================
jdelard
===============================================================================================================================================
Showing rows 0 - 1 (2 total, Query took 0.0208 sec)
SELECT p.num, p.title
FROM projects_to_tags pt1, projects_to_tags pt2, projects p
WHERE pt1.project_id =1
AND pt2.project_id !=1
AND pt1.tag_id = pt2.tag_id
AND p.num = pt2.project_id
GROUP BY pt2.project_id
LIMIT 0 , 30

id  select_type table   type    possible_keys   key key_len ref rows    Extra
1   SIMPLE  pt1 ref project_id  project_id  4   const   1   Using index; Using temporary; Using filesort
1   SIMPLE  pt2 index   project_id  project_id  8   NULL    75001   Using where; Using index
1   SIMPLE  p   eq_ref  PRIMARY PRIMARY 4   test.pt2.project_id 1    




===============================================================================================================================================
OMG Ponies - Using IN
===============================================================================================================================================
Showing rows 0 - 2 (3 total, Query took 0.2777 sec)
SELECT p . * 
FROM projects p
JOIN projects_to_tags pt ON pt.project_id = p.num
WHERE pt.tag_id
IN (

SELECT x.tag_id
FROM projects_to_tags x
WHERE x.project_id =1
)
LIMIT 0 , 30

id  select_type table   type    possible_keys   key key_len ref rows    Extra
1   PRIMARY pt  index   project_id  project_id  8   NULL    100001  Using where; Using index
1   PRIMARY p   eq_ref  PRIMARY PRIMARY 4   test.pt.project_id  1    
2   DEPENDENT SUBQUERY  x   ref project_id  project_id  8   const,func  12000   Using where; Using index




===============================================================================================================================================
OMG Ponies - Using EXISTS
===============================================================================================================================================
Showing rows 0 - 2 (3 total, Query took 0.2581 sec)
SELECT p . * 
FROM projects p
JOIN projects_to_tags pt ON pt.project_id = p.num
WHERE EXISTS (

SELECT NULL 
FROM projects_to_tags x
WHERE x.project_id = 1
AND x.tag_id = pt.tag_id
)
LIMIT 0 , 30




===============================================================================================================================================
OMG Ponies - Using JOINS
===============================================================================================================================================
Showing rows 0 - 2 (3 total, Query took 0.0160 sec)
SELECT DISTINCT p . * 
FROM projects p
JOIN projects_to_tags pt ON pt.project_id = p.num
JOIN projects_to_tags x ON x.tag_id = pt.tag_id
AND x.project_id = 1
LIMIT 0 , 30

id  select_type table   type    possible_keys   key key_len ref rows    Extra
1   SIMPLE  x   ref project_id  project_id  4   const   1   Using index; Using temporary
1   SIMPLE  pt  index   project_id  project_id  8   NULL    75001   Using where; Using index
1   SIMPLE  p   eq_ref  PRIMARY PRIMARY 4   test.pt.project_id  1   

---

SQL 代码，用于复制/粘贴和混乱。

CREATE TABLE IF NOT EXISTS `projects` (
  `num` int(2) NOT NULL auto_increment,
  `title` varchar(30) NOT NULL,
  PRIMARY KEY  (`num`)
) ENGINE=MyISAM  DEFAULT CHARSET=latin1 AUTO_INCREMENT=5 ;


INSERT INTO `projects` (`num`, `title`) VALUES(1, 'project 1'),(2, 'project 2'),(3, 'project 3'),(4, 'project 4');


CREATE TABLE IF NOT EXISTS `projects_to_tags` (
  `project_id` int(2) NOT NULL,
  `tag_id` int(2) NOT NULL,
  KEY `project_id` (`project_id`,`tag_id`)
) ENGINE=MyISAM DEFAULT CHARSET=latin1;


INSERT INTO `projects_to_tags` (`project_id`, `tag_id`) VALUES(1, 1),(1, 2),(1, 3),(2, 1),(2, 2),(3, 1),(4, 3);


CREATE TABLE IF NOT EXISTS `tags` (
  `tag_id` int(2) NOT NULL auto_increment,
  `tag` varchar(30) NOT NULL,
  PRIMARY KEY  (`tag_id`),
  UNIQUE KEY `tag` (`tag`)
) ENGINE=MyISAM  DEFAULT CHARSET=latin1 AUTO_INCREMENT=4 ;


INSERT INTO `tags` (`tag_id`, `tag`) VALUES(1, 'tag a'),(2, 'tag b'),(3, 'tag c');

Answer 1

在以下任何一种情况下，如果您不知道 PROJECT.num/PROJECT_TO_TAGS.project_id，则必须加入 PROJECTS code> table 来获取 id 值，以找出它关联的标签。

使用 IN

SELECT p.*
  FROM PROJECTS p
  JOIN PROJECTS_TO_TAGS pt ON pt.project_id = p.num
 WHERE pt.tag_id IN (SELECT x.tag_id
                       FROM PROJECTS_TO_TAGS x
                      WHERE x.project_id = 4)

使用 EXISTS

SELECT p.*
  FROM PROJECTS p
  JOIN PROJECTS_TO_TAGS pt ON pt.project_id = p.num
 WHERE EXISTS (SELECT NULL
                 FROM PROJECTS_TO_TAGS x
                WHERE x.project_id = 4
                  AND x.tag_id = pt.tag_id)

使用 JOINS（这是最有效的！）

DISTINCT 是必要的，因为 JOIN 存在在结果集中出现重复数据的风险...

SELECT DISTINCT p.*
  FROM PROJECTS p
  JOIN PROJECTS_TO_TAGS pt ON pt.project_id = p.num
  JOIN PROJECTS_TO_TAGS x ON x.tag_id = pt.tag_id
                         AND x.project_id = 4

Answer 2

怎么样...（项目 1 的示例）

SELECT p.num, p.title
FROM projects_to_tags pt1, projects_to_tags pt2, projects p
where pt1.project_id = 1 and 
      pt2.project_id != 1 and 
      pt1.tag_id = pt2.tag_id and 
      p.num = pt2.project_id 
group by pt2.project_id

也许可以在projects_to_tags 中为tag_id 添加一个单独的索引，这样您就可以单独使用它，而不是组合使用它。不再需要输入 ALL。 （表格扫描）
将 1 替换为 4 也能得到所需的结果。

Answer 3

像这样的东西...？

SELECT *
FROM projects AS L
WHERE
   EXISTS (
      SELECT 1
      FROM
         projects_to_tags PT
         INNER JOIN projects_to_tags PT2 ON PT.tag_id = PT2.tag_id
      WHERE
         L.num = PT.project_id
         AND PT2.project_id = 4
         AND PT2.project_id <> L.num
   )

这是 2 次搜索和一次扫描。

更新

从 jdelard 的书中摘取一页，一个微小的修改使我的查询优于他的查询（当然我是在 SQL Server 上执行此操作，这意味着我取出了他的 GROUP BY 并放入了 DISTINCT，所以 YMMV在 MySQL 上）：

SELECT *
FROM projects AS L
WHERE
   L.num != 4 -- instead of <> PT2.project_id inside
   AND EXISTS (
      SELECT 1
      FROM
         projects_to_tags PT
         INNER JOIN projects_to_tags PT2 ON PT.tag_id = PT2.tag_id
      WHERE
         L.num = PT.project_id
         AND PT2.project_id = 4
   )

对其查询的改进来自于不执行 DISTINCT 或聚合，并使用半连接而不是完全连接，因此不必连接每一行。否则，在语义上它们基本上是相同的。

我必须记住 jdelard 的技巧，因为它是一个非常有用的工具。由于某种原因，查询引擎不够智能，无法计算给定的 {a = 4, a != b} 然后 {b != 4}。