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选择具有匹配标记的所有项目

join left-join tags mysql sql

Frankie · 技术社区 · 15 年前

我正试图找到最有效的方法来处理这件事,但我必须告诉你,我把事情搞得一团糟。环顾四周,没有发现任何相关的东西,就这样。

如何选择与所需项目具有相似标记的所有项目?

_{(下面是重新创建表的sql代码)}

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

选择项目4只应返回项目1

到目前为止,我的查询非常依赖于左联接,可以肯定的是,有更好的方法可以做到这一点:

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

谢谢大家的意见。我想我会和你们分享一个100k项目数据库的所有查询的平均结果,100k标签数据库有100k项目\u到\u标签的关系。所有查询都已更改为请求项目1。

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

非常感谢你们。我会相应地更新我所有的查询。

下面是所有的查询和相应的MySQL解释,以及时间

===============================================================================================================================================
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

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');

3 回复 | 直到 15 年前

Frankie 15 年前

在下列任何情况下,如果你不知道 PROJECT.num / PROJECT_TO_TAGS.project_id ,你必须加入 PROJECTS

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)

使用存在

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)

使用连接(这是最有效的连接!)

DISTINCT 是必需的,因为在结果集中出现重复数据。。。

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

jdelard 15 年前

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

或者在项目中为标记添加一个单独的索引,这样您就可以单独使用它,而不是复合索引。没有更多的类型全部。(表格扫描) 1 具有 4 同时给出期望的结果。

momo 15 年前

像这样的?

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
   )

那是两次搜索和一次扫描。

以jdelard的书为例,一个小小的修改就将我的查询转换为优于他的查询(当然,我是在sqlserver上这样做的,这意味着我将他的GROUP BY去掉,并在MySQL上放入一个独特的YMMV):

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
   )

对他的查询的改进来自于不执行DISTINCT或aggregate,并且使用半连接而不是完全连接,这样就不必连接每一行。否则,它们在语义上基本相同。

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