Question

我正在努力找到解决这个问题的最有效方法，但我必须告诉你，我已经把它搞得一团糟。环顾四周，没有发现任何相关性，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：
_{（用于重新创建表格的sql代码）}

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

选择项目1应该返回所有项目选择项目4应仅返回项目项目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可以执行此操作：

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

谢谢大家的意见。我虽然与大家分享一下100k项目数据库的所有查询的平均结果，100k标签数据库与100k projects_to_tags关系。所有查询都已更改为要求project_1。

甜蜜与短暂：

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。要相应地更新我的所有查询。

此处查看所有查询及相应的MySQL EXPLAIN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

复制/粘贴和混乱的SQL代码。

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表以获取用于查找的ID值输出与之相关的标签。

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会冒重复数据出现在结果集中......

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

如何......（项目1的例子）

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

并且可能在projects_to_tags中为tag_id添加单独的索引，因此您可以单独使用它，而不是复合。没有更多类型的ALL。（表扫描）将 1 替换为 4 也可以获得所需的结果。

Answer 3

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
   )

那是2寻找和扫描。

<强>更新

从jdelard的书中读取一个页面，一个微小的修改将我的查询切换为优于他（当然我在SQL Server上执行此操作意味着我取出他的GROUP BY并放入DISTINCT，因此在MySQL上使用YMMV）：< / p>

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
   )

对查询的改进来自于不进行DISTINCT或聚合，并使用半连接而不是完整连接，因此不是每行都必须连接。否则，在语义上它们大致相同。

我必须记住jdelard的诀窍，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{a = 4，a！= b}然后{b！= 4}。

选择具有匹配标记的所有项目

3 个答案:

使用IN

使用EXISTS

使用JOINS（这是最有效的！）