Question

我有一个存储用户兴趣的数据库表：

CREATE TABLE `ade39_findme_settings_words` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `user_id` int(11) DEFAULT NULL,
  `word` varchar(255) DEFAULT NULL,
  `language` varchar(2) DEFAULT NULL,
  `timestamp` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP ON UPDATE CURRENT_TIMESTAMP,
  PRIMARY KEY (`id`),
  KEY `clickeditem` (`user_id`),
  KEY `count` (`word`)
) ENGINE=MyISAM AUTO_INCREMENT=134 DEFAULT CHARSET=utf8;

word = interest

现在我想在此表中搜索具有一个或多个相同兴趣的用户，并按相同兴趣的数量对此列表进行排序。知道sql查询的样子吗？

喜欢“与user_id = 62具有相似兴趣的所有用户”

示例：

 USER  INTEREST
 62    riding
 62    reading
 62    dancing
 75    riding
 75    dancing
 80    riding

输出：

 USER  SAME_COUNT 
  75   2
  80   1

感谢。

Answer 1

您可能遇到性能问题，因此您可以尝试使用EXISTS

进行改进

SELECT t.user_id, COUNT(*) as num_interests, GROUP_CONCAT(word) as interests
FROM ade39_findme_settings_words t
WHERE EXISTS
(   SELECT 1
    FROM ade39_findme_settings_words t1
    WHERE t1.user_id = 62
      AND t1.word = t.word
)
AND user_id <> 62
GROUP BY t.user_id
ORDER BY num_interests DESC;

最快的方法是使用排除JOIN，但是这样

SELECT t.user_id, COUNT(*) as num_interests, GROUP_CONCAT(t.word) as interests
FROM ade39_findme_settings_words t
LEFT JOIN ade39_findme_settings_words t1 ON t1.user_id = 62 AND t1.word = t.word
WHERE t.user_id <> 62
GROUP BY user_id
ORDER BY num_interests DESC;

Answer 2

SELECT `user_id`, COUNT(*) as same_count
FROM `ade39_findme_settings_words`
WHERE `word` IN (SELECT `word` FROM `ade39_findme_settings_words` WHERE `user_id` = 62)
GROUP BY `user_id`
ORDER BY same_count DESC;

Answer 3

这对我也很有用：

select temp.id_new , count(temp.intrst) 
from 
(select a.user_id id_new, a.word intrst 
from   ade39_findme_settings_words a 
where a.word in (select b.word  from ade39_findme_settings_words b where b.user_id=62) 
and a.user_id<>62 ) temp
group by temp.id_new;

Mysql命令来自表中的相同项目

3 个答案: