Question

我的数据库中有模型（表格），表格和字段名称如

tags                 (id, name)
taggings             (id, tag_id, taggable_id, taggable_type, context)
employment_histories (id, user_id, grades, subjects, my_interests )
users                (id)

taggable_id实际上是employment_histories_id，上下文可以是成绩或科目或my_interests

现在我有一系列标签，例如G = { “第九”， “10日”} 我希望获得用户，只有其标签与上述g数组匹配。

我写了下面的查询：

SELECT DISTINCT users.* FROM `users` 
LEFT OUTER JOIN `employment_histories` 
  ON `employment_histories`.`user_id` = `users`.`id` 
LEFT OUTER JOIN `taggings` 
  ON `employment_histories`.`id` = `taggings`.`taggable_id` 
  AND `taggings`.`taggable_type` = 'EmploymentHistory' 
LEFT OUTER JOIN `tags` ON taggings.context = 'subjects' 
WHERE tags.name='9th' OR tags.name='10th'

但是它也给了我那些与任何标签匹配的用户，但我希望它只返回匹配所有两个标签的用户

假设标签9和10具有标签id 9和10然后我想要它只返回taggable_id（对于jobshistories.id），这对于这两个tag_id（即9和10）在标签中有共同的taggable_id表

例如我有两个用户tariq和kamal，这两个用户都有第9个标签常见但kamal没有标记10th所以想要查询如果通过这两个标签应该只返回tariq，其标签都是这两个标签但是用户喜欢与任何标签匹配的kamal也应该过滤

Answer 1

SELECT users.* FROM users 
INNER JOIN employment_histories 
  ON employment_histories.user_id = users.id 
INNER JOIN taggings 
  ON employment_histories.id = taggings.taggable_id 
  AND taggings.taggable_type = 'EmploymentHistory' 
  AND taggings.context = 'subjects'
INNER JOIN tags ON tags.id = taggings.tag_id
WHERE tags.name IN ('9th','10th')
GROUP BY users.id
HAVING COUNT(DISTINCT(tags.name)) = 2;

Answer 2

来自php chat room：

SELECT
users.* ,
count(*) AS count
FROM users
    LEFT JOIN employment_histories ON  users.id = employment_histories.user_id
    LEFT JOIN tagging ON tagging.taggable_id = employment_histories.id
    LEFT JOIN tags ON tags.id = tagging.tag_id
WHERE tags.name = "9th"
       OR tags.name = "10th"
GROUP BY users.id
HAVING count = 2

Answer 3

我重新编写了查询。

几乎没有变化：

加入tags.id = taggings.tag_id
从where子句中删除OR，并使用in，提高性能。

   SELECT DISTINCT users.*, count(*) as totRow FROM `users` 
   LEFT OUTER JOIN `employment_histories`
   ON `employment_histories`.`user_id` =
   `users`.`id`  LEFT OUTER JOIN
   `taggings`    ON
   `employment_histories`.`id` =
   `taggings`.`taggable_id`    AND
   `taggings`.`taggable_type` =
  'EmploymentHistory'    AND
   `taggings`.`context` = 'subjects' 
   LEFT OUTER JOIN `tags` ON `tags`.`id` = `taggings`.`tag_id` 
  WHERE tags.name = '9th' or tags.name = '10th'
  GROUP BY `users`.`id`

mysql查询中的复杂搜索问题

3 个答案: