Question

我有两张桌子：

用户：

ID  |   title
0   |   Adam
1   |   Ben
2   |   Sara
3   |   Camille

users_meta：

ID  |   user_ID |   meta_key    |   meta_value
0   |       0   |   'weight'    |   '80'
1   |       0   |   'sex'       |   'male'
2   |       1   |   'weight'    |   '93'
3   |       1   |   'sex'       |   'male'
4   |       2   |   'weight'    |   '57'
5   |       2   |   'sex'       |   'female'
6   |       3   |   'weight'    |   '71'
7   |       4   |   'sex'       |   'female'

现在我想让所有男性和体重都超过90的用户，我的查询：

SELECT u.* FROM users AS u
RIGHT JOIN users_meta AS weight on(u.ID=weight.user_ID)
RIGHT JOIN users_meta AS sex on(u.ID=sex.user_ID)
WHERE
    weight.meta_key='weight' AND
    weight.meta_key>90 AND
    sex.meta_key='sex' AND
    sex.meta_value ='male'

以上查询对我来说很好，但我很好奇是否有更好的方法可以做到这一点。如果每个用户都有大量的元数据，那么最终会在同一个表上使用多个联接来检查不同的值。

Answer 1

您可以轻松地按users_meta聚合user_id表，并仅通过应用适当的HAVING子句保留符合条件的用户ID。然后为这组所需的用户ID选择用户。

select * 
from users
where user_id in
(
  select user_id
  from users_meta
  where meta_key in ('weight', 'sex') -- not necessary, but probably faster
  group by user_id
  having sum(meta_key = 'weight' and meta_value > 90) > 0
     and sum(meta_key = 'sex' and meta_value = 'male') > 0
);

Answer 2

您可以使用单join，例如：

SELECT DISTINCT u.id, u.title
FROM users u JOIN users_meta m ON u.id = m.user_id
WHERE (u.meta_key = 'weight' AND u.meta_value > 90)
OR (u.meta_key = 'sex' AND u.meta_value > 'male');

更新

如果你想让所有体重超过90的男性，你可以做两个join查询，例如：

SELECT * 
FROM (
  SELECT id, title FROM user WHERE id in (SELECT user_ID FROM user_meta WHERE meta_key = 'weight' AND meta_value > 90)
) a JOIN (
  SELECT id, title FROM user WHERE id in (SELECT user_ID FROM user_meta WHERE meta_key = 'sex' AND meta_value > 'male')
) b ON a.id = b.id;

MySQL查询连接在同一个表上

2 个答案: