select 如何从 user 和 usermeta 表中获取最后的用户信息?

How select last user information from user and usermeta tables?

我正在使用 mysql,我有两个表格如下:

User:
user_id - user_name - phone

UserMeta:
user_meta_id - user_id - meta_key - meta_value - meta_date

我有如下一些记录:

user_id: 23  
meta_key: gender  
meta_value: male  
meta_date: 1534533650



user_id: 23  
meta_key: city  
meta_value: london  
meta_date: 1534533650



user_id: 23  
meta_key: name  
meta_value: jack  
meta_date: 1534533650  



user_id: 25  
meta_key: name  
meta_value: Jamie 
meta_date: 1534593881  

user_id: 25  
meta_key: gender  
meta_value: male 
meta_date: 1534593881  

user_id: 23  
meta_key: gender  
meta_value: female  
meta_date: 1534595971

user_id: 23  
meta_key: city  
meta_value: liverpool  
meta_date: 1534595971 

And ...

我需要获取所有用户信息 (user_id = 23),并注册了 最新 更改,例如:

user_id: 23  
meta_key: name  
meta_value: Jamie  
meta_date: 1534533650 

user_id: 23  
meta_key: city  
meta_value: liverpool  
meta_date: 1534595971  

user_id: 23  
meta_key: gender  
meta_value: female  
meta_date: 1534595971  

这个操作的查询很复杂,我很困惑,请帮助我,
我用了这个但没有得到正确的结果:

    "SELECT kmu.*
            FROM user_meta kmu
            INNER JOIN
            (SELECT `meta_key`, `meta_value`, MAX(`meta_date`) AS MaxDateTime
            FROM user_meta
            GROUP BY meta_key) groupedtt
            ON user_id=:id
            AND kmu.meta_key = groupedtt.meta_key
            AND kmu.meta_date = groupedtt.MaxDateTime";

您可以使用以下内容:

SELECT `user`.*, user_meta.meta_key, user_meta.meta_value, user_meta.meta_date 
FROM `user` INNER JOIN (
    SELECT user_id, meta_key, MAX(meta_date) AS meta_date 
    FROM user_meta 
    GROUP BY user_id, meta_key
) meta_select ON `user`.user_id = meta_select.user_id 
INNER JOIN user_meta ON meta_select.meta_key = user_meta.meta_key 
    AND meta_select.user_id = user_meta.user_id 
    AND meta_select.meta_date = user_meta.meta_date
WHERE `user`.user_id = 23

demo: https://www.db-fiddle.com/f/euqvGNW6PguCG495Yi9dTa/1

虽然我强烈反对这样的实施,但我不会试图改变您的想法。 虽然有很多方法可以实现这一点,但我相信下面的查询对性能的影响最小(不使用聚合)。

SELECT 
  user.*, meta.meta_key, meta.meta_value, meta.meta_date FROM user 
LEFT JOIN 
  (
    SELECT * FROM 
      (SELECT * FROM user_meta WHERE user_id = 23 ORDER BY meta_date DESC) sub 
    GROUP BY 
      sub.meta_key
  ) meta 
ON 
  user.user_id = meta.user_id
ORDER BY 
  meta.meta_key;

为了便于阅读,我将查询格式化为这样。您可能会看到我使用了 2 个嵌套子查询。该要求来自这样一个事实,即当您使用 GROUP BY 子句时,ORDER BY 无效。因此,首先我们 select 行并按日期 desc 对它们进行排序,然后我们进行分组。这样它就保留了排序顺序。

请参阅工作解决方案 here,另请注意,我在左侧窗格中的示例数据库创建脚本中添加了 2 个 INDEX CREATE 语句。随着数据库的增长,这些都是最低限度的,以确保至少有足够的性能。

我想提出一个基于 ROW_NUMBER function. This solution has been broadly using in other databases. For example, look at here. But your version of MySQL doesn't support the ROW_NUMBER function and i used its emulation described here. To check my solution, click here 的解决方案。

SELECT
  *
FROM `user` AS u
LEFT JOIN (
  SELECT
    t.*
  FROM (
    SELECT
      @rn := IF(@uid = user_id AND @mk = meta_key, @rn + 1, 1) AS rn,
      @uid := user_id AS uid,
      @mk := meta_key AS mk,
      um.*
    FROM `user_meta` AS um
    CROSS JOIN (SELECT @uid := 0, @mk := 0, @rn := 0) AS i
    ORDER BY um.user_id, um.meta_key, um.meta_date DESC
  ) AS t
  WHERE t.rn = 1
) AS r ON u.user_id = r.user_id
WHERE u.user_id = 23