Question

我有两个桌子

表A具有usr_id , usr_name , usr_status [1 or 0]

表B具有id, usr_id,tracked_date [Each user can track multiple time]

需要获取每个用户拥有usr_status = 1的最后一个曲目

如何通过MySQL查询实现？

尝试过：

SELECT 
    *
FROM
    tableb b
        LEFT JOIN
    tablea a ON a.usr_id = b.usr_id
GROUP BY a.usr_id

请咨询？

Answer 1

在派生表中，为每个tracked_date获取usr_id的最大值
将此结果集重新加入主表，以获取相应的行。

尝试：

SELECT 
  ta.*, tb.* 
FROM 
  tablea AS ta 
JOIN tableb AS tb 
  ON tb.user_id = ta.usr_id 
JOIN (
      SELECT 
        user_id, 
        MAX(tracked_date) AS max_tracked_date 
      FROM tableb
      GROUP BY user_id
     ) AS dt 
  ON dt.usr_id = tb.usr_id AND 
     dt.max_tracked_date = tb.tracked_date 
WHERE ta.usr_status = 1

在MySQL版本（8.0.2及更高版本）中，我们可以使用Window Functions。使用Row_Number()，我们可以将tracked_date分区中具有最高usr_id值的行的行号设置为1。现在，我们可以将其用作派生表，并仅考虑行号等于1的那些行。

SELECT 
  dt.* 
FROM 
(
  SELECT
    ta.*, 
    tb.*, 
    ROW_NUMBER() OVER (PARTITION BY ta.usr_id 
                       ORDER BY tb.tracked_date DESC) AS row_no 
  FROM 
    tablea AS ta 
  JOIN tableb AS tb ON tb.usr_id = ta.usr_id 
  WHERE ta.usr_status = 1
) AS dt 
WHERE dt.row_no = 1

其他：如果您希望获得所有用户（status = 1），但还没有tracked_date条目；表和派生表之间不能使用直线LEFT JOIN。

我们必须首先在单独的“派生表”中获得与最大值tracked_date相对应的 complete 行；然后从用户表中进行LEFT JOIN，以获取所有用户。

SELECT
ta.*, dt2.* 
FROM
  tablea AS ta
LEFT JOIN 
(SELECT tb.*, 
        dt.max_tracked_date
 FROM tableb AS tb
 JOIN (
       SELECT
         usr_id,
         MAX(tracked_date) AS max_tracked_date
       FROM tableb
       GROUP BY usr_id
      ) AS dt
   ON (dt.usr_id = tb.usr_id AND
       dt.max_tracked_date = tb.tracked_date) 
) AS dt2 ON ta.usr_id = dt2.usr_id  
WHERE ta.usr_status = 1

结果

| usr_id | usr_name | usr_status | id  | usr_id | stage | tracked_date        | max_tracked_date    |
| ------ | -------- | ---------- | --- | ------ | ----- | ------------------- | ------------------- |
| 1      | john     | 1          | 2   | 1      | 2     | 2018-11-12 13:12:12 | 2018-11-12 13:12:12 |
| 2      | dave     | 1          | 3   | 2      | 1     | 2018-11-12 13:12:12 | 2018-11-12 13:12:12 |
| 3      | smith    | 1          |     |        |       |                     |                     |

View on DB Fiddle

MYSQL-两个表从每个用户的第二个表的最后一个跟踪行获取记录

1 个答案: