我试图为每个唯一的用户 ID 选择最新的行,其中 pid = 50 且 active = 1。我一直无法弄清楚。
这是一个示例表
+-----+----------+-------+-----------------------+---------+
| id | userid | pid | start_date | active |
+-----+----------+-------+-----------------------+---------+
| 1 | 4 | 50 | 2015-05-15 12:00:00 | 1 |
| 2 | 4 | 50 | 2015-05-16 12:00:00 | 1 |
| 3 | 4 | 50 | 2015-05-17 12:00:00 | 0 |
| 4 | 4 | 51 | 2015-06-29 12:00:00 | 1 |
| 5 | 4 | 51 | 2015-06-30 12:00:00 | 1 |
| 6 | 5 | 50 | 2015-07-05 12:00:00 | 1 |
| 7 | 5 | 50 | 2015-07-06 12:00:00 | 1 |
| 8 | 5 | 51 | 2015-07-08 12:00:00 | 1 |
+-----+----------+-------+-----------------------+---------+
期望的结果
+-----+----------+-------+-----------------------+---------+
| id | userid | pid | start_date | active |
+-----+----------+-------+-----------------------+---------+
| 2 | 4 | 50 | 2015-05-16 12:00:00 | 1 |
| 7 | 5 | 50 | 2015-07-06 12:00:00 | 1 |
+-----+----------+-------+-----------------------+---------+
我已经尝试了很多东西,这是我得到的最接近的东西,但不幸的是它并没有停止。
SELECT *
FROM mytable t1
WHERE
(
SELECT COUNT(*)
FROM mytable t2
WHERE
t1.userid = t2.userid
AND t1.start_date < t2.start_date
) < 1
AND pid = 50
AND active = 1
ORDER BY start_date DESC
最佳答案
计划
<小时/>
- get last record grouping by userid where pid is 50 and is active
- inner join to mytable to get the record info associated with last
查询
select
my.*
from
(
select userid, pid, active, max(start_date) as lst
from mytable
where pid = 50
and active = 1
group by userid, pid, active
) maxd
inner join mytable my
on maxd.userid = my.userid
and maxd.pid = my.pid
and maxd.active = my.active
and maxd.lst = my.start_date
;
输出
+----+--------+-----+------------------------+--------+
| id | userid | pid | start_date | active |
+----+--------+-----+------------------------+--------+
| 2 | 4 | 50 | May, 16 2015 12:00:00 | 1 |
| 7 | 5 | 50 | July, 06 2015 12:00:00 | 1 |
+----+--------+-----+------------------------+--------+
注释
按照@Strawberry 的建议,更新为也在 pid 和 active 上加入。这将避免出现不活动或 pid 不是 50 但具有完全相同日期的记录也被渲染的可能性。
关于mysql 按日期为每个用户选择最近的行,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/31758064/