我已经阅读了很多关于与这个非常相似的问题的问题和答案,主要问题是他们使用 LIMIT
作为返回最新记录的解决方案,该记录将无法使用这个查询是因为我需要返回多个结果。
逻辑如下:
- 每条记录都有一个
effective_date
,即记录值生效的时间。 - 仅提取
effective_date
是今天或已经过去的记录 - 在我撤回的记录中,根据该记录的“
date_entered
”给我最新的记录
到目前为止,我已经能够使用此查询完成除最后一点逻辑之外的所有操作。我不能在 GROUP BY
之后使用 ORDER BY
,因为结果已经分组所以我不能使用 MAX()
日期。我需要在 WHERE
语句中执行逻辑。
SELECT
aos_products.`name`,
cac_customize_agent_comp_cstm.commission_percentage_c,
cac_customize_agent_comp.date_entered
FROM
wn_writing_number
LEFT OUTER JOIN wn_writing_number_cac_customize_agent_comp_1_c ON wn_writing_number.id = wn_writing_number_cac_customize_agent_comp_1_c.wn_writing946b_number_ida
LEFT OUTER JOIN cac_customize_agent_comp_cstm ON wn_writing_number_cac_customize_agent_comp_1_c.wn_writing3148nt_comp_idb = cac_customize_agent_comp_cstm.id_c
LEFT OUTER JOIN cac_customize_agent_comp ON cac_customize_agent_comp_cstm.id_c = cac_customize_agent_comp.id
LEFT OUTER JOIN aos_products_cac_customize_agent_comp_1_c ON cac_customize_agent_comp_cstm.id_c = aos_products_cac_customize_agent_comp_1_c.aos_produca2b8nt_comp_idb
LEFT OUTER JOIN aos_products ON aos_products_cac_customize_agent_comp_1_c.aos_products_cac_customize_agent_comp_1aos_products_ida = aos_products.id
WHERE
wn_writing_number.id = 'b556d816-f2e6-b78f-c776-576c0542585c'
AND wn_writing_number_cac_customize_agent_comp_1_c.deleted = '0'
AND cac_customize_agent_comp_cstm.effective_date_c <= CURDATE()
GROUP BY aos_products.`name`
没有 GROUP BY,它返回:
(varchar) (varchar) (datetime)
+---------+-------------------------+----------------------+
| name | commission_percentage_c | date_entered |
+---------+-------------------------+----------------------+
| Val 1 | 15.25 | 2016-07-06 23:57:28 |
+---------+-------------------------+----------------------+
| Val 1 | 15.75 | 2016-07-07 00:03:03 |
+---------+-------------------------+----------------------+
| Val 1 | 16 | 2016-07-07 00:31:08 |
+---------+-------------------------+----------------------+
| Val 2 | 14.75 | 2016-07-07 15:04:02 |
+---------+-------------------------+----------------------+
它返回 GROUP BY
+---------+-------------------------+----------------------+
| name | commission_percentage_c | date_entered |
+---------+-------------------------+----------------------+
| Val 2 | 14.75 | 2016-07-07 15:04:02 |
+---------+-------------------------+----------------------+
| Val 1 | 15.25 | 2016-07-06 23:57:28 |
+---------+-------------------------+----------------------+
我想得到的是:
+---------+-------------------------+----------------------+
| name | commission_percentage_c | date_entered |
+---------+-------------------------+----------------------+
| Val 2 | 14.75 | 2016-07-07 15:04:02 |
+---------+-------------------------+----------------------+
| Val 1 | 16 | 2016-07-07 00:31:08 |
+---------+-------------------------+----------------------+
因为我需要返回最近的 VAL 1 记录,比如
WHERE the datediff BETWEEN NOW() and date_entered = the smallest amount of time compared with other records
最佳答案
试试这个
SELECT *
FROM
(SELECT
aos_products.`name` as name,
cac_customize_agent_comp_cstm.commission_percentage_c as commission_percentage_c,
cac_customize_agent_comp.date_entered as date_entered
FROM
wn_writing_number
LEFT OUTER JOIN wn_writing_number_cac_customize_agent_comp_1_c ON wn_writing_number.id = wn_writing_number_cac_customize_agent_comp_1_c.wn_writing946b_number_ida
LEFT OUTER JOIN cac_customize_agent_comp_cstm ON wn_writing_number_cac_customize_agent_comp_1_c.wn_writing3148nt_comp_idb = cac_customize_agent_comp_cstm.id_c
LEFT OUTER JOIN cac_customize_agent_comp ON cac_customize_agent_comp_cstm.id_c = cac_customize_agent_comp.id
LEFT OUTER JOIN aos_products_cac_customize_agent_comp_1_c ON cac_customize_agent_comp_cstm.id_c = aos_products_cac_customize_agent_comp_1_c.aos_produca2b8nt_comp_idb
LEFT OUTER JOIN aos_products ON aos_products_cac_customize_agent_comp_1_c.aos_products_cac_customize_agent_comp_1aos_products_ida = aos_products.id
WHERE
wn_writing_number.id = 'b556d816-f2e6-b78f-c776-576c0542585c'
AND wn_writing_number_cac_customize_agent_comp_1_c.deleted = '0'
AND cac_customize_agent_comp_cstm.effective_date_c <= CURDATE()
ORDER BY cac_customize_agent_comp.date_entered DESC) t
GROUP BY t.name
这样做的想法是使用 group by
按 date_entered
对记录进行排序。来自 this post , groupby
将使用与正常 SELECT
相同的顺序选择第一个。
关于MYSQL - 返回 GROUP BY 之前的最新记录,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/38255694/