MYSQL - 返回GROUP BY前的最近记录

我已经阅读了很多与这个问题非常类似的问题和答案，主要问题是他们使用LIMIT作为解决方案来返回最新的记录，吨与这个查询工作，因为我需要几个结果回来。MYSQL - 返回GROUP BY前的最近记录

逻辑如下为：

每个记录具有effective_date，当记录值生效。
只有拉回来的记录谁是effective_date的是今天或已通过
我拉回来的记录，给我基础上，最近的记录“date_entered”该记录

到目前为止通过此查询，我能够完成除了最后一点逻辑之外的所有内容。我不能在GROUP BY之后使用ORDER BY，因为结果已经被分组，因此我无法使用MAX()日期。我需要做WHERE声明中的逻辑。

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，则返回：

+---------+-------------------------+----------------------+ 
    | 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 | 
    +---------+-------------------------+----------------------+

：

 (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 | 
    +---------+-------------------------+----------------------+

与返回

+---------+-------------------------+----------------------+ 
    | 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 | 
    +---------+-------------------------+----------------------+

我试图找回是集团

因为我需要最新的VAL 1记录d回来，像使用

WHERE the datediff BETWEEN NOW() and date_entered = the smallest amount of time compared with other records

来源

2016-07-07 Josh Whitlow

使用'max（）'为date_entered – CSK

[检索每个组中的最后一条记录]的可能重复（http://stackoverflow.com/questions/1313120/retrieving-the-last-record-in-each-group ） – imtheman

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

的这个想法是使用group by前，向date_entered记录进行排序。从this post，groupby将使用挑选第一个在正常的SELECT相同的顺序。

来源

2016-07-08 00:18:44 r44

这工作！非常感谢！我一定会保留这个答案书签为将来使用。 –

max()为最新的日期时间

SELECT 
aos_products.`name`, 
cac_customize_agent_comp_cstm.commission_percentage_c, 
max(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() 
GROUP BY aos_products.`name`

来源

2016-07-07 21:40:16 CSK

这将返回最近的date_entered列，但其他2列保持不变。换句话说，它与数据不匹配。我相信这是因为列在不同的表中，但我可能是错的 –

MYSQL - 返回GROUP BY前的最近记录

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