今天工作中遇到一点小难题,场景是在所有商家的所有会员的交易记录累计的那张表,只查询每个会员最新的那条记录,返回的是jsonArr
一开始,想到是的group by来让会员手机号去重,再循环所有已去重的手机号取最新的那一条limit 1 order date desc,这样能实现我的需求。
let ret = await this.model('tablename').field('ANY_VALUE(id), card_id, ANY_VALUE(total_fill), ANY_VALUE(create_at)').group('card_id').order('ANY_VALUE(create_at) desc').select()
但始终发现不优雅,然后在DDDM的stackoverflow上找到了更优雅的实现方法:
Sample -
ID Name Date Marks .. .. .. 1 XY 4/3/2017 27 1 fv 4/3/2014 98 1 jk 4/3/2016 09 2 RF 4/12/2015 87 2 kk 4/3/2009 56 2 PP 4/3/2011 76 3 ee 4/3/2001 12 3 ppp 4/3/2003 09 3 lll 4/3/2011 23
The Answer should be
ID Name Date Marks .. .. .. 1 XY 4/3/2017 27 2 RF 4/12/2015 87 3 lll 4/3/2011 23
某大牛的方案:
SELECT tt.* FROM myTable tt INNER JOIN (SELECT ID, MAX(Date) AS MaxDateTime FROM myTable GROUP BY ID) groupedtt ON tt.ID = groupedtt.ID AND tt.Date = groupedtt.MaxDateTime
https://stackoverflow.com/questions/45381743/select-rows-in-sql-with-latest-date-for-each-id-repeated-multiple-times
相当完美,一条语句搞掂,就是有点复杂,看得晕头转向的
最后附上我follow的代码
SELECT tt.id, tt.merchant_id, tt.card_id, tt.create_at, tt.surplus_score FROM tablename AS tt INNER JOIN (SELECT card_id, MAX(create_at) AS MaxDateTime FROM tablename GROUP BY card_id) groupedtt ON tt.card_id = groupedtt.card_id AND tt.create_at = groupedtt.MaxDateTime
感谢大牛们