查询中经常遇到这种查询,分组后取每组第一条.分享下一个SQL语句: --根据 x 分组后.根据 y 排序后取第一条 select * from ( select ROW_NUMBER() over(partition by x order by y desc) RowNum ,testTable.* 注:我使用MS SQL 08 R2
之前在一个项目的开发中,有遇到要根据分类来分组获取每组一条按某个条件字段排序的数据结果,于是先自己写了一条语句: select * from `表A` GROUP BY `c`; 上面这个语句有可以根据分类分组获得数据,但是无法对获得的数据进行排序,so 继续完善: select * from `表A` where `del`=0 and `markbok`=1 and `id` in(select SUBSTRING_INDEX(group_concat(`id` order by `add_
我要实现的功能是统计订单日志表中每一个订单的前三条日志记录,表结构如下: 一个订单在定点杆日志表中有多条记录,要根据时间查询出每一个订单的前三条日志记录,sql如下: select b.OrderNumber,b.creationtime,b.remark FROM ( SELECT a.OrderNumber,a.CreationTime,a.Remark FROM [FortuneLabFord].[dbo].[SO_Log] a where a.SysId IN ( SysId from
首先,将按条件查询并排序的结果查询出来. mysql order by accepttime desc; +---------------------+------+-----+ | accepttime | user | job | +---------------------+------+-----+ :: | :: | :: | :: | +---------------------+------+-----+ rows in set 然后,从中分组选出最新一条记录. mysql ord
select a.lng,a.lat from (select row_number() over ( partition by uid,grid_id) as rnum,weighted_centroid_lon as lng,weighted_centroid_lat lat from resultcccc)a where a.rnum = 1; 返回每组的第一条记录,速度贼溜
查询username,根据fundcode分组,按照date倒序,取date最大的一条数据 select * from ( select username, row_number() over(partition by fundcode, order by date desc) rn from usertable ) t -----------------------------------------------------------------------------感谢打赏!
sql中group by后,获取每组中的前N行数据,目前我知道的有2种方法 比如有个成绩表: 里面有字段学生ID,科目,成绩.我现在想取每个科目的头三名. 1. 子查询 select * from score s where StudentName in (select top 3 StudentName from score where s.Subjects = Subjects group by Subjects,StudentName,Score order by Score desc
想实现如下效果,就是分组后时间最大的那一条数据: 1.SQL SELECT * FROM ( SELECT * , ROW_NUMBER() OVER ( PARTITION BY RIP_GUID ORDER BY RU_CreatedTime DESC ) rn FROM RIP_FlowInfo ) t WHERE t.rn <= 1; 2.LINQ var groupQuery = from t in query group t by t.RIP_Guid into g select n