以前在开发的时候遇到过一个需求,就是要按照某一列进行分组后取前几条数据,今天又有同事碰到了,帮解决了之后顺便写一篇博客记录一下. 首先先建一个基础数据表,代码如下: IF OBJECT_ID(N'Test') IS NOT NULL BEGIN DROP TABLE Test END CREATE TABLE Test(ID bigint IDENTITY(1,1),Name nvarchar(50),Department nvarchar(50)) INSERT IN
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
如下图, 计划实现 :按照 parent_code 分组, 取组中code最大值所在的整条记录,如红色部分.(类似hive中: row_number() over(partition by)) select c.* from ( end) as sort_num,(@key_i:=parent_code) as tmp ,@key_i:='') b order by parent_code,code desc) c ; 个人理解, mysql 运行顺序: from >> where >
获取分组后取某字段最大一条记录 方法一:(效率最高) select * from test as a where typeindex = (select max(b.typeindex) from test as b where a.type = b.type ); 方法二:(效率次之) select a.* from test a, (select type,max(typeindex) typeindex from test group by type) b where a.type = b
select * from ( select last_comment, row_number() over(partition by employeeid,roadline,stationname order by logindate desc) rn from reocrd ) t where t.rn <=1 这段的意思是,将reocrd表根据员工工号( employeeid),线路(,roadline),站点名称(stationname)分组后,取登录日期(logindate) 最大的那
一.问题 groupBY分组后取最新一条记录的SQL的解决方案. 二.解决方案 select Message,EventTime from PT_ChildSysAlarms as a where EventTime = (select max(b.EventTime) from PT_ChildSysAlarms as b where a.PtName = b.PtName ) group by Message,EventTime order by EventTime desc
select * from bdcdj.lqentry1 a where 顺序号 in (select max(顺序号) from bdcdj.lqentry1 b WHERE b.archival_code IS NOT NULL group by archival_code): 通过archival_code分组 ,取顺序号的最大值.
MySQL中GROUP BY分组取前N条记录实现 mysql分组,取记录 GROUP BY之后如何取每组的前两位下面我来讲述mysql中GROUP BY分组取前N条记录实现方法. 这是测试表(也不知道怎么想的,当时表名直接敲了个aa,汗~~~~): 结果: 方法一: SELECT a.id,a.SName,a.ClsNo,a.Score FROM aa a LEFT JOIN aa b ON a.ClsNo=b.ClsNo AND a.Score<b.Score group by a.id,a.
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!
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