需求:查询各个设备存在未消缺的缺陷,取缺陷等级最高的,作为这个设备当前的缺陷等级: 数据源: select t.device_id, t.defect_level from sp_pd_defect t ', ', ', ', ') group by t.device_id, t.defect_level order by t.device_id, t.defect_level asc; 如图:     要求: 需求获取红色部分的内容,,那么对应的SQL语句为 select device_id

-- 根据编号分组取第一条数据 select * from table t where t.no=(select max(no) from table t1 where t1.no=t.no) -- 根据编号分组后取第一条数据 SELECT * FROM (SELECT ROW_NUMBER() OVER (partition BY no ORDER BY no) rowId,* from table) t WHERE rowId=1

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

resultset 对象获取行字段数据时报:java.sql.SQLException: Column 'id' not found. 代码: String sql="SELECT d.content,c.name AS categoryName FROM news_detail d,news_category c WHERE d.categoryId=c.id"; Object[] params ={}; System.out.println(this.executeQuery(sq

--有id,name,createDate的一张表testTable--根据name分组,获取每组中createDate最大的那条记录(整条)查询出来---------------------------------------------- 创建一张表,语句如下: CREATE TABLE [dbo].[testTable] ( [id] [int] NOT NULL IDENTITY(1, 1), [name] [nchar] (10) COLLATE Chinese_PRC_CI_AS N

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

数据格式: 分组取第一条的效果: sql语句: SELECT * FROM ( ;

‘数据格式 分组取第一条的效果 sql SELECT * FROM (SELECT ROW_NUMBER() OVER(PARTITION BY x ORDER BY y DESC) rn, test1.* FROM test1) WHERE rn = 1 ;

select gr,num,dt,(select bys from test where gr=b.gr and dt=b.dt) bys from ( select gr,count(0) num,max(dt) dt from test group by gr ) b //如果有重复项,可用如下语句(针对Mysql的limit,Oracle 可用 rownum<2) select gr,num,dt,(select bys from test where gr=b.gr and dt=b.d

SELECT * FROM( SELECT id,name,counts,createDate,row_number() OVER(partition BY name ORDER BY createDate DESC) rowno FROM testTable ) t

大家都知道用group by的话,select 后面指定的字段必须与group by后面的一致.group by 只有个别字段,如果拿出其他未分组的字段信息呢?在网上搜了下, 总结如下: 使用了group by 之后,就要求select后面的字段包含在group by 或聚合函数里面,这时如果想读取其它字段则无法实现. 将你需要的字段放进max或min函数中,max:支持字符类型.数字类型.select max(id) as id,username,password from usersgrou

使用Group By来实现取最新记录,需要注意一个问题,如果最大时间相同的数据都会被取出来. PS:即使数据字段类型是timestamp,也会登录相同的时间的数据. select A.* from A inner join ( select C,Max(Time) from A group by C ) B on A.C = B.C and A.Time = B.Time

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!

MySQL语句: SELECT GROUP_CONCAT(DISTINCT transaction_no) FROM `lm_wh_trans` GROUP BY staff_code; 如果transaction_no 在staff_code 分组下有多个不同的值的时候,将会将不同的值进行拼接上去

SELECT * FROM (SELECT ROW_NUMBER() OVER(PARTITION BY cc.queuename ORDER BY cc.enroldate DESC) rn, cc.* FROM (select * from (select a.patname,a.queuesign,a.queuename,a.status,a.checkroom,a.areapart,a.enroldate from qs_register a where a.status='就诊中' g

场景:sql server 2008 drop table ID CREATE TABLE ID ( id ,) not null, code int , D date, PRIMARY KEY (id) ) ,getdate()) ,getdate()) ,getdate()) ,getdate()) ,'2017-08-02') ,'2017-08-01') select * from ID 目标: select COUNT(*) from ID group by code 产生code列唯

查询 每个班级英语成绩最高的前两名的记录 原文:https://www.cnblogs.com/hxfcodelife/p/10226934.html select a.Classid,a.English from (select Classid,English,row_number() over(partition by Classid order by English desc) as n from CJ) a where n<=2 另外一种情况:取组内最新的数据select max(cre

分享一个朋友的人工智能教程.零基础!通俗易懂!风趣幽默!大家可以看看是否对自己有帮助,点击查看教程. 比如查询用户某一天最后一笔交易后的账户余额 SELECT *( SELECT *, row_number () OVER ( partition BY username ORDER BY tradeTime DESC ) rowid FROMtable WHEREtradeDate = '2017-08-31' ) t WHERE rowid = 1;

背景:  A表.B表两表关联,关联出来的结果里B表有不止一条,需求是只要B表结果中的某一条(按某字段排序) 经过百度,发现 row_number() over(partition by a order by b desc)函数可用(需要说明下,order by 必须有缺少会报错),以下是数据环境及结果. 创建数据环境 create table A(ANAME varchar(20)); insert into A values('alan'); insert into A values('Ale

SELECT uploaddate ,ptnumber ,instcount FROM ( SELECT ROW_NUMBER() OVER( PARTITION BY uploaddate ORDER BY uploaddate DESC ,instcount DESC ) AS row ,* FROM ( SELECT uploaddate ,ptnumber ,COUNT(*) instcount FROM LtblLog.dbo.instlog201408 GROUP BY upload

oracle分组后取每组第一条数据   '数据格式     分组取第一条的效果   [sql] SELECT * FROM (SELECT ROW_NUMBER() OVER(PARTITION BY x ORDER BY y DESC) rn, test1.* FROM test1) WHERE rn = 1 ;