不啰嗦,直接上图,大概实现效果如下: 有上面这样一份数据,将他们按照userAccount和submitTime进行分组,然后提前每组数据的前两条记录 提取后数据如下: 实现的SQL如下: select t.* from (select *,row_number() over(partition by userAccount, submitTime order by submitTime) rn from demoTable) t @_@! 结束啦~~
一.对分组的记录取前N条记录:例子:取前 2条最大(小)的记录 .用子查询: SELECT * FROM right2 a WHERE > (SELECT COUNT(*) FROM right2 b WHERE b.id=a.id AND b.account>a.account) ORDER BY a.id,a.account DESC .用exists半连接: SELECT * FROM right2 a WHERE EXISTS (SELECT COUNT(*) FROM right2
表结构如下: id varchar(32) info json 数据: id = info = {"age": "18","disname":"小明"} -------------------------------------------- 现在我需要获取info中disanme的值,查询方法有: 1. 结果: id = 1, disname=“小明” 以上sql查出的disname值是带有双引号的,有
测试表如下: /* Navicat Premium Data Transfer Source Server : Source Server Type : MySQL Source Server Version : 50717 Source Host : localhost:3306 Source Schema : ceshi Target Server Type : MySQL Target Server Version : 50717 File Encoding : 65001 Date: 1
按分组排序,并查出每个分组的前3名 单表 SELECT * FROM ( SELECT ZONEID, uid, NAME, fight, IF ( , ) AS rank, ( @zone := zoneid ) AS zone FROM `user` u, ( ) t, ( ) z ORDER BY zoneid DESC, fight DESC ) a WHERE rank <= 3 ORDER BY zoneid DESC, rank DESC
select * from (select *,row_number() over(partition by PropertyRoofBeamID order by PropertyRoomID desc) as Sequencecount from PropertyRoom)t1 where Sequencecount<=3 partition by 具有分组的作用row_number排序进行分组.
select * from ( select *, ROW_NUMBER() over(partition by IPAddress order by recordtime desc) as rowNum from MonitoringSystem ) A order by A.IPAddress, A.recordtime desc 如果表中的数据是以秒记录的 但是显示的时候想以分钟显示 怎么办 select * from MonitoringSystem where RecordTime
mysql分组取每组前几条记录(排名) 附group by与order by的研究,需要的朋友可以参考下 --按某一字段分组取最大(小)值所在行的数据 复制代码代码如下: /* 数据如下: name val memo a 2 a2(a的第二个值) a 1 a1--a的第一个值 a 3 a3:a的第三个值 b 1 b1--b的第一个值 b 3 b3:b的第三个值 b 2 b2b2b2b2 b 4 b4b4 b 5 b5b5b5b5b5 */ --创建表并插入数据: 复制代码代码如下: create