1.GROUP BY 与聚合函数 2.GROUP BY 与 HAVING 3.GROUP BY 扩展分组 3.1.GROUP BY ROLLUP 3.2.GROUP BY CUBE 3.3.GROUP BY GROUPING SETS 4.GROUP BY 扩展函数 4.1.GROUPING 函数 4.2.GROUPING_ID 函数 5.本文小结 1.GROUP BY 与聚合函数 GROUP BY 是一种能将查询结果划分为多个行组的查询语句的子句,其目的通常是为了在每个组上执行一个或多个聚合运…
方法一 select t1.a,t1.b,t1.c from test t1 inner join (seelct a,max(b) as b from test group by a) t2 on t1.a=t2.a and t1.b=t2.b 方法二 select * from (select t.*, row_number() over(partition by 分组字段 order by 排序字段 desc ) rnfrom tablename t )where rn=1 方法三 (不一…
继上篇<SQL 列转行 合并多条记录>后,有网友反馈新的需求还是不太会用. 现举例说明 一,网友需要如下的效果: 其实,这个需求依然可以我上篇的方法进行解答,但为了实现分组,需要distinct group1,同时,为了根据key1,key2,key3是否相同进行分组,所以要用where进行连接. 语法格式:select .....from t1 where key1=a.key1 and key2=a.key2... for xml path('') 二,进入正题,代码如下: select…
由于本人并未对oracle数据库进行深入了解,但是工作中又需要知道一些基础的sql,所以记录下操作的sql语句.方便日后查看 1.将序列号作为分组查询的条件,再将查询出来的结果进行筛选. select t.serial_number,count(t.serial_number) from stamp_info t group by t.serial_number ;…
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 ( select mp.MsgID,m.Content,m.CreatorID,m.CreateTime,ROW_NUMBER() over(partition by m.CreatorID order by m.CreateTime desc) as new_index from U_Account_WX_MsgProperty mp join U_Account_WX_Messages m on mp.MsgID=m.MsgID where mp.Us…
这是CSDN问答里面有人提出的一道问题,题目如下. 表格如下: 得出结果如下: 求精简的SQL语句. SQL查询语句: with a as( select rank() over (partition by 商店 order by 商店,sum(价钱) desc) as rowid, 商店,sum(价钱) as 价钱和,备注 from 产品 group by 备注,商店), b as( ), c as( select 商店,sum(价钱和) as 卖出价钱和 from a group by 商店…
select * from(select row_number() over(partition by IDCARD order by DATATM desc) as rownum,* from (SELECT * FROM TABLENAME)as H1 ) as H where H.rownum = 1 解释:红色为以什么分组 蓝色为以什么排序 紫色为表名 目前是获取表中每个 IDCARD中时间最大的一条…
$k=M('order a'); $bj=$k->join("left join __CHANGE__ b on b.tb_name='order'and a.order_id=b.tb_id ") ->join("left join __USERS__ c on c.user_id=a.user_id") ->join("left join __NATION__ e on e.code=a.city") ->join(&…
引言: 为什么会引入这个问题,在程序中遇到这样的问题,在某个数据表中,相同的AID(项目ID)被多次添加到数据表中,所以对应于不同的时间,只想选取添加时间最早的哪一条记录. 参考:红黑联盟 所用到的数据表: 想实现的效果: 解释:相同的ID,由不同的人(Name)添加,我们选取值最小(Val)的那一行记录. 方法如下: 方法一: select a.* from T1 a , (select id,min(val) as val from T1 group by id) b where a.ID=…
SELECT B.id , LEFT(tempname, LEN(tempname) - 1) AS name FROM ( SELECT id , ( SELECT name + ',' FROM tablename WHERE id = A.id FOR XML PATH('') ) AS tempname FROM tablename A GROUP BY id ) B…
with MiPriceTopOne as (select classid,max(dataTime) dataTime,max(id) as id from MiPrice group by classid)select * from MiPrice where id in (select id from MiPriceTopOne)…
语句1: SELECT TypeID, COUNT(*) AS [count] FROM GoodsInfo GROUP BY TypeID 得到结果 解析结果:GoodsInfo表有 4条记录, 分别统计每个TypeID 有多少: 例子中说明 TypeID=1 的有2个,TypeID=2 的有1个,TypeID=3 有1个 语句2: SELECT COUNT(*) AS [count] FROM ( SELECT TypeID FROM GoodsInfo GROUP BY Type…
select DATE_FORMAT(f.upload_time,'%Y%u') weeks,count(*),sum(p.download_times),sum(p.collection_times),sum(p.click_times) from file_base f left join file_property p on f.property_id = p.id group by weeks; select DATE_FORMAT(f.upload_time,'%Y-%m-%d') d…