有一个数据表,id user_id score 三个字段,计算总成绩最高的前两名 SELECT * FROM (SELECT user_name,SUM(score) AS score FROM user GROUP BY user_name) AS t ORDER BY t.score DESC LIMIT 2
实际应用中经常需要统计单位时间的记录数: 一个小时: SELECT ),createdate,) AS 日期 ,DATEPART(hh,createdate) AS 小时,COUNT(*)AS 记录 --varchar(10)得到结果为日期('2016-01-01') FROM log --或者用DATE格式提取日期(VARCHAR(10)换成DATE) where createdate>=CAST(CONVERT(VARCHAR(10),createdate,120))+' 00:00:00'
尊重原创:http://blog.csdn.net/love_java_cc/article/details/78268326 有如下一张表score: 建表语句: CREATE TABLE `score` (`id` int NOT NULL AUTO_INCREMENT ,`name` varchar(20) NOT NULL ,`category` varchar(20) NOT NULL ,`score` int NOT NULL ); 现在要去查询学生的各科成绩和总分,效果如下
按用户注册时间统计每天注册的不同来源.不同状态的用户数量: ), RegisterTime, ) RDate ,--DATEPART(YEAR, RegisterTime) 年 ) END 'AWaitForAudit' , ) END 'AAudited' , ) END 'BAudited' INTO #Temp FROM dbo.Customer WHERE RegisterTime BETWEEN '2015-11-23 00:00:00' AND '2015-11-25 23:59:5
--语法形式: ROW_NUMBER() OVER(PARTITION BY COL1 ORDER BY COL2) --解释: 根据COL1分组,在分组内部根据 COL2排序,而此函数计算的值就表示每组内部排序后的顺序编号(组内连续的唯一的) --常用的使用场景: 取每个学科的前3名 --1.创建测试表 create table #score ( name varchar(20), subject varchar(20), score int ) --2.插入测试数据 insert into
select sClass 班级,count(*) 班级学生总人数, sum(case when sGender=0 then 1 else 0 end) 女生人数, sum(case when sGender=0 then 1 else 0 end)*1.0/count(*)女生所占比例, sum(case when sGender=1 then 1 else 0 end) 男生人数, sum(case when sGender=1 then 1 else 0 end)*1.0 /coun
select NVL(sum(case when create_date_time>=to_date('2014-11-24 00:00:00','yyyy-mm-dd hh24:mi:ss') and create_date_time<=to_date('2014-11-24 23:59:59','yyyy-mm-dd hh24:mi:ss') then amount end),0) today ,NVL(sum(case when create_date_time>=to_date(
) FROM ((SELECT BaseID FROM dbo.Life_cheliang WHERE BaseCreateDate BETWEEN '2015-6-5' AND '2015-6-11') UNION (SELECT BaseID FROM dbo.Life_fangchan WHERE BaseCreateDate BETWEEN '2015-6-5' AND '2015-6-11') UNION (SELECT BaseID FROM dbo.Life_tiaozao WHE
sql在不同数据库查询前几条数据 1. ORACLE SELECT * FROM TABLE_NAME WHERE ROWNUM <= N; HQL: from table_name t order by t.create_time desc where rownum<=n; 2. postgres查询前几条记录SQL SELECT * FROM TABLE_NAME LIMIT N; 3. DB2 SELECT * ROW_NUMBER() OVER(ORDER BY COL1 DESC