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orAClE 分组统计

CREATE TABLE info ( users varchar(100));INSERT INTO info VALUES('userA@userB@userC');INSERT INTO info VALUES('userB@userC@userD');INSERT INTO info VALUES('userC@userD@userE');COLUMN "用户" FORMAT A15SELECT to_char(strvalue) as ...

SELECT ord.ORDER_TYPE as 订单类型, COUNT(ord.ORDER_ID) as 订单总数, orde.succCount as 交易成功订单总数 FROM ORDER ord LEFT JOIN ( SELECT ORDER_TYPE, COUNT(ORDER_ID) AS succCount FROM ORDER WHERE STATUS_ID = 'ORDER_SUCCESS' GROU...

select count(*) from tb group by id,name

直接写sql的话我能想到的就是直接group by case when ALTER session SET nls_date_format = 'yyyy-mm-dd';WITH tmp_sleest AS ( SELECT to_date('2016-01-26','yyyy-mm-dd') AS my_date FROM dual UNION SELECT to_date('2016-02-25','yyyy-mm-d...

select * from (select a.*, row_number() over (partition by a.col_a order by create_ts desc) as rn from table1 a ) where rn = 1; -- table1 换成你的表, col_a 换成你的分组列, create_ts 是你的时间戳字段, 如果是字符串需转换为date

对于你的要求,应该是这样 select count(id) from 表 where id = '1' 但是如果你需要查询表id字段中所有数据出现的次数,那需要进行分组查询: select id,count(id) as 出现次数 from 表 group by id

oracle数据库的分组查询语句,主要是根据一个字段,使用关键字group by来分组,如下代码: select to_char(date_column, 'yyyy-Q'),count(*) from xxx where date_column between '01-Jan-2007' and '31-Dec-2009' group by to_char(date_column...

SELECT TO_CHAR(TO_DATE(20160316,'YYYYMMDD'),'W') FROM DUAL; TO_CHAR函数可以获取某一天是在该月中的第几周,然后可以按照这个函数来分组了

创建测试表 create table test(姓名 varchar2(10),访问时间 date,进入时间 date,离开时间 date); insert into test values ('张三',to_date('2013-02-03 15:23:22','yyyy-mm-dd hh24:mi:ss'),to_date('2013-02-03 15:23:22','yyyy-mm-dd hh24:mi...

条件 同样的姓名和客户号,有多个订单,支付的金额不等于退款的金额 执行结果如下 姓名 客户号 差异金额 JACK 001 40 执行语句参考 create table test_001 as select 'JACK' as 姓名 ,'001' AS 客户号 ,'10000' AS 订单号 ,'支付' AS 类型 ,100 ...

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