select * from (select row_number() over(partition by id order by time desc) rn, a.* from table a) where rn = 1以上。

--用外链接把AA表和CC表链接起来查询就可以了SELECT AA.CODE, AA.NUM, AA.PRICE, CC.DATETIME, CC.SUMNUM FROM test AA --下面的BB表查询的是根据code分组查询num求和及最大datetime，之后作为一个表CC LEFT JOIN (SELECT BB.CODE, SUM(NUM) SUMNUM,MAX(BB.DATETIME) DATETIME FROM ...

where 条件后加 GROUP BY MONTHOFYEAR (WEEK STARING)

decode(tl, '00:00——00:30', to_char(a-1,'yyyymmdd'), to_char(a,'yyyymmdd')) as 天,decode(tl, '00:00——00:30', '16:31——00:30', '16:31——00:00', '16:31——00:30', t1) as 时间段,sum(b)from (select a,case when to_char(a, 'hhmi') <='0030' ...

oracle进行分组后，取出每组的前几条数据SELECT *FROM (SELECT 分组的字段名,ROW_NUMBER() OVER(PARTITION BY 分组的字段名 ORDER BY 排序的字段名) AS RNFROM 表名)WHERE RN <= 10�0�2

---补充--- 问了你半天也不答，按startdate的时间为准了 --建表如下 create table test(startdate date,enddate date,autualdate date,"desc" varchar2(10),"value" int)--插入数据 insert into test values (to_date('2013-05-17','yyyy-mm-dd'),to_date('2013-05-17','yyyy-mm-d...

oracle的date时间类型是 '2011-7-30 14:31'这种形式 因此就算你用extract()或者to_char()截取时间段也没有没办法区分周 我建议的方法是新建一个字段week，从时间里截取月和日去计算周，可以从每月的第一天00:00开始算，完成7天算一周。也就是说这个字段week下要有至少4条记录，分别是每一个月的...

'yyyy-mm-dd') FROM dual UNION SELECT to_date('2016-01-27','yyyy-mm-dd') FROM dual UNION SELECT to_date('2015-12-27','yyyy-mm-dd') FROM dual UNION SELECT to_date('2016-11-29','yyyy-mm-dd') FROM dual UNION SELECT to_date('2015-12-26','yyyy-mm-dd...

T2.*, ROW_NUMBER()OVER(PARTITION BY MAE_NAME ORDER BY TIME_STAMP_END DESC) AS RN2 FROM (SELECT T1.*,ROW_NUMBER() OVER(PARTITION BY MAE_NAME, PRODUCT, OP, VERSION ORDER BY TIME_STAMP_END DESC) AS RN1 FROM CONDITION_CSV T1) WHERE RN1 = 1) T2 WHERE RN2 < 11 ...

告诉你个高级点的方法，用开窗函数 select * from (select id,fid,time,name,rank()over(partition by fid order by time desc) mm from a )where mm=1