日期:2014-05-17  浏览次数:21018 次

请教Oracle窗口函数求当前时间向前推30天的平均值?急急!
如题。数据大致如下:
  日期 数值
20121129 2
20121129 1
20121129 3
20121130 4
20121130 3
20121130 1
20121130 2
20121201 3
20121201 2
20121201 1
20121202 3
20121202 4
20121202 2
20121202 1
20121211 3
20121211 2
20121211 1
20121212 1
20121212 2
20121212 3
20121212 4
使用窗口函数写法如下:
SELECT tef.日期,AVG(SUM(tef.数值)) OVER (PARTITION BY tef.日期 ORDER BY TRUNC(TO_DATE(tef.日期,'yyyy-mm-dd'))
RANGE BETWEEN INTERVAL '30' DAY preceding AND INTERVAL '0' DAY following ) tstst
 FROM 表 tef 
GROUP BY tef.日期
;
使用这种写法向前30天由于日期不连续,只能获取连续日期的数据,比如20121212只能汇总20121211+20121212的数据无法汇总到20121202的数据,请教如何修改语句获取连续日期的数据汇总满足需求?

------解决方案--------------------
探讨

测试数据:
SQL code

CREATE TABLE T168
(
MyDate DATE,
MyNum NUMBER(4)
);
INSERT INTO T168 VALUES(to_date('20121129', 'YYYYMMDD'), 1);
INSERT INTO T168 VALUES(to_date('20121129', 'YYYYMMDD'), 2)……

------解决方案--------------------
比较好懂的方法 获取前30天每天的平均值
SQL code


with t1 as 
(
select '1' 编号,'20121127' 日期,'2' 数值 from dual union all
select '2',  '20121128','1' from dual union all
select '3',  '20121129','3' from dual union all
select '4',  '20121130','4' from dual union all
select '5',  '20121130','3' from dual union all
select '6',  '20121130','1' from dual union all
select '7',  '20121130','2' from dual union all
select '8',  '20121201','3' from dual union all
select '9',  '20121201','2' from dual union all
select '10', '20121202','2' from dual union all
select '11', '20121203','2' from dual union all
select '12', '20121204','2' from dual union all
select '13', '20121205','2' from dual union all
select '14', '20121206','4' from dual union all
select '15', '20121207','3' from dual union all
select '16', '20121208','6' from dual union all
select '17', '20121209','7' from dual union all
select '18', '20121210','3' from dual union all
select '19', '20121211','5' from dual union all
select '20', '20121212','2' from dual union all
select '21', '20121213','2' from dual union all
select '22', '20121214','2' from dual union all
select '23', '20121215','2' from dual union all
select '24', '20121216','2' from dual union all
select '25', '20121217','2' from dual union all
select '26', '20121218','2' from dual union all
select '27', '20121219','2' from dual union all
select '28', '20121220','2' from dual union all
select '29', '20121221','2' from dual union all
select '30', '20121222','2' from dual union all
select '31', '20121223','2' from dual union all
select '32', '20121224','2' from dual union all
select '33', '20121225','2' from dual union all
select '34', '20121226','2' from dual union all
select '35', '20121227','2' from dual union all
select '36', '20121228','2' from dual union all
select '37', '20121228','2' from dual union all
select '38', '20121228','2' from dual union all
select '39', '20121228','2' from dual
)

select t1.日期,avg(sum(t1.数值)/count(t1.数值)) over(partition by t1.日期 order by t1.日期) a_count from t1,
(
    select 日期,rownum from 
    (
        select distinct 日期
        from t1
        order by 日期
    )
    where rownum<=30
) t2
where t1.日期 in t2.日期
group by t1.日期
order by t1.日期

        日期      a_count
-------------------------------
1    20121127    2
2    20121128    1
3    20121129    3
4    20121130    2.5
5    20121201    2.5
6    20121202    2
7    20121203    2
8    20121204    2
9    20121205    2
10    20121206    4
11    20121207    3
12    20121208    6
13    20121209    7
14    20121210    3
15    20121211    5
16    20121212    2
17    20121213    2
18    20121214    2
19    20121215    2
20    20121216    2
21    20121217    2
22    20121218    2
23    20121219    2
24    20121220    2
25    2012122