4

我有一个简化表xx如下:

rdate  date
rtime  time
rid    integer
rsub   integer
rval   integer
primary key on (rdate,rtime,rid,rsub)

我想获得值的总和(所有 id)的平均值(所有时间)。

通过示例表,我有(为了便于阅读,连续的相同值被空白):

rdate       rtime     rid  rsub  rval
-------------------------------------
2010-01-01  00.00.00    1     1    10
                              2    20
                        2     1    30
                              2    40
            01.00.00    1     1    50
                              2    60
                        2     1    70
                              2    80
            02.00.00    1     1    90
                              2   100
2010-01-02  00.00.00    1     1   999

我可以得到我想要的总和:

select rdate,rtime,rid, sum(rval) as rsum
from xx
where rdate = '2010-01-01'
group by rdate,rtime,rid

这给了我:

rdate       rtime     rid  rsum
-------------------------------
2010-01-01  00.00.00    1    30  (10+20)
                        2    70  (30+40)
            01.00.00    1   110  (50+60)
                        2   150  (70+80)
            02.00.00    1   190  (90+100)

正如预期的那样。

现在我想要的是查询也将在时间维度上平均这些值,给我:

rdate       rtime    ravgsum
----------------------------
2010-01-01  00.00.00      50  ((30+70)/2)
            01.00.00     130  ((110+150)/2)
            02.00.00     190  ((190)/1)

我正在使用 DB2 for z/OS,但如果可能的话,我更喜欢标准 SQL。

4

2 回答 2

4
select rdate,rtime,avg(rsum) as ravgsum from (
    select rdate,rtime,rid, sum(rval) as rsum
    from xx
    where rdate = '2010-01-01'
    group by rdate,rtime,rid
) as subq
group by rdate,rtime
于 2010-06-10T03:34:11.040 回答
1

怎么样

select rdate,rtime, sum(rsum) / count(rsum) as sumavg
from
(select rdate, rtime, rid, sum(rval) as rsum
from xx
where rdate = '2010-01-01'
group by rdate,rtime,rid) as subq
group by rdate,rtime
于 2010-06-10T03:35:16.377 回答