Oracle 10g：如何按日期差异聚合答案

【问题标题】：Oracle 10g: howto aggregate by date differencesOracle 10g：如何按日期差异聚合
【发布时间】：2015-02-02 12:43:35
【问题描述】：

您有以下数据：

KEY                 VALUE               TIMESTAMP
-------------- ---------- -----------------------
0F8CE962              900          20141124054503
0F8CE962              900          20141124082431
0F8CE962                0          20141124083808
0F8CE962                0          20141124104408
0F8CE962                0          20141124105009
0F8CE962                0          20141124110213
0F8CE962              900          20141124110720
0F8CE962              900          20141125051641
0F8CE962                0          20141125054112

每一行都与接下来的大约 15 分钟“相距”，最后两行除外。如果我执行：

select KEY, sum(VALUE), min(TIMESTAMP), max(TIMESTAMP)
from myTable
group by KEY

我明白了（当然）

KEY            sum(VALUE)         min(TIMESTAMP)          max(TIMESTAMP)
-------------- ---------- ----------------------- -----------------------
0F8CE962             3600          20141124054503          20141125054112

我需要的是只聚合不同的行，一旦订购，最多 15 分钟。这就是我想要的：

select KEY, sum(VALUE), min(TIMESTAMP), max(TIMESTAMP)
from myTable
group by KEY
some_magic_function(max(15 minutes))

KEY            sum(VALUE)         min(TIMESTAMP)          max(TIMESTAMP)
-------------- ---------- ----------------------- -----------------------
0F8CE962             2700          20141124054503          20141124110720
0F8CE962              900          20141125051641          20141125054112

有可能吗？

【问题讨论】：

TIMESTAMP的格式是什么？如果是 YYYYMMDDHH24MISS 则 05:45:03
是的，它是 YYYYMMDDHH24miss。请注意最后两行是 11/25，前两行是 11/24 :-)

标签： sql oracle column-aggregation

【解决方案1】：

您可以使用其他地方演示的“组开始”方法。应用于您的示例并假设时间戳是或可以转换为数值：

with mytable1 as
  (select mytable.*
        , case
            when lag(timestamp, 1, timestamp-150001) over
                   (partition by key order by timestamp) < timestamp-150000
            then 1
            else 0
          end start_of_group
    from mytable)
, mytable2 as
  (select mytable1.*
        , sum(start_of_group) over (partition by key order by timestamp) grp
   from  mytable1)
select key
     , sum(value)
     , min(timestamp)
     , max(timestamp)
from mytable2
group by key
       , grp
order by key
       , min(timestamp)

【讨论】：