限制oracle中over语句内的数据

Question

我想根据时间戳聚合列。

举个例子：

Table 包含 col1、col2、...、col_ts（时间戳列）等列。

SELECT
SUM(col1) OVER (ORDER BY col_ts ROWS BETWEEN 2 PRECEDING AND 2 FOLLOWING) SUM1,
SUM(col2) OVER (ORDER BY col_ts ROWS BETWEEN 2 PRECEDING AND 2 FOLLOWING) SUM2
FROM ...

现在，当时间戳之间的差异 <= 5 分钟时，我只需要 2 PRECEDING 和 2 FOLLOWING ROWS SUMMED。

例如，让我们看一下这些时间戳值：

14.09.15 15:44:00
14.09.15 15:50:00
14.09.15 15:51:00
14.09.15 15:52:00
14.09.15 15:53:00

当我们位于时间戳值“14.09.15 15:51:00”的行时，我希望 SUM OVER 从 15:50 到 15:53 的值，因为两者之间的差异15:50 和 15:44 超过 5 分钟。

有没有办法在over子句中写这样的条件？

或者有人对此有好的、高效的解决方案吗？

Answer 1

我认为 sql 太多了。您可以限制 window 中的数量或元素，您可以以某种方式（见下文）限制值，但不能同时限制两者。

drop table fg_test;
create table fg_test(col_ts timestamp, n number);

insert into fg_test values (systimestamp, 1);
insert into fg_test values (systimestamp+4/1440/60, 2);
insert into fg_test values (systimestamp+5/1440/60, 3);
insert into fg_test values (systimestamp+6/1440/60, 4);
insert into fg_test values (systimestamp+7/1440/60, 5);
insert into fg_test values (systimestamp+13/1440/60, 6);

select col_ts, n, 
  SUM(n) OVER (ORDER BY col_ts ROWS BETWEEN 1 PRECEDING AND 1 FOLLOWING) SUM1,
  SUM(n) OVER (ORDER BY col_ts RANGE BETWEEN current row AND interval '5' second FOLLOWING) SUMNEW
from fg_test;

结果：

COL_TS                                   N       SUM1       SUM2
------------------------------- ---------- ---------- ----------
14-SEP-15 06.16.28.825395000 PM          1          3          3 
14-SEP-15 06.16.33.000000000 PM          2          6         14 
14-SEP-15 06.16.34.000000000 PM          3          9         12 
14-SEP-15 06.16.35.000000000 PM          4         12          9 
14-SEP-15 06.16.36.000000000 PM          5         15          5 
14-SEP-15 06.16.42.000000000 PM          6         11          6

（很抱歉没有按照你的问题举出确切的例子）

另一种方法是写一些PL/SQL（打开游标并进行一些处理）。

Answer 2

好的，我看到这里的问题了。谢谢弗罗林。那么一些预处理呢？我可以找到解决方案，但我不确定是否有更快的解决方案：

select col_ts, 
       n, 
       SUM(n) OVER (ORDER BY col_ts ROWS BETWEEN LEFT_VALUE PRECEDING AND RIGHT_VALUE FOLLOWING) MY_SUM,
       SUM(n) OVER (ORDER BY col_ts RANGE BETWEEN interval '5' second PRECEDING AND interval '5' second FOLLOWING) OLD_SUM
from (
       select col_ts,
              n,
              CASE
              WHEN (LEAD(col_ts,1) OVER (ORDER BY col_ts ) - col_ts) <= INTERVAL '5' second 
              THEN 
                   CASE
                   WHEN (LEAD(col_ts,2) OVER (ORDER BY col_ts ) - LEAD(col_ts,1) OVER (ORDER BY col_ts )) <= INTERVAL '5' second 
                   THEN 2 
                   ELSE 1
                   END
             ELSE 0
             END AS RIGHT_VALUE,
             CASE 
             WHEN (col_ts - LAG(col_ts,1) OVER (ORDER BY col_ts ) ) <= INTERVAL '5' second 
             THEN 
                  CASE 
                  WHEN (LAG(col_ts,1) OVER (ORDER BY col_ts ) - LAG(col_ts,2) OVER (ORDER BY col_ts )) <= INTERVAL '5' second 
                  THEN 2 
                  ELSE 1
                  END
            ELSE 0
            END AS LEFT_VALUE
      from fg_test
  );

结果：

COL_TS                           N   MY_SUM      OLD_SUM
---------------------------  -----  -------  -----------
15.09.15 09:34:24,069000000      1        6            6
15.09.15 09:34:28,000000000      2       10           15
15.09.15 09:34:29,000000000      3       15           15
15.09.15 09:34:30,000000000      4       14           14
15.09.15 09:34:31,000000000      5       12           14
15.09.15 09:34:37,000000000      6        6            6

你怎么看？

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