Oracle MIN作为分析函数-ORDER BY的异常行为？

Question

这个特殊情况是从一个示例中提炼出来的，在该示例中，程序员假定两次装罐车，首先要装载1号线。我更正了此问题，以允许以任何顺序执行加载-
但是，我发现在Oracle中MIN() OVER (PARTITION BY)允许使用a ORDER BY（在SQL
Server中不允许这样做），此外，它更改了函数的行为，导致ORDER BY显然是已添加到中PARTITION BY。

WITH data AS (
SELECT 1 AS SHIPMENT_ID, 1 AS LINE_NUMBER, 2 AS TARE, 3 AS GROSS FROM DUAL
UNION ALL
SELECT 1 AS SHIPMENT_ID, 2 AS LINE_NUMBER, 1 AS TARE, 2 AS GROSS FROM DUAL
)
SELECT MIN(tare) OVER (PARTITION BY shipment_id) first_tare
,MAX(gross) OVER (PARTITION BY shipment_id) last_gross
,FIRST_VALUE(tare) OVER (PARTITION BY shipment_id ORDER BY LINE_NUMBER) first_tare_incorrect
,FIRST_VALUE(gross) OVER (PARTITION BY shipment_id ORDER BY LINE_NUMBER DESC) last_gross_incorrect
,MIN(tare) OVER (PARTITION BY shipment_id ORDER BY LINE_NUMBER) first_tare_incorrect_still
,MAX(gross) OVER (PARTITION BY shipment_id ORDER BY LINE_NUMBER DESC) last_gross_incorrect_still
,MIN(tare) OVER (PARTITION BY shipment_id, LINE_NUMBER) first_tare_incorrect_still2
,MAX(gross) OVER (PARTITION BY shipment_id, LINE_NUMBER) last_gross_incorrect_still2
 FROM data

一个SQL Server示例（注释掉了不适用的代码）：

WITH data AS (
SELECT 1 AS SHIPMENT_ID, 1 AS LINE_NUMBER, 2 AS TARE, 3 AS GROSS -- FROM DUAL
UNION ALL
SELECT 1 AS SHIPMENT_ID, 2 AS LINE_NUMBER, 1 AS TARE, 2 AS GROSS -- FROM DUAL
)
SELECT MIN(tare) OVER (PARTITION BY shipment_id) first_tare
,MAX(gross) OVER (PARTITION BY shipment_id) last_gross
-- ,FIRST_VALUE(tare) OVER (PARTITION BY shipment_id ORDER BY LINE_NUMBER) first_tare_incorrect
-- ,FIRST_VALUE(gross) OVER (PARTITION BY shipment_id ORDER BY LINE_NUMBER DESC) last_gross_incorrect
-- ,MIN(tare) OVER (PARTITION BY shipment_id ORDER BY LINE_NUMBER) first_tare_incorrect_still
-- ,MAX(gross) OVER (PARTITION BY shipment_id ORDER BY LINE_NUMBER DESC) last_gross_incorrect_still
,MIN(tare) OVER (PARTITION BY shipment_id, LINE_NUMBER) first_tare_incorrect_still2
,MAX(gross) OVER (PARTITION BY shipment_id, LINE_NUMBER) last_gross_incorrect_still2
 FROM data

那么问题来了：Oracle在做什么？为什么，对吗？

Answer 1

如果ORDER BY在MIN分析函数中添加，则将其变为“到目前为止的最小”函数，而不是总的最小值。对于您要进行分区的最后一行，结果将是相同的。但是，前面的行可能有一个不同于整体最小值的“到目前为止”。

以该EMP表为例，您可以看到到目前为止该部门的最低工资最终收敛于该部门的整体最低工资。您会看到，任何给定部门的“到目前为止”的最小值都会随着遇到更低的值而减小。

SQL> ed
Wrote file afiedt.buf

  1  select ename,
  2         deptno,
  3         sal,
  4         min(sal) over (partition by deptno order by ename) min_so_far,
  5         min(sal) over (partition by deptno) min_overall
  6    from emp
  7*  order by deptno, ename
SQL> /

ENAME          DEPTNO        SAL MIN_SO_FAR MIN_OVERALL
---------- ---------- ---------- ---------- -----------
CLARK              10       2450       2450        1300
KING               10       5000       2450        1300
MILLER             10       1300       1300        1300
ADAMS              20       1110       1110         800
FORD               20       3000       1110         800
JONES              20       2975       1110         800
SCOTT              20       3000       1110         800
smith              20        800        800         800
ALLEN              30       1600       1600         950
BLAKE              30       2850       1600         950
MARTIN             30       1250       1250         950
SM0                30        950        950         950
TURNER             30       1500        950         950
WARD               30       1250        950         950
BAR
PAV

16 rows selected.

当然，当您尝试进行诸如计算个人最佳成绩之类的事情以供将来比较时，使用这种形式的分析函数会更有意义。如果您要跟踪某人的高尔夫得分，英里数或体重的下降，则显示个人最佳成绩可能是一种激励。

SQL> ed
Wrote file afiedt.buf

  1  with golf_scores as
  2  (  select 1 golfer_id, 80 score, sysdate dt from dual union all
  3     select 1, 82, sysdate+1 dt from dual union all
  4     select 1, 72, sysdate+2 dt from dual union all
  5     select 1, 75, sysdate+3 dt from dual union all
  6     select 1, 71, sysdate+4 dt from dual union all
  7     select 2, 74, sysdate from dual )
  8  select golfer_id,
  9         score,
 10         dt,
 11         (case when score=personal_best
 12               then 'New personal best'
 13               else null
 14           end) msg
 15    from (
 16  select golfer_id,
 17         score,
 18         dt,
 19         min(score) over (partition by golfer_id
 20                              order by dt) personal_best
 21    from golf_scores
 22*        )
SQL> /

 GOLFER_ID      SCORE DT        MSG
---------- ---------- --------- -----------------
         1         80 12-SEP-11 New personal best
         1         82 13-SEP-11
         1         72 14-SEP-11 New personal best
         1         75 15-SEP-11
         1         71 16-SEP-11 New personal best
         2         74 12-SEP-11 New personal best

6 rows selected.