8

这个特殊情况是从一个例子中提炼出来的，在这个例子中，程序员假设将两批货物装入一辆油罐车，第 1 行将首先装载。我更正了这一点以允许以任何顺序执行加载 - 但是，我发现它MIN() OVER (PARTITION BY)允许ORDER BY在 Oracle 中（这在 SQL Server 中是不允许的），此外，它改变了函数的行为，导致ORDER BY显然是添加到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

所以问题是：甲骨文在做什么，为什么这样做，是否正确？

4

1 回答 1

12

如果ORDER BY向MIN分析函数添加一个，则将其转换为“到目前为止的最小值”函数，而不是整体最小值。对于您分区的最后一行，结果将是相同的。但是之前的行可能具有与整体最小值不同的“最小值”。

以该EMP表为例，您可以看到该部门迄今为止的最低工资最终收敛于该部门的总体最低工资。您可以看到任何给定部门的“到目前为止的最小值”值会随着遇到较低的值而减小。

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，当您尝试执行诸如计算个人最好成绩之类的事情时，使用这种形式的分析函数会更有意义，您可以将其用作未来时期的比较。如果您正在跟踪个人不断下降的高尔夫分数、英里时间或体重，那么展示个人最好成绩可能是一种激励形式。

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.
于 2011-09-12T21:09:16.487 回答