这个特殊案例是从一个例子中提炼出来的，程序员假设两辆货物装入油罐车，首先装载＃1号线。我更正了这一点以允许以任何顺序执行加载 - 但是，我发现MIN() OVER (PARTITION BY)允许Oracle中的ORDER BY（SQL Server中不允许这样做），此外，它还会更改该函数的行为导致ORDER BY显然被添加到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
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：Oracle在做什么以及为什么这是对的？
答案 0 :(得分：11)
如果您向ORDER BY分析函数添加MIN，则将其转换为“最小到目前为止”的函数，而不是整体最小值。对于您要分区的最后一行，结果将是相同的。但是，之前的行可能具有与总体最小值不同的“最小值”。
使用EMP表作为示例，您可以看到该部门到目前为止的最低工资最终会收敛到该部门的总体最低工资。并且您可以看到，任何给定部门的“最小到目前为止”值会随着遇到较低值而降低。
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，当你尝试做一些像计算个人最好的东西时，使用这种形式的分析函数会更有意义，你可以将它用作未来时期的比较。如果你追踪一个人的高尔夫球得分，英里时间或体重正在下降，那么展示个人成绩可能是一种动力。
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.