这个特殊情况是从一个示例中提炼出来的,其中程序员假设对于两次装运到 jar 车的 cargo ,第 1 行将首先装载。我更正了这一点,以允许按任何顺序执行加载 - 但是,我发现 MIN() OVER (PARTITION BY)
允许在 Oracle 中使用 ORDER BY
(这SQL Server 中不允许),此外,它还改变了函数的行为,导致 ORDER BY 显然被添加到 PARTITION BY 中。
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(已注释掉不适用的代码):
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,为什么这样做,这样做是否正确?
最佳答案
如果将 ORDER BY
添加到 MIN
分析函数,则会将其转换为“到目前为止的最小值”函数,而不是总体最小值。对于您分区所依据的最后一行,结果将是相同的。但前面的行可能具有与总体最小值不同的“到目前为止最小值”。
以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“min so far”值都会随着遇到较低值而减小。
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佳成绩之类的操作以便在未来期间进行比较时,使用这种形式的分析函数会更有意义。如果您正在跟踪个人高尔夫成绩、英里时间或体重的下降情况,那么展示个人最佳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.
关于sql - Oracle MIN 作为分析函数 - ORDER BY 的奇怪行为?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/7393974/