SQL 声明 return 给定月份每篇文章的最小和最大库存量

Question

我有一个 table，我试图从中 return 文章在系统中的每天数量。

示例在 table Bestand 中，有多个不同物品的托盘，每个托盘都有一个预订日期和预订日期；我试图找出系统中每篇文章和每个月的最小和最大库存量。

我的想法是，如果我可以return每天的库存数量，然后读出最小值和最大值。

时间跨度将设置在运行和 SQL 的时间，文章将被固定。

为了找出每天的数量，我使用了以下 SQL：

SELECT DISTINCT 
    a.artbez1 AS Artikelbezeichnung, 
    b.artikelnr AS Artikelnummer,
    SUM(CASE WHEN TO_DATE('2019-11-01 00:00:00', 'YYYY-MM-DD HH24:MI:SS') BETWEEN b.neu_datum AND b.aender_datum THEN 1 * b.menge_ist ELSE 0 END) AS "01 Nov 2019"
FROM
    artikel a, bestand b
WHERE 
    b.artikelnr IN ('273632002', .... (huge long list of numbers) ....)
    AND b.artikelnr = a.artikelnr
GROUP BY 
    a.artbez1, b.artikelnr;

这个return例如：

ARTIKELBEZEICHNUNG	ARTIKELNUMMER	01 Nov 2019
SC-4400.CW	220450002	39
S-320.FK120	220502004	0
H-595.FK120	220800004	35
AC-548.FK209	220948032	0
AS-6800.CW	221355002	20

我想要 return 这个月的每一天，然后 return 每篇文章的最小值和最大值

我有以下 SQL 到 return 给定月份的天数，想知道是否有人对如何组合它们有任何想法（如果可能的话）：

SELECT to_date('01.11.2019','dd.mm.yyyy')+LEVEL-1
FROM dual 
CONNECT BY LEVEL <= TO_CHAR(LAST_DAY(to_date('01.11.2019','dd.mm.yyyy')),'DD')

DATES
2019-11-01 00:00:00
2019-11-02 00:00:00
2019-11-03 00:00:00
2019-11-04 00:00:00
2019-11-05 00:00:00
2019-11-06 00:00:00
2019-11-07 00:00:00

我尝试得到的结果类似于：

ARTIKELBEZEICHNUNG	ARTIKELNUMMER	Nov 19 Min	Nov 19 Max
SC-4400.CW	220450002	5	39
S-320.FK120	220502004	0	15
H-595.FK120	220800004	2	35
AC-548.FK209	220948032	0	0
AS-6800.CW	221355002	10	20

这在 SQL 中完全可行吗？

感谢您花时间阅读我的post。

杰日

Answer 1

您可以使用分区外连接：

WITH calendar ( day ) AS (
  SELECT DATE '2019-11-01'
  FROM   DUAL
UNION ALL
  SELECT day + INTERVAL '1' DAY
  FROM   calendar
  WHERE  day < LAST_DAY( DATE '2019-11-01' )
),
daily_totals ( artbez1, Artikelnr, Day, total_menge_ist ) AS (
  SELECT MAX( ab.artbez1 ), 
         ab.artikelnr,
         c.day,
         COALESCE( SUM( ab.menge_ist ), 0 )
  FROM   calendar c
         LEFT OUTER JOIN 
         ( SELECT a.artikelnr,
                  a.artbez1,
                  b.neu_datum,
                  b.aender_datum,
                  b.menge_ist
           FROM   artikel a
                  LEFT JOIN bestand b
                  ON ( a.artikelnr = b.artikelnr )
           -- WHERE  b.artikelnr IN ('273632002', .... (huge long list of numbers) ....)
         ) ab
         PARTITION BY ( ab.artikelnr, ab.artbez1 )
         ON ( c.day BETWEEN ab.neu_datum AND ab.aender_datum )
  GROUP BY ab.artikelnr, c.day
)
SELECT MAX( artbez1 ) AS Artikelbezeichnung,
       artikelnr AS Artikelnummer,
       TRUNC( day, 'MM' ) AS month,
       MIN( total_menge_ist ) AS min_total_menge_ist,
       MAX( total_menge_ist ) AS max_total_menge_ist
FROM   daily_totals
GROUP BY artikelnr, TRUNC( day, 'MM' );

其中，对于示例数据：

CREATE TABLE artikel ( artikelnr, artbez1 ) AS
SELECT 220450002, 'SC-4400.CW'   FROM DUAL UNION ALL
SELECT 220502004, 'S-320.FK120'  FROM DUAL UNION ALL
SELECT 220800004, 'H-595.FK120'  FROM DUAL UNION ALL
SELECT 220948032, 'AC-548.FK209' FROM DUAL UNION ALL
SELECT 221355002, 'AS-6800.CW'   FROM DUAL;

CREATE TABLE bestand ( artikelnr, neu_datum, aender_datum, menge_ist ) AS
SELECT 220450002, DATE '2019-10-30', DATE '2019-11-01', 20 FROM DUAL UNION ALL
SELECT 220450002, DATE '2019-11-01', DATE '2019-11-05', 19 FROM DUAL UNION ALL
SELECT 220502004, DATE '2019-11-05', DATE '2019-11-03',  5 FROM DUAL UNION ALL
SELECT 220800004, DATE '2019-11-01', DATE '2019-11-15', 35 FROM DUAL UNION ALL
SELECT 221355002, DATE '2019-10-20', DATE '2019-11-05',  5 FROM DUAL UNION ALL
SELECT 221355002, DATE '2019-10-25', DATE '2019-11-10',  5 FROM DUAL UNION ALL
SELECT 221355002, DATE '2019-10-28', DATE '2019-11-13',  5 FROM DUAL UNION ALL
SELECT 221355002, DATE '2019-10-30', DATE '2019-11-15',  5 FROM DUAL UNION ALL
SELECT 221355002, DATE '2019-11-05', DATE '2019-11-20',  5 FROM DUAL;

输出：

ARTIKELBEZEICHNUNG | ARTIKELNUMMER | MONTH               | MIN_TOTAL_MENGE_IST | MAX_TOTAL_MENGE_IST
:----------------- | ------------: | :------------------ | ------------------: | ------------------:
SC-4400.CW         |     220450002 | 2019-11-01 00:00:00 |                   0 |                  39
S-320.FK120        |     220502004 | 2019-11-01 00:00:00 |                   0 |                   0
AC-548.FK209       |     220948032 | 2019-11-01 00:00:00 |                   0 |                   0
H-595.FK120        |     220800004 | 2019-11-01 00:00:00 |                   0 |                  35
AS-6800.CW         |     221355002 | 2019-11-01 00:00:00 |                   0 |                  25

db<>fiddle here

SQL 声明 return 给定月份每篇文章的最小和最大库存量

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