生成用于在 TSQL 中递增日期的结果集并使用最后一个已知值作为数量列

问题描述

使用开始日期和结束日期，我们需要生成一个天数列表以及每天的手头数量余额.当一天的手头数量记录不存在时，结果集应使用手头数量的最新已知值.

Using start and end dates, we need to generate a list of days in between with the quantity on hand balance for each day. When the quantity on hand record for a day does not exist, the result set should use the most recent known value for the quantity on hand.

手头上.

例如，使用此数据作为我的 qoh 表

for example, using this data as my qoh table

create table #t1
(postdate date,
qoh int)

insert #t1 select '1/1/2014', 10
insert #t1 select '1/5/2014', 30
insert #t1 select '1/9/2014', 50
insert #t1 select '1/11/2014', 60

我想要选择的结果是

2014-01-01  10
2014-01-02  10
2014-01-03  10
2014-01-04  10
2014-01-05  30
2014-01-06  30
2014-01-07  30
2014-01-08  30
2014-01-09  50
2014-01-10  50
2014-01-11  60

我已经试过了

WITH dates AS
(
    SELECT CAST('1/1/2014' AS DATE) 'date'
    UNION ALL
    SELECT DATEADD(day, 1, t.date)
    FROM dates t
    WHERE DATEADD(dd, 1, t.date) <= '3/1/2014'
)

SELECT dates.date,
    (SELECT TOP 1 qoh FROM #t1
     WHERE #t1.postdate = dates.date
     ORDER BY date ASC)
FROM dates

返回这些结果.我想用最后一个已知值替换 NULLS.

which returns these results. I want to replace the NULLS with with last known values.

date    (No column name)
2014-01-01  10
2014-01-02  NULL
2014-01-03  NULL
2014-01-04  NULL
2014-01-05  30
2014-01-06  NULL
2014-01-07  NULL
2014-01-08  NULL
2014-01-09  50
2014-01-10  NULL
2014-01-11  60

推荐答案

只需稍作调整即可工作:

It works with just minor adjustments:

WITH dates AS
(
    SELECT CAST('20140101' AS DATE) 'date'
 UNION ALL
    SELECT DATEADD(day, 1, D.date)
    FROM dates D
    WHERE DATEADD(dd, 1, D.date) <= '20140301'
)
SELECT
    D.date
    ,(  SELECT TOP 1 qoh
        FROM #t1 T
        WHERE T.postdate <= D.[date]
        ORDER BY T.postdate DESC
    )
FROM
    dates D

返回

2014-01-01  10
2014-01-02  10
2014-01-03  10
2014-01-04  10
2014-01-05  30
2014-01-06  30
2014-01-07  30
2014-01-08  30
2014-01-09  50
2014-01-10  50
2014-01-11  60
2014-01-12  60
2014-01-13  60
... ...

http://sqlfiddle.com/#!6/79578/1

只是一个旁注:我更喜欢使用日历表(如果可能的话).只需在接下来的十年中每天投入，它仍然是一张非常小的桌子，您可以加入它以每天返回一行.它非常方便，查询也比这种递归更容易阅读.

Just a sidenote: I prefer to use a Calendar-table (if possible). Just throw in each day for the next ten years and it's still a very small table and you can join on it to return one row per day. It's quite convenient and the queries are easier to read than such recursion.

设置日历表:

CREATE table tblCalendar ([date] date PRIMARY KEY);

DECLARE @n int;
SET @n = 0;

WHILE @n < (365*5)
BEGIN
  INSERT INTO tblCalendar([date]) VALUES(DATEADD(day, @n, '20140101'));
  SET @n = @n +1;
END

减少查询:

SELECT
    C.[date]
    ,(  SELECT TOP 1 qoh
        FROM @t1 T1
        WHERE T1.postdate <= C.[date]
        ORDER BY T1.postdate DESC
    )
FROM
    tblCalendar C
WHERE
    C.date BETWEEN '20140101' AND '20140301'

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