具有可变种子的DATEDIFF

问题描述 投票：2回答：3

我有下表

<< img src =“ https://image.soinside.com/eyJ1cmwiOiAiaHR0cHM6Ly9pLnN0YWNrLmltZ3VyLmNvbS9UOXRQVi5wbmcifQ==” alt =“在此处输入图像描述”>

第一列是按日期DESC排序的交易日期。第二列是每个交易的二进制状态，可以为0或1。第三列是最小日期与每个日期相比的datediff。产生该表的代码是这样的

DECLARE @date DATE 
SELECT @Date = MIN(CONVERT(DATE,Transaction_Created)) FROM #dates
SELECT      CONVERT(DATE,Transaction_Created)                       AS Date
    ,       MAX(Is_Deposit)                                         AS Is_Deposit
    ,       DATEDIFF(dd,@Date,CONVERT(DATE,Transaction_Created))    AS Datedif  

FROM        #DATES
GROUP BY    CONVERT(DATE,Transaction_Created)
order by 1 desc

我的问题是，当is_Deposit = 1时，我需要为DATEDIFF重置种子，而IS_Deposit = 1的日期成为我的新MIN日期，依此类推，对于表中找到的每个IS_Deposit = 1，依此类推。例如，从2015-12-04到2015-12-16，一切正常。但是在746 2015-12-17行中，我希望再次设为1而不是13，因为我们已经达到下一个IS_Deposit = 1且必须重新设置。

现在的结果集：

Date    Is_Deposit  Datedif

2016-02-12  0   70
2016-02-11  0   69
2016-02-10  0   68
2016-02-09  0   67
2016-02-08  0   66
2016-02-07  0   65
2016-02-06  0   64
2016-02-05  0   63
2016-02-04  0   62
2016-02-03  0   61
2016-02-02  0   60
2016-02-01  0   59
2016-01-31  0   58
2016-01-30  0   57
2016-01-29  0   56
2016-01-28  0   55
2016-01-27  0   54
2016-01-26  0   53
2016-01-25  0   52
2016-01-24  0   51
2016-01-23  0   50
2016-01-22  0   49
2016-01-21  0   48
2016-01-20  1   47
2016-01-17  0   44
2016-01-16  0   43
2016-01-15  0   42
2016-01-14  0   41
2016-01-13  0   40
2016-01-12  0   39
2016-01-11  0   38
2016-01-10  0   37
2016-01-09  0   36
2016-01-08  0   35
2016-01-07  0   34
2016-01-06  0   33
2016-01-05  0   32
2016-01-04  0   31
2016-01-03  0   30
2016-01-02  0   29
2016-01-01  0   28
2015-12-31  0   27
2015-12-30  0   26
2015-12-29  0   25
2015-12-28  0   24
2015-12-27  0   23
2015-12-26  0   22
2015-12-25  1   21
2015-12-20  0   16
2015-12-19  0   15
2015-12-18  0   14
2015-12-17  0   13
2015-12-16  1   12
2015-12-14  0   10
2015-12-13  0   9
2015-12-12  0   8
2015-12-11  0   7
2015-12-10  0   6
2015-12-09  0   5
2015-12-08  0   4
2015-12-07  0   3
2015-12-05  0   1
2015-12-04  1   0

需要结果集：

2016-02-12  0   23

2016-02-11  0   22
2016-02-10  0   21
2016-02-09  0   20
2016-02-08  0   19
2016-02-07  0   18
2016-02-06  0   17
2016-02-05  0   16
2016-02-04  0   15
2016-02-03  0   14
2016-02-02  0   13
2016-02-01  0   12
2016-01-31  0   11
2016-01-30  0   10
2016-01-29  0   9
2016-01-28  0   8
2016-01-27  0   7
2016-01-26  0   6
2016-01-25  0   5
2016-01-24  0   4
2016-01-23  0   3
2016-01-22  0   2
2016-01-21  0   1
2016-01-20  1   26
2016-01-17  0   23
2016-01-16  0   22
2016-01-15  0   21
2016-01-14  0   20
2016-01-13  0   19
2016-01-12  0   18
2016-01-11  0   17
2016-01-10  0   16
2016-01-09  0   15
2016-01-08  0   14
2016-01-07  0   13
2016-01-06  0   12
2016-01-05  0   11
2016-01-04  0   10
2016-01-03  0   9
2016-01-02  0   8
2016-01-01  0   7
2015-12-31  0   6
2015-12-30  0   5
2015-12-29  0   4
2015-12-28  0   3
2015-12-27  0   2
2015-12-26  0   1
2015-12-25  1   9
2015-12-20  0   4
2015-12-19  0   3
2015-12-18  0   2
2015-12-17  0   1
2015-12-16  1   12
2015-12-14  0   10
2015-12-13  0   9
2015-12-12  0   8
2015-12-11  0   7
2015-12-10  0   6
2015-12-09  0   5
2015-12-08  0   4
2015-12-07  0   3
2015-12-05  0   1
2015-12-04  1   0
sql sql-server database tsql datediff
3个回答
0
投票

您可以通过几个窗口功能来做到这一点。首先是运行总计以构建累积序列ID，然后在每个序列中挑选出最小值：

With 
DatesSequence AS
(
SELECT Date,Is_Deposit
, SUM(Is_Deposit) OVER(ORDER BY Date DESC) AS Sequence
FROM #Dates
)
SELECT 
Date
,Is_Deposit
, DATEDIFF(day,
            MIN(Date) OVER (PARTITION BY Sequence),
            Date) + 1 AS DateDif
FROM DatesSequence
0
投票

简单的子查询可以解决此问题。首先，创建并填充示例表（请在您将来的问题中为我们保存此步骤）：

DECLARE @T AS TABLE
(
    Date date,
    Is_Deposit bit,
    Datedif int
);

INSERT INTO @T (Date, Is_Deposit, Datedif) VALUES
('2016-02-12', 0, 70),
('2016-02-11', 0, 69),
('2016-02-10', 0, 68),
('2016-02-09', 0, 67),
('2016-02-08', 0, 66),
('2016-02-07', 0, 65),
('2016-02-06', 0, 64),
('2016-02-05', 0, 63),
('2016-02-04', 0, 62),
('2016-02-03', 0, 61),
('2016-02-02', 0, 60),
('2016-02-01', 0, 59),
('2016-01-31', 0, 58),
('2016-01-30', 0, 57),
('2016-01-29', 0, 56),
('2016-01-28', 0, 55),
('2016-01-27', 0, 54),
('2016-01-26', 0, 53),
('2016-01-25', 0, 52),
('2016-01-24', 0, 51),
('2016-01-23', 0, 50),
('2016-01-22', 0, 49),
('2016-01-21', 0, 48),
('2016-01-20', 1, 47),
('2016-01-17', 0, 44),
('2016-01-16', 0, 43),
('2016-01-15', 0, 42),
('2016-01-14', 0, 41),
('2016-01-13', 0, 40),
('2016-01-12', 0, 39),
('2016-01-11', 0, 38),
('2016-01-10', 0, 37),
('2016-01-09', 0, 36),
('2016-01-08', 0, 35),
('2016-01-07', 0, 34),
('2016-01-06', 0, 33),
('2016-01-05', 0, 32),
('2016-01-04', 0, 31),
('2016-01-03', 0, 30),
('2016-01-02', 0, 29),
('2016-01-01', 0, 28),
('2015-12-31', 0, 27),
('2015-12-30', 0, 26),
('2015-12-29', 0, 25),
('2015-12-28', 0, 24),
('2015-12-27', 0, 23),
('2015-12-26', 0, 22),
('2015-12-25', 1, 21),
('2015-12-20', 0, 16),
('2015-12-19', 0, 15),
('2015-12-18', 0, 14),
('2015-12-17', 0, 13),
('2015-12-16', 1, 12),
('2015-12-14', 0, 10),
('2015-12-13', 0, 9 ),
('2015-12-12', 0, 8 ),
('2015-12-11', 0, 7 ),
('2015-12-10', 0, 6 ),
('2015-12-09', 0, 5 ),
('2015-12-08', 0, 4 ),
('2015-12-07', 0, 3 ),
('2015-12-05', 0, 1 ),
('2015-12-04', 1, 0 );

查询：

SELECT    Date
        , Is_Deposit
        , DATEDIFF(
            DAY, 
            (
                SELECT TOP 1 Date 
                FROM @T AS t1 
                WHERE t1.Date <= t0.Date 
                AND Is_Deposit = 1
            ), 
            Date
        ) As Datedif
FROM @T As t0

结果：

Date        Is_Deposit  Datedif
2016-02-12  0           23
2016-02-11  0           22
2016-02-10  0           21
2016-02-09  0           20
2016-02-08  0           19
2016-02-07  0           18
2016-02-06  0           17
2016-02-05  0           16
2016-02-04  0           15
2016-02-03  0           14
2016-02-02  0           13
2016-02-01  0           12
2016-01-31  0           11
2016-01-30  0           10
2016-01-29  0           9
2016-01-28  0           8
2016-01-27  0           7
2016-01-26  0           6
2016-01-25  0           5
2016-01-24  0           4
2016-01-23  0           3
2016-01-22  0           2
2016-01-21  0           1
2016-01-20  1           0
2016-01-17  0           23
2016-01-16  0           22
2016-01-15  0           21
2016-01-14  0           20
2016-01-13  0           19
2016-01-12  0           18
2016-01-11  0           17
2016-01-10  0           16
2016-01-09  0           15
2016-01-08  0           14
2016-01-07  0           13
2016-01-06  0           12
2016-01-05  0           11
2016-01-04  0           10
2016-01-03  0           9
2016-01-02  0           8
2016-01-01  0           7
2015-12-31  0           6
2015-12-30  0           5
2015-12-29  0           4
2015-12-28  0           3
2015-12-27  0           2
2015-12-26  0           1
2015-12-25  1           0
2015-12-20  0           4
2015-12-19  0           3
2015-12-18  0           2
2015-12-17  0           1
2015-12-16  1           0
2015-12-14  0           10
2015-12-13  0           9
2015-12-12  0           8
2015-12-11  0           7
2015-12-10  0           6
2015-12-09  0           5
2015-12-08  0           4
2015-12-07  0           3
2015-12-05  0           1
2015-12-04  1           0

如果要更新表，可以将查询更改为cte并像这样进行更新：

WITH CTE AS
(
    SELECT    Date
            , Is_Deposit
            , Datedif
            , DATEDIFF(
                DAY, 
                (
                    SELECT TOP 1 Date 
                    FROM @T AS t1 
                    WHERE t1.Date <= t0.Date 
                    AND Is_Deposit = 1
                ), 
                Date
            ) As NewDatedif
    FROM @T As t0
)

UPDATE CTE 
SET dateDif = NewDateDif;
0
投票

我认为您可以只使用累积条件max()。我无法确定是否确实需要聚合，因此这可以满足您的要求：

SELECT Date, Is_Deposit,
       COALESCE(DATEDIFF(day,
                         date,
                         MAX(CASE WHEN Is_Deposit = 1 THEN DATE END) OVER
                             (ORDER BY Date
                              ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW
                             )
                        ), 0
               )
FROM #Dates
最新问题
© www.soinside.com 2019 - 2024. All rights reserved.