DATEPART修复日

Question

CASE WHEN DAY % 2 = 0 AND POL = 'SUUA'
THEN CONVERT(VARCHAR(15),DATEPART(DAY,5)) + ' TO ' + CONVERT(VARCHAR(15),DATEPART(DAY,3),103)

我把日期放在第一位，但我不能改进它。我很抱歉信息少了。我在使用SQL Server。我上传了.img作为例子。

CASE WHEN DAY % 2 = 0 AND POL = 'SUUA' THEN CONVERT(VARCHAR(15),DATEPART(DAY,5)) + ' TO ' + CONVERT(VARCHAR(15),DATEPART(DAY,3),103)

我想表现得像这样

PORT            COLLECT             DEPARTURE

MANAUS      07/12 TO 12/12     15/12/2018(ODD DAY)

Answer 1

为此，我强烈建议使用日历表。日历表保存有附加信息的日期值，因此更容易找到特定的日期(如本例所示的工作日或工作日)。

下面的解决方案使用一个日历表和2个CROSS APPLY操作符来获取以前的收集日期。

您可以这样创建日历表(递归CTE)：

SET DATEFIRST 1 -- 1: Monday, 7: Sunday

-- Create a Calendar Table
IF OBJECT_ID('tempdb..#CalendarTable') IS NOT NULL
    DROP TABLE #CalendarTable

;WITH CalendarTable AS
(
    SELECT
        Date = CONVERT(DATE, '2016-01-01'),
        Weekday = DATEPART(WEEKDAY, '2016-01-01')

    UNION ALL

    SELECT
        Date = DATEADD(DAY, 1, C.Date),
        Weekday = DATEPART(WEEKDAY, DATEADD(DAY, 1, C.Date))
    FROM
        CalendarTable AS C
    WHERE
        C.Date <= '2020-01-01'
)
SELECT
    C.Date,
    C.Weekday
INTO
    #CalendarTable
FROM
    CalendarTable AS C
OPTION
    (MAXRECURSION 0)

该表如下：

SELECT * FROM #CalendarTable ORDER BY Date DESC

Date        Weekday
2020-01-02  4
2020-01-01  3
2019-12-31  2
2019-12-30  1
2019-12-29  7
2019-12-28  6
2019-12-27  5
2019-12-26  4
2019-12-25  3
2019-12-24  2
2019-12-23  1
2019-12-22  7
2019-12-21  6
2019-12-20  5
2019-12-19  4
2019-12-18  3
2019-12-17  2
2019-12-16  1
2019-12-15  7
2019-12-14  6
2019-12-13  5
2019-12-12  4
2019-12-11  3

我们将用这个来找最近的星期三和星期一，就在一个特定的出发日期之前。我们使用CROSS APPLY将DepartureDate作为更高的限制，然后搜索特定的工作日(周一为1，周三为3)。然后使用TOP 1和ORDER BY Date DESC，在出发日期之前的周一/周三获得最高值。

-- Build your Collect periods
;WITH SampleData AS
(
    SELECT
        V.Departure
    FROM
        (VALUES
            ('2018-12-01'),
            ('2018-12-09'),
            ('2018-12-25'),
            ('2018-12-29'),
            ('2019-01-02'),
            ('2019-01-07'),
            ('2019-01-10')) AS V(Departure)
)
SELECT
    V.Departure,

    -- Friday to Wednesday
    ClosestWednesdayBeforeDeparture = W.Date,
    PreviousFridayOfThatWednesday = DATEADD(DAY, -5, W.Date),

    -- Wednesday to Monday
    ClosestMondayBeforeDeparture = M.Date,
    PreviousWednesdayOfThatMonday = DATEADD(DAY, -5, M.Date),

    -- Check for odd/even
    IsOdd = CASE WHEN DATEPART(DAY, V.Departure) % 2 = 1 THEN 1 ELSE 0 END,

    -- Use previous expressions to build your collect periods
    Collect = CASE
        WHEN 
            DATEPART(DAY, V.Departure) % 2 = 1 -- IsOdd
        THEN
            CONVERT(VARCHAR(100), DATEADD(DAY, -5, W.Date), 120) -- PreviousFridayOfThatWednesday
            + ' TO '
            + CONVERT(VARCHAR(100), W.Date, 120) -- ClosestWednesdayBeforeDeparture

        ELSE -- IsEven
            CONVERT(VARCHAR(100), DATEADD(DAY, -5, M.Date), 120) -- PreviousWednesdayOfThatMonday
            + ' TO '
            + CONVERT(VARCHAR(100), M.Date, 120) -- ClosestMondayBeforeDeparture
        END
FROM
    SampleData AS V
    CROSS APPLY (
        SELECT TOP 1
            C.Date
        FROM
            #CalendarTable AS C
        WHERE
            C.Date < V.Departure AND
            C.Weekday = 3 -- 3: Wednesday
        ORDER BY
            C.Date DESC) AS W
    CROSS APPLY (
        SELECT TOP 1
            C.Date
        FROM
            #CalendarTable AS C
        WHERE
            C.Date < V.Departure AND
            C.Weekday = 1 -- 1: Monday
        ORDER BY
            C.Date DESC) AS M
ORDER BY
    V.Departure

从周三找到前一个星期五就像往后退5天一样简单，从周一到周三也是如此。

结果：

Departure   IsOdd   Collect                     ClosestWednesdayBeforeDeparture     PreviousFridayOfThatWednesday   ClosestMondayBeforeDeparture    PreviousWednesdayOfThatMonday
2018-12-01  1       2018-11-23 TO 2018-11-28    2018-11-28                          2018-11-23                      2018-11-26                      2018-11-21
2018-12-09  1       2018-11-30 TO 2018-12-05    2018-12-05                          2018-11-30                      2018-12-03                      2018-11-28
2018-12-25  1       2018-12-14 TO 2018-12-19    2018-12-19                          2018-12-14                      2018-12-24                      2018-12-19
2018-12-29  1       2018-12-21 TO 2018-12-26    2018-12-26                          2018-12-21                      2018-12-24                      2018-12-19
2019-01-02  0       2018-12-26 TO 2018-12-31    2018-12-26                          2018-12-21                      2018-12-31                      2018-12-26
2019-01-07  1       2018-12-28 TO 2019-01-02    2019-01-02                          2018-12-28                      2018-12-31                      2018-12-26
2019-01-10  0       2019-01-02 TO 2019-01-07    2019-01-09                          2019-01-04                      2019-01-07                      2019-01-02

这是一个很好的SQL练习。

Answer 2

非常感谢埃兹洛，那真是太棒了。这是一个很好的SQL练习。我认为这很适合我的工作。很抱歉看上去很无聊，我有一些问题：

编辑：.我注意到V(出发)的日期是固定的。我想把日期做得很棒。因为当我把时间安排在

CROSS APPLY(SELECT TOP 1 
            C.DATE
        FROM 
            #CALENDARTABLE C, SAMPLEDATA V
        WHERE
            C.DATE < V.DEPARTURE AND
            C.WEEKDAY = 1 
        ORDER BY
            C.DATE DESC) AS W

CROSS APPLY (SELECT TOP 1
            C.DATE
        FROM
            #CALENDARTABLE C,SAMPLEDATA V
        WHERE
            C.DATE < V.DEPARTURE AND
            C.WEEKDAY = 7 
        ORDER BY
            C.DATE DESC) AS M

所有港口的收货日期。抱歉如果我无聊的话。你们帮了很多忙。