Question

用户将选择前端和灵活日期的日期，例如，如果他们选择'2014-07-17'作为日期，灵活日期为2，那么我们需要显示前2个和下2个工作日，如下所示，

2014年7月15日
2014年7月16日
2014年7月17日
2014年7月20日
2014年7月21日

不包括周末（周五和周六），周末是星期五和星期六。

我使用了以下查询

DECLARE @MinDate DATE, @MaxDate DATE;
    SELECT @MinDate = DATEADD(Day, -@inyDays ,@dtDate), @MaxDate = DATEADD(Day,@inyDays ,@dtDate)

    DECLARE @DayExclusionValue VARCHAR(20)
    SELECT @DayExclusionValue = dbo.UDF_GetConfigSettingValue('DaysToExclude')

    DECLARE @NumOfWeekends INT
    SELECT @NumOfWeekends= (DATEDIFF(wk, @MinDate, @MaxDate) * 2) +(CASE WHEN DATENAME(dw, @MinDate) = 'Friday' THEN 1 ELSE 0 END)      +(CASE WHEN DATENAME(dw, @MaxDate) = 'Saturday' THEN 1 ELSE 0 END)

    SET @MaxDate = DATEADD(Day,@inyDays + @NumOfWeekends ,@dtDate)

    ;WITH CalculatedDates AS
    (
        SELECT dates = @MinDate
        UNION ALL
        SELECT DATEADD(day, 1, dates)
        FROM CalculatedDates
        WHERE DATEADD(day, 1, dates)  <= @MaxDate

    )
    SELECT dates FROM CalculatedDates 
    WHERE dates >= CAST(GETDATE() AS DATE)
    AND DATENAME(DW, dates) NOT IN (SELECT Value FROM UDF_GetTableFromString(@DayExclusionValue))
    OPTION (MAXRECURSION 0);

但上述查询无效。

你能不能给我建议任何其他解决方案。

Answer 1

此示例适用于Oracle，您没有说明您使用的是什么数据库。如果您有假期列表，您需要按照指示加入。它需要是一个外连接，你需要添加一个案例或某些东西，以便度假表排除＆＃39;天覆盖生成的天数。

我也随机选择了乘数。当只处理周末8时绰绰有余，但如果你的假期表包括很多连续的休假日，它可能不再是。

select d from(
select rownum nn, d, sysdate - d, first_value (rownum) over (order by abs(sysdate-d)) zero_valu
from (
select sysdate+n d, to_char(sysdate+n,'DAY'), CASE to_char(sysdate+n,'D') WHEN '6' THEN 'exclude' WHEN '7' THEN 'exclude' ELSE 'include' END e_or_i  from
(SELECT ROWNUM-9 n -- 9=flexibleday*8/2 +1
FROM   ( SELECT 1 just_a_column
         FROM   dual
         CONNECT BY LEVEL <= 16 -- 8=flexibleday * 8
       ) 
)
) where e_or_i = 'include'  -- in this step you need to join in a table of holidays or such if you need that.
) where abs(nn-7) <= 2 -- 2=flexiday
order by d

Answer 2

DECLARE @StartDate DATE = '2014-07-17';

SELECT *
FROM
(
    --Show Closest Previous 2 Days Not In Friday or Saturday
    SELECT TOP 2
        DATEADD(DAY, -nr, @StartDate) CheckDate, 
        DATENAME(DW, DATEADD(DAY, -nr, @StartDate)) CheckName, 
        -nr CheckCount
    FROM (VALUES(1),(2),(3),(4)) AS Numbers(nr)
    WHERE DATENAME(DW, DATEADD(DAY, -nr, @StartDate)) NOT IN ('Friday','Saturday')

    UNION ALL

    --Show Todays Date If Not Friday or Saturday
    SELECT TOP 1
        DATEADD(DAY, +nr, @StartDate) CheckDate, 
        DATENAME(DW, DATEADD(DAY, +nr, @StartDate)) CheckName, 
        nr CheckCount
    FROM (VALUES(0)) AS Numbers(nr)
    WHERE DATENAME(DW, DATEADD(DAY, +nr, @StartDate)) NOT IN ('Friday','Saturday')

    UNION ALL

    --Show Closest Next 2 Days Not In Friday or Saturday
    SELECT TOP 2
        DATEADD(DAY, +nr, @StartDate) CheckDate, 
        DATENAME(DW, DATEADD(DAY, +nr, @StartDate)) CheckName, 
        nr CheckCount
    FROM (VALUES(1),(2),(3),(4)) AS Numbers(nr)
    WHERE DATENAME(DW, DATEADD(DAY, +nr, @StartDate)) NOT IN ('Friday','Saturday')
) d
ORDER BY d.CheckDate

我将其分为3个部分，前2天，今天（如果适用）和接下来的2天这是输出：

CheckDate   CheckName  CheckCount
2014-07-15  Tuesday    -2
2014-07-16  Wednesday  -1
2014-07-17  Thursday   0
2014-07-20  Sunday     3
2014-07-21  Monday     4

我使用日期名称，因为不确定服务器设置为@@ datefirst的是什么。 values（）部分只是一个数字表（您应该创建一个数字表，与您想要返回的记录数量以及您要越过的任何周末一样大）然后将替换第一部分和最后部分中的TOP 2以及您希望在之前和之后返回的天数。

****添加了通用数字表功能的更新：

我们在此声明我们想要提取的开始日期和上一天和下一天的数量：

DECLARE @StartDate DATE = '2014-07-20';
DECLARE @MaxBusDays INT = 5

下一节创建一个数字表（可以通过谷歌轻松找到）

DECLARE @number_of_numbers INT = 100000;
;WITH
    a AS (SELECT 1 AS i UNION ALL SELECT 1),
    b AS (SELECT 1 AS i FROM a AS x, a AS y),
    c AS (SELECT 1 AS i FROM b AS x, b AS y),
    d AS (SELECT 1 AS i FROM c AS x, c AS y),
    e AS (SELECT 1 AS i FROM d AS x, d AS y),
    f AS (SELECT 1 AS i FROM e AS x, e AS y),
    numbers AS 
(
    SELECT TOP(@number_of_numbers)
    ROW_NUMBER() OVER (ORDER BY (SELECT NULL)) AS number
    FROM f
)

现在我们使用数字表和row_number设置来仅提取工作日之前和之后的行数（加上所需的日期，如果它不是fri / sat）（不是fri / sat）

SELECT *
FROM
(
    --Show Closest Previous x Working Days (Not Friday or Saturday)
    SELECT * FROM
    (
        SELECT
            DATEADD(DAY, -number, @StartDate) CheckDate, 
            DATENAME(DW, DATEADD(DAY, -number, @StartDate)) CheckName, 
            -number CheckCount,
          ROW_NUMBER() OVER (ORDER BY number ASC) AS RowCounter
        FROM Numbers
        WHERE DATENAME(DW, DATEADD(DAY, -number, @StartDate)) NOT IN ('Friday','Saturday')
    ) a
    WHERE a.RowCounter <= @MaxBusDays

    UNION ALL

    --Show Todays Date If Working Day (Not Friday or Saturday)
    SELECT TOP 1
        @StartDate CheckDate, 
        DATENAME(DW, @StartDate) CheckName, 
        0 CheckCount,
      0 RowCounter
    WHERE DATENAME(DW, @StartDate) NOT IN ('Friday','Saturday')

    UNION ALL

    --Show Closest Next x Working Days (Not Friday or Saturday)
    SELECT * FROM
    (
        SELECT
            DATEADD(DAY, +number, @StartDate) CheckDate, 
            DATENAME(DW, DATEADD(DAY, +number, @StartDate)) CheckName, 
            number CheckCount,
          ROW_NUMBER() OVER (ORDER BY number ASC) AS RowCounter
        FROM Numbers
        WHERE DATENAME(DW, DATEADD(DAY, +number, @StartDate)) NOT IN ('Friday','Saturday')
    ) b
    WHERE b.RowCounter <= @MaxBusDays

) c
ORDER BY c.CheckDate

以下是输出：（2014-07-20是中间行）

CheckDate     CheckName   CheckCount  RowCounter
2014-07-13    Sunday      -7          5
2014-07-14    Monday      -6          4
2014-07-15    Tuesday     -5          3
2014-07-16    Wednesday   -4          2
2014-07-17    Thursday    -3          1
2014-07-20    Sunday      0           0
2014-07-21    Monday      1           1
2014-07-22    Tuesday     2           2
2014-07-23    Wednesday   3           3
2014-07-24    Thursday    4           4
2014-07-27    Sunday      7           5

从给定日期开始的工作日

2 个答案: