Question

我从以下问题中得到以下代码： Passing in Week Day name to get nearest date in SQL

我需要根据我的表格中相应的星期几的今天日期找到接下来的4个工作日，即如果今天为2015-01-24，1/24, 1/31, 2/7, 2/14的结果应为Saturdays 1}}。

表格

enter image description here

样本查询

create table #t
(
    jobId int,
    personId int,
    frequencyVal varchar(10)
);

insert into #t values (1,100,'Mondays'),(2,101,'Saturdays');

WITH cte(n) AS
(
    SELECT 0
    UNION ALL
    SELECT n+1 FROM cte WHERE n < 3
)

select #t.jobId, #t.personId, #t.frequencyVal, STUFF(a.d, 1, 1, '') AS FutureDates
from #t
cross apply (SELECT CASE #t.frequencyVal
                         WHEN 'SUNDAYS'    THEN 1 
                         WHEN 'MONDAYS'    THEN 2 
                         WHEN 'TUESDAYS'   THEN 3 
                         WHEN 'WEDNESDAYS' THEN 4 
                         WHEN 'THURSDAYS'  THEN 5 
                         WHEN 'FRIDAYS'    THEN 6
                         WHEN 'SATURDAYS'  THEN 7 
                    END)tranlationWeekdays(n)
cross apply (select ',' +  CONVERT(varchar(10),  CONVERT(date,dateadd(WEEK, cte.n,CONVERT(DATE, DATEADD(DAY, (DATEPART(WEEKDAY, GETDATE()) + tranlationWeekdays.n) % 7, GETDATE()))))) from cte FOR XML PATH('')) a(d);

drop table #t;

预期结果

enter image description here

Answer 1

获取当月的第一天。

DECLARE @FIRSTDAY DATE=DATEADD(month, DATEDIFF(month, 0, GETDATE()), 0)

创建表并插入值

create table #t
(
    jobId int,
    personId int,
    frequencyVal varchar(10)
);

insert into #t values (1,100,'Mondays'),(2,101,'Saturdays');

您可以根据自己的情况使用以下任一查询。

查询1：选择特定工作日当前月份的前4周

 -- Gets the first day of current month
 DECLARE @FIRSTDAY DATE=DATEADD(month, DATEDIFF(month, 0, GETDATE()), 0)

;WITH  CTE as
(
     -- Will find all dates in current month
     SELECT @FIRSTDAY as DATES
     UNION ALL
     SELECT DATEADD(DAY,1,DATES)    
     FROM    CTE
     WHERE   DATES < DATEADD(MONTH,1,@FIRSTDAY)
 )
,CTE2 AS
(
   -- Join the #t table with  CTE on the datename+'s' 
   SELECT jobId,personId,frequencyVal, DATES,
   ROW_NUMBER() OVER(PARTITION BY DATENAME(WEEKDAY,CTE.DATES) ORDER BY CTE.DATES) DATECNT
   FROM CTE
   JOIN #t ON DATENAME(WEEKDAY,CTE.DATES)+'s' = #t.frequencyVal
   WHERE MONTH(DATES)= MONTH(GETDATE())   
)
-- Converts to CSV and make sure that only 4 days are generated for month
SELECT  DISTINCT C2.jobId,C2.personId,frequencyVal,   
        SUBSTRING(
        (SELECT ', ' + CAST(DATEPART(MONTH,DATES) AS VARCHAR(2)) + '/'  + 
                       CAST(DATEPART(DAY,DATES) AS VARCHAR(2))
        FROM CTE2 
        WHERE C2.jobId=jobId AND C2.personId=personId AND DATECNT<5
        ORDER BY CTE2.DATES
        FOR XML PATH('')),2,200000) futureDates
        FROM CTE2 C2

的 SQL FIDDLE

例如，在 Query1 中最近的日期（这里我们以星期六为例）

2015-Jan-10将为01/03,01/10,01/17,01/24

2015-Jan-24将为01/03,01/10,01/17,01/24

查询2：选择当月最近4周的特定工作日

-- Gets the first day in current month DECLARE @FIRSTDAY DATE=DATEADD(month, DATEDIFF(month, 0, GETDATE()), 0) ;WITH CTE as ( -- Will find all dates in current SELECT CAST(@FIRSTDAY AS DATE) as DATES UNION ALL SELECT DATEADD(DAY,1,DATES) FROM CTE WHERE DATES < DATEADD(MONTH,1,@FIRSTDAY) ) ,CTE2 AS ( -- Join the #t table with CTE on the datename+'s' SELECT jobId,personId,frequencyVal,DATES, -- Get week difference for each weekday DATEDIFF(WEEK,DATES,GETDATE()) WEEKDIFF, -- Count the number of weekdays in a month COUNT(DATES) OVER(PARTITION BY DATENAME(WEEKDAY,CTE.DATES)) WEEKCOUNT FROM CTE JOIN #t ON DATENAME(WEEKDAY,CTE.DATES)+'s' = #t.frequencyVal WHERE MONTH(DATES)= MONTH(GETDATE()) ) -- Converts to CSV and make sure that only nearest 4 week of days are generated for month SELECT DISTINCT C2.jobId,C2.personId,frequencyVal, SUBSTRING( (SELECT ', ' + CAST(DATEPART(MONTH,DATES) AS VARCHAR(2)) + '/' + CAST(DATEPART(DAY,DATES) AS VARCHAR(2)) FROM CTE2 WHERE C2.jobId=jobId AND C2.personId=personId AND C2.frequencyVal=frequencyVal AND ((WEEKDIFF<3 AND WEEKDIFF>-3 AND WEEKCOUNT = 5) OR WEEKCOUNT <= 4) ORDER BY CTE2.DATES FOR XML PATH('')),2,200000) futureDates FROM CTE2 C2

的 SQL FIDDLE

例如，在 Query2 中最近的日期（此处我们以星期六为例）

2015-Jan-10将为01/03,01/10,01/17,01/24

2015-Jan-24将为01/10,01/17,01/24,01/31

QUERY 3：选择与月份无关的特定工作日的下一个4周的日期

;WITH CTE as ( -- Will find all dates in current month SELECT CAST(GETDATE() AS DATE) as DATES UNION ALL SELECT DATEADD(DAY,1,DATES) FROM CTE WHERE DATES < DATEADD(DAY,28,GETDATE()) ) ,CTE2 AS ( -- Join the #t table with CTE on the datename+'s' SELECT jobId,personId,frequencyVal, DATES, ROW_NUMBER() OVER(PARTITION BY DATENAME(WEEKDAY,CTE.DATES) ORDER BY CTE.DATES) DATECNT FROM CTE JOIN #t ON DATENAME(WEEKDAY,CTE.DATES)+'s' = #t.frequencyVal ) -- Converts to CSV and make sure that only 4 days are generated for month SELECT DISTINCT C2.jobId,C2.personId,frequencyVal, SUBSTRING( (SELECT ', ' + CAST(DATEPART(MONTH,DATES) AS VARCHAR(2)) + '/' + CAST(DATEPART(DAY,DATES) AS VARCHAR(2)) FROM CTE2 WHERE C2.jobId=jobId AND C2.personId=personId AND C2.frequencyVal=frequencyVal AND DATECNT < 5 ORDER BY CTE2.DATES FOR XML PATH('')),2,200000) futureDates FROM CTE2 C2

的 SQL FIDDLE

如果GETDATE()（如果是星期六）
，则输出如下
2015-01-05 - 1/10, 1/17, 1/24, 1/31 2015-01-24 - 1/24, 1/31, 2/7, 2/14

Answer 2

我认为这是一种更简单的方式，我认为它符合您的要求请注意，我已将您的frequency_val列更改为一个整数，该整数表示从SQL Server角度来看星期几，并添加了一个计算列，以说明如何从中轻松派生日期名称。

/*************************************************/
--Set up our sample table
/*************************************************/
declare @t table
(
    jobId int,
    personId int,
    --frequencyVal varchar(10) -- why store a string when a tiny int will do.
    frequency_val tinyint,
    frequency_day as datename(weekday,frequency_val -1) + 's'
)

insert into @t 
values 
    (1,100,1),--'Mondays'),
    (2,101,6),--'Saturdays');
    (3,101,7),--'Sundays');
    (4,100,2)--'Tuesdays'),
--select * from @t

/*************************************************/
--Declare & initialise variables 
/*************************************************/
declare @num_occurances int = 4
declare @from_date date = dateadd(dd,3,getdate()) -- this will allow you to play with the date simply by changing the increment value

/*************************************************/
-- To get a row for each occurance
/*************************************************/
;with r_cte (days_ahead, occurance_date)
as (select 0, convert(date,@from_date,121)
    union all
    select r_cte.days_ahead +1, convert(date,dateadd(DD, r_cte.days_ahead+1, @from_date),121)
    from r_cte
    where r_cte.days_ahead < (7 * @num_occurances) -1
    )
select t.*, r_cte.occurance_date
from
    @t t
    inner join r_cte
        on DATEPART(WEEKDAY, dateadd(dd,@@DATEFIRST - 1 ,r_cte.occurance_date)) = t.frequency_val

/*************************************************/
--To get a single row with a CSV of every occurance
/*************************************************/
;with r_cte (days_ahead, occurance_date)
as (select 0, convert(date,@from_date,121)
    union all
    select r_cte.days_ahead +1, convert(date,dateadd(DD, r_cte.days_ahead+1, @from_date),121)
    from r_cte
    where r_cte.days_ahead < (7 * @num_occurances) -1
    )
select 
    t.*, 
    STUFF(  (select ', '
                + convert(varchar(2),datepart(month,occurance_date),0) + '/' 
                + convert(varchar(2),datepart(day,occurance_date),0) as occurance
            from r_cte
            where DATEPART(WEEKDAY, dateadd(dd,@@DATEFIRST - 1 ,r_cte.occurance_date)) = t.frequency_val
            FOR XML PATH (''),TYPE).value('.','varchar(30)') 
        ,1,2,'') occurance_date -- rest of STUFF() function
from
    @t t

找到下一周的最佳方法是什么？

2 个答案: