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sql - 在SQL中当前选择后如何查看值是否在表中再次出现

问题描述

我有以下数据表 [CRA_feasibility_test]

atmid       CRA     monday  tuesday wednesday   thursday    friday  saturday
DA0068C1    ABC       Y      N      Y           N           N        Y

我想根据我的日期的工作日选择数据。如果它是 Y，那么想要获得最接近的“Y”，即从@date 开始的 Y 天。

我也想得到下一个

我正在做以下事情

declare @date datetime set @date ='2018-06-15' 
Select  @dateT as indentdate, DATENAME(dw,@dateT) as weekdayname,

          case when DATENAME(dw,@dateT)  ='monday'        
                                                 then  monday

                  when DATENAME(dw,@dateT)  ='tuesday'           then  
                                                            tuesday
                  when DATENAME(dw,@dateT)  ='wednesday'      then  
                                                         wednesday
                  when DATENAME(dw,@dateT)  ='Thursday'         then  
                                                       thursday
                 when DATENAME(dw,@dateT)  ='Friday'               then  
                                                   friday
                 when DATENAME(dw,@dateT)  ='Saturday'           then  
                                                 saturday       


end   feasible
from [CRA_feasibility_test]

标签： sqlcase-when

解决方案

如果我理解正确，您可以尝试为接下来的 7 天 by 创建日历表cte recursive，然后使用CROSS APPLYwith valuesmake unpivot 获取Y最近的日期。

declare @dateT datetime
set @dateT ='2018-06-15'

;with calendar as (
  select @dateT startDt,dateadd(day,7,@dateT) endDt, 1 rn
  union all
  select dateadd(day,1,startDt) , endDt,rn +1
  from calendar
  WHERE  dateadd(day,1,startDt)<endDt
), cte as (
select 
        atmid,v.*
from CRA_feasibility_test cross apply (
    values ('monday',monday), ('tuesday',tuesday), ('wednesday',wednesday), ('thursday',thursday), ('friday',friday), ('saturday',saturday)
) v(name,flag)

)


select atmid,
    @dateT today,
    min(case when flag = 'Y' then startDt end) nextYDay,
    datename(dw,min(case when flag = 'Y' then startDt end) ) nextYname
from calendar t1 
join cte t2 on datename(dw,startDt) = t2.name
group by atmid

sqlfiddle

结果

atmid       today               nextYDay            nextYname
DA0068C1    15/06/2018 00:00:00 16/06/2018 00:00:00 Saturday

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