Question

我有一张名为“Days”的桌子

Day 1    Day 2    Day 3    Day 4    Day 5    Day 6    Day 7
1        1        1        1        0        0        0
1        1        1        1        0        0        0

具有相应的细节。其中1为'true'且0为false。

DaY 1 - sunday
day 2- monday
day 3 - tuesday
day 4- wednesday
day5 - thursday
day 6 - friday
dayy 7- saturday

1 - denotes there is work on the given day
0 - denotes there is no work on the given day.

鉴于上述细节，我需要进行列到表的映射并且需要生成一个这样的表，我需要只获取具有标志1

的数据

1   Sunday 
2   Monday 
3   Tuesday     
4   Wednesday

第一张唱片。

Answer 1

WITH    [days] (id, day1, day2, day3, day4, day5, day6, day7) AS
        (
        SELECT  1, 1, 1, 1, 1, 1, 0, 0
        UNION ALL
        SELECT  2, 1, 1, 1, 1, 1, 0, 0
        )
SELECT  id, DATENAME(dw, DATEADD(d, CAST(SUBSTRING(wd, 4, 1) AS INT), '2005-01-01')), work
FROM    [days]
UNPIVOT
        (
        work FOR wd IN
        (day1, day2, day3, day4, day5, day6, day7)
        ) AS up
WHERE   work = 1

Answer 2

我相信你正在寻找以下内容：

select 
case when day1=1 then '1 Sunday' end Sunday,
case when day2=1 then '2 Monday' end Monday,
case when day3=1 then '3 Tuesday' end Tuesday,
case when day4=1 then '4 Wednesday' end Wednesday,
case when day5=1 then '5 Thursday' end Thursday,
case when day6=1 then '6 Friday' end Friday, 
case when day7=1 then '7 Saturday' end Saturday
from [table]
where day1+day2+day3+day4+day5+day6+day7 <> 7

那些应该是行？然后，您需要首先unpivot源数据：

select     
cast(substring(dayname, 4, 1) as tinyint) as DayNumber, 
case cast(substring(dayname, 4, 1) as tinyint) 
when 1 then 'Sunday'
when 2 then 'Monday'
when 3 then 'Tuesday'
when 4 then 'Wednesday'
when 5 then 'Thursday'
when 6 then 'Friday'
when 7 then 'Saturday' end [DayName]
, DayFlag 
from (select * from [table] t where day1+day2+day3+day4+day5+day6+day7 <> 7) f
unpivot 
(
DayFlag for DayName in ([day1], [day2], [day3], [day4], [day5], [day6], [day7])
) unpvt
where DayFlag = 1 --is this what you need?

Answer 3

免责声明 - 我知道它并不能解决您的问题，但是如果将表格反规格化为以下内容并不是更好：

ID  | WeekNo  |  DayNo
----------------------
1     1          1
2     1          2
3     1          3
4     2          6
5     2          7

因此，您实际上只会在当天进行工作的表格中添加天数（即第1周的第4-7天没有工作，第2周仅在第6天和第7天进行工作）。 ??然后你可以在你的.net应用程序中找到一些逻辑，这些逻辑“知道”本周完整的日期，并向你提供了一个例外列表（即任何给定的WeekNo表中没有的日期）

这将是我对这场'辩论'的第一个选择......

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3 个答案: