Question

我有一张表Holiday_Details

下表中的列

CALENDAR_DATE, Day1,Day2,Day3,Day4,Day5,Day6,Day7,Day8,Day9,Day10,Day11,Day12,Day13,    Day14,Day15,Day16,Day17,Day18,Day19,Day20,Day21,Day22,Day23,Day24,Day25,Day26,Day27,    Day28,Day29,Day30,Day31 the columns from Day1 to Day31 are having null values.

Year    Month     Days
2014    1   01-01-2014
2014    1   02-02-2014
2014    1   03-03-2014
2014    1   04-03-2014
2014    1   05-03-2014

现在我想要的结果如下：

结果

Year    Month   Day1    Day2    Day3    Day4    Day5    Day6    Day31   
2014    1   1   1   1   0   0   ……. 1   
2014    2   0   0   1   0   0   ……. 1   
2014    3   0   0   1   0   0   ……. 1

工作日= 1   （周一，周二，周三，周四，周五）
  假日= 0（周六，周日）

请告诉我如何将列CALENDAR_DATES转换为与每个月和每年相对应的行。请帮助我，因为我无法获得如上所示的此订单。

Answer 1

测试表：

create table dates(Year int, Month int, Days int, date date)

;with cte as
(
select 0 x
union all
select x + 1
from cte 
where x < 364
)
insert dates
select 2014, month(dateadd(d, x, '20140101')), day(dateadd(d, x, '20140101')), dateadd(d, x, '20140101')
from cte
option (maxrecursion 0)

查询：

select * 
from 
(
  select year, month, days, case when datediff(d, 0, date)%7 in (5,6) 
  then 0 else 1 end status 
  from dates
) x
pivot (max(status) for Days in (
 [1], [2], [3], [4], [5], [6], [7], [8], [9],[10],
[11],[12],[13],[14],[15],[16],[17],[18],[19],[20],
[21],[22],[23],[24],[25],[26],[27],[28],[29],[30],[31]
)) as pvt

Fiddle

Answer 2

试试这个：

;with CalendarDate as 
(
Select 1 as Level , convert(date,'20140101') as myDate 
union all 
Select Level +1 , dateadd(day,1,myDate)  from CalendarDate
where Level<100
)
Select Year,Month,
[Day1],[Day2],[Day3],[Day4],[Day5],[Day6],[Day7],[Day8],[Day9],[Day10],[Day11],[Day12],   [Day13],[Day14],[Day15],[Day16],[Day17],[Day18],[Day19],[Day20],[Day21],[Day22],[Day23],   [Day24],[Day25],[Day26],[Day27],[Day28],[Day29],[Day30],[Day31] 
from 
(
Select myDate,
       year(myDate) as Year,
       Month(myDate) as Month,
       case when datename(Weekday,myDate) in ('Saturday','Sunday')
            then 'Weekend'
            ELSE
           'Day'+convert(varchar(3),day(myDate))
      END as Col1
from CalendarDate 
 ) pv
pivot 
(
 count(myDate)
 for Col1 in 
 ([Day1],[Day2],[Day3],[Day4],[Day5],[Day6],[Day7],[Day8],[Day9],[Day10],[Day11],[Day12],[Day13],[Day14],[Day15],[Day16],[Day17],[Day18],[Day19],[Day20],[Day21],[Day22],[Day23],[Day24],[Day25],[Day26],[Day27],[Day28],[Day29],[Day30],[Day31] 
 )
) 
pvt

结果： - SQL Fiddle

日历日期转换为具有状态更新假期的行

2 个答案: