如何将一年中的双周时间添加到表中？ （SQL Server）

Question

此SQL语句不足以添加正确的双周期。 我想要的是这样的：

Column 1 (period) / Column 2 (start period) / Column 3 (end period)
20160115 / 2016-01-01 / 2016-01-15
20160131 / 2016-01-15 / 2016-01-31
20160215 / 2016-02-01 / 2016-02-15
20160229 / 2016-02-16 / 2016-02-29

依旧......

这就是我现在所拥有的，这是错误的/不完整的。 该值正在表格列中正确插入，但差距是错误的，因为几个月没有相同的天数

有人能帮助我吗？非常感谢你！

    DECLARE @d date= '20020101'
WHILE @d<'20030101'
    BEGIN
        INSERT INTO [TABLE]
        VALUES ((SELECT CONVERT(VARCHAR(8), @d, 112) AS [YYYYMMDD]), @d, @d, 'Fechado', '1')
        SET @d=DATEADD(DAY,15,@d)
    END

Answer 1

使用common table expression表示特殊数字表，union all表示使用dateadd()进行日期数学计算的两个查询：

declare @startdate date = '2016-01-01'
declare @enddate   date = '2016-12-31'

;with cte as (
select top (datediff(month,@startdate,@enddate)+1)
  Month = convert(date,dateadd(month,row_number() over (order by (select 1))-1,@startdate))
from master..spt_values
order by month
)
select 
    Period = convert(char(8), dateadd(day,14,month), 112)
  , PeriodStart = month
  , PeriodEnd = dateadd(day,14,month)
from cte
union all 
select 
    Period = convert(char(8), eomonth(month), 112)
  , PeriodStart = dateadd(day,14,month)
  , PeriodEnd = eomonth(month)
from cte
order by period

rextester演示：http://rextester.com/BSYIXW7864

返回：

+----------+-------------+------------+
|  Period  | PeriodStart | PeriodEnd  |
+----------+-------------+------------+
| 20160115 | 2016-01-01  | 2016-01-15 |
| 20160131 | 2016-01-15  | 2016-01-31 |
| 20160215 | 2016-02-01  | 2016-02-15 |
| 20160229 | 2016-02-15  | 2016-02-29 |
| 20160315 | 2016-03-01  | 2016-03-15 |
| 20160331 | 2016-03-15  | 2016-03-31 |
| 20160415 | 2016-04-01  | 2016-04-15 |
| 20160430 | 2016-04-15  | 2016-04-30 |
| 20160515 | 2016-05-01  | 2016-05-15 |
| 20160531 | 2016-05-15  | 2016-05-31 |
| 20160615 | 2016-06-01  | 2016-06-15 |
| 20160630 | 2016-06-15  | 2016-06-30 |
| 20160715 | 2016-07-01  | 2016-07-15 |
| 20160731 | 2016-07-15  | 2016-07-31 |
| 20160815 | 2016-08-01  | 2016-08-15 |
| 20160831 | 2016-08-15  | 2016-08-31 |
| 20160915 | 2016-09-01  | 2016-09-15 |
| 20160930 | 2016-09-15  | 2016-09-30 |
| 20161015 | 2016-10-01  | 2016-10-15 |
| 20161031 | 2016-10-15  | 2016-10-31 |
| 20161115 | 2016-11-01  | 2016-11-15 |
| 20161130 | 2016-11-15  | 2016-11-30 |
| 20161215 | 2016-12-01  | 2016-12-15 |
| 20161231 | 2016-12-15  | 2016-12-31 |
+----------+-------------+------------+

Answer 2

创建一个函数，然后调用您的查询：

功能：

urlpatterns = [
    url(r'^(?P<username>[\w.@+-]+)/$', UserDetailAPIVIew.as_view(), 
        name='user-posts-api'),
    url(r'^(?P<username>[\w.@+-]+)/follow/$', 
        FollowToggleAPIView.as_view(), name='follow-api'),
]

代码：

create FUNCTION advance_biweekly
(
    @current date
)
RETURNS date
AS
BEGIN
    --if it's the first day of month, advance two weeks
    if (DAY(@current) = 1)
        return DATEADD(WEEK, 2, @current)
    else
    --if its last day of month, advance one day
    if (DAY(DATEADD(d, -1, DATEADD(m, DATEDIFF(m, 0, @current) + 1, 0)))  = DAY(@current))
        return DATEADD(DAY, 1, @current)
    else
    --else, it's the middle of the month, go to end
        return dateadd(month,((YEAR(@current)-1900)*12)+MONTH(@current)-1,DAY(DATEADD(d, -1, DATEADD(m, DATEDIFF(m, 0, @current) + 1, 0)))-1)

    return null

END
GO

结果：

DECLARE @d date= '20020101'
WHILE @d<'20030101'
begin
    set @d = dbo.advance_biweekly(@d)
    print @d
end