假设我想在表格中存储数千天，我将如何从日历中检索它？
答案 0 :(得分：42)
这是一个可以在SQL Server中使用的通用脚本。只需修改开始和结束日期：
IF EXISTS (SELECT * FROM information_schema.tables WHERE Table_Name = 'Calendar' AND Table_Type = 'BASE TABLE')
BEGIN
DROP TABLE [Calendar]
END
CREATE TABLE [Calendar]
(
[CalendarDate] DATETIME
)
DECLARE @StartDate DATETIME
DECLARE @EndDate DATETIME
SET @StartDate = GETDATE()
SET @EndDate = DATEADD(d, 365, @StartDate)
WHILE @StartDate <= @EndDate
BEGIN
INSERT INTO [Calendar]
(
CalendarDate
)
SELECT
@StartDate
SET @StartDate = DATEADD(dd, 1, @StartDate)
END
如果你想要一个更高级的日历，这是我刚才在网上找到的日历：
CREATE SCHEMA Auxiliary
-- We put our auxiliary tables and stuff in a separate schema
-- One of the great new things in SQL Server 2005
go
CREATE FUNCTION Auxiliary.Computus
-- Computus (Latin for computation) is the calculation of the date of
-- Easter in the Christian calendar
-- http://en.wikipedia.org/wiki/Computus
-- I'm using the Meeus/Jones/Butcher Gregorian algorithm
(
@Y INT -- The year we are calculating easter sunday for
)
RETURNS DATETIME
AS
BEGIN
DECLARE
@a INT,
@b INT,
@c INT,
@d INT,
@e INT,
@f INT,
@g INT,
@h INT,
@i INT,
@k INT,
@L INT,
@m INT
SET @a = @Y % 19
SET @b = @Y / 100
SET @c = @Y % 100
SET @d = @b / 4
SET @e = @b % 4
SET @f = (@b + 8) / 25
SET @g = (@b - @f + 1) / 3
SET @h = (19 * @a + @b - @d - @g + 15) % 30
SET @i = @c / 4
SET @k = @c % 4
SET @L = (32 + 2 * @e + 2 * @i - @h - @k) % 7
SET @m = (@a + 11 * @h + 22 * @L) / 451
RETURN(DATEADD(month, ((@h + @L - 7 * @m + 114) / 31)-1, cast(cast(@Y AS VARCHAR) AS Datetime)) + ((@h + @L - 7 * @m + 114) % 31))
END
GO
CREATE TABLE [Auxiliary].[Calendar] (
-- This is the calendar table
[Date] datetime NOT NULL,
[Year] int NOT NULL,
[Quarter] int NOT NULL,
[Month] int NOT NULL,
[Week] int NOT NULL,
[Day] int NOT NULL,
[DayOfYear] int NOT NULL,
[Weekday] int NOT NULL,
[Fiscal_Year] int NOT NULL,
[Fiscal_Quarter] int NOT NULL,
[Fiscal_Month] int NOT NULL,
[KindOfDay] varchar(10) NOT NULL,
[Description] varchar(50) NULL,
PRIMARY KEY CLUSTERED ([Date])
)
GO
ALTER TABLE [Auxiliary].[Calendar]
-- In Celkoish style I'm manic about constraints (Never use em ;-))
-- http://www.celko.com/
ADD CONSTRAINT [Calendar_ck] CHECK ( ([Year] > 1900)
AND ([Quarter] BETWEEN 1 AND 4)
AND ([Month] BETWEEN 1 AND 12)
AND ([Week] BETWEEN 1 AND 53)
AND ([Day] BETWEEN 1 AND 31)
AND ([DayOfYear] BETWEEN 1 AND 366)
AND ([Weekday] BETWEEN 1 AND 7)
AND ([Fiscal_Year] > 1900)
AND ([Fiscal_Quarter] BETWEEN 1 AND 4)
AND ([Fiscal_Month] BETWEEN 1 AND 12)
AND ([KindOfDay] IN ('HOLIDAY', 'SATURDAY', 'SUNDAY', 'BANKDAY')))
GO
SET DATEFIRST 1;
-- I want my table to contain datedata acording to ISO 8601
-- http://en.wikipedia.org/wiki/ISO_8601
-- thus first day of a week is monday
WITH Dates(Date)
-- A recursive CTE that produce all dates between 1999 and 2020-12-31
AS
(
SELECT cast('1999' AS DateTime) Date -- SQL Server supports the ISO 8601 format so this is an unambigious shortcut for 1999-01-01
UNION ALL -- http://msdn2.microsoft.com/en-us/library/ms190977.aspx
SELECT (Date + 1) AS Date
FROM Dates
WHERE
Date < cast('2021' AS DateTime) -1
),
DatesAndThursdayInWeek(Date, Thursday)
-- The weeks can be found by counting the thursdays in a year so we find
-- the thursday in the week for a particular date
AS
(
SELECT
Date,
CASE DATEPART(weekday,Date)
WHEN 1 THEN Date + 3
WHEN 2 THEN Date + 2
WHEN 3 THEN Date + 1
WHEN 4 THEN Date
WHEN 5 THEN Date - 1
WHEN 6 THEN Date - 2
WHEN 7 THEN Date - 3
END AS Thursday
FROM Dates
),
Weeks(Week, Thursday)
-- Now we produce the weeknumers for the thursdays
-- ROW_NUMBER is new to SQL Server 2005
AS
(
SELECT ROW_NUMBER() OVER(partition by year(Date) order by Date) Week, Thursday
FROM DatesAndThursdayInWeek
WHERE DATEPART(weekday,Date) = 4
)
INSERT INTO Auxiliary.Calendar
SELECT
d.Date,
YEAR(d.Date) AS Year,
DATEPART(Quarter, d.Date) AS Quarter,
MONTH(d.Date) AS Month,
w.Week,
DAY(d.Date) AS Day,
DATEPART(DayOfYear, d.Date) AS DayOfYear,
DATEPART(Weekday, d.Date) AS Weekday,
-- Fiscal year may be different to the actual year in Norway the are the same
-- http://en.wikipedia.org/wiki/Fiscal_year
YEAR(d.Date) AS Fiscal_Year,
DATEPART(Quarter, d.Date) AS Fiscal_Quarter,
MONTH(d.Date) AS Fiscal_Month,
CASE
-- Holidays in Norway
-- For other countries and states: Wikipedia - List of holidays by country
-- http://en.wikipedia.org/wiki/List_of_holidays_by_country
WHEN (DATEPART(DayOfYear, d.Date) = 1) -- New Year's Day
OR (d.Date = Auxiliary.Computus(YEAR(Date))-7) -- Palm Sunday
OR (d.Date = Auxiliary.Computus(YEAR(Date))-3) -- Maundy Thursday
OR (d.Date = Auxiliary.Computus(YEAR(Date))-2) -- Good Friday
OR (d.Date = Auxiliary.Computus(YEAR(Date))) -- Easter Sunday
OR (d.Date = Auxiliary.Computus(YEAR(Date))+39) -- Ascension Day
OR (d.Date = Auxiliary.Computus(YEAR(Date))+49) -- Pentecost
OR (d.Date = Auxiliary.Computus(YEAR(Date))+50) -- Whitmonday
OR (MONTH(d.Date) = 5 AND DAY(d.Date) = 1) -- Labour day
OR (MONTH(d.Date) = 5 AND DAY(d.Date) = 17) -- Constitution day
OR (MONTH(d.Date) = 12 AND DAY(d.Date) = 25) -- Cristmas day
OR (MONTH(d.Date) = 12 AND DAY(d.Date) = 26) -- Boxing day
THEN 'HOLIDAY'
WHEN DATEPART(Weekday, d.Date) = 6 THEN 'SATURDAY'
WHEN DATEPART(Weekday, d.Date) = 7 THEN 'SUNDAY'
ELSE 'BANKDAY'
END KindOfDay,
CASE
-- Description of holidays in Norway
WHEN (DATEPART(DayOfYear, d.Date) = 1) THEN 'New Year''s Day'
WHEN (d.Date = Auxiliary.Computus(YEAR(Date))-7) THEN 'Palm Sunday'
WHEN (d.Date = Auxiliary.Computus(YEAR(Date))-3) THEN 'Maundy Thursday'
WHEN (d.Date = Auxiliary.Computus(YEAR(Date))-2) THEN 'Good Friday'
WHEN (d.Date = Auxiliary.Computus(YEAR(Date))) THEN 'Easter Sunday'
WHEN (d.Date = Auxiliary.Computus(YEAR(Date))+39) THEN 'Ascension Day'
WHEN (d.Date = Auxiliary.Computus(YEAR(Date))+49) THEN 'Pentecost'
WHEN (d.Date = Auxiliary.Computus(YEAR(Date))+50) THEN 'Whitmonday'
WHEN (MONTH(d.Date) = 5 AND DAY(d.Date) = 1) THEN 'Labour day'
WHEN (MONTH(d.Date) = 5 AND DAY(d.Date) = 17) THEN 'Constitution day'
WHEN (MONTH(d.Date) = 12 AND DAY(d.Date) = 25) THEN 'Cristmas day'
WHEN (MONTH(d.Date) = 12 AND DAY(d.Date) = 26) THEN 'Boxing day'
END Description
FROM DatesAndThursdayInWeek d
-- This join is for getting the week into the result set
inner join Weeks w
on d.Thursday = w.Thursday
OPTION(MAXRECURSION 0)
GO
CREATE FUNCTION Auxiliary.Numbers
(
@AFrom INT,
@ATo INT,
@AIncrement INT
)
RETURNS @RetNumbers TABLE
(
[Number] int PRIMARY KEY NOT NULL
)
AS
BEGIN
WITH Numbers(n)
AS
(
SELECT @AFrom AS n
UNION ALL
SELECT (n + @AIncrement) AS n
FROM Numbers
WHERE
n < @ATo
)
INSERT @RetNumbers
SELECT n from Numbers
OPTION(MAXRECURSION 0)
RETURN;
END
GO
CREATE FUNCTION Auxiliary.iNumbers
(
@AFrom INT,
@ATo INT,
@AIncrement INT
)
RETURNS TABLE
AS
RETURN(
WITH Numbers(n)
AS
(
SELECT @AFrom AS n
UNION ALL
SELECT (n + @AIncrement) AS n
FROM Numbers
WHERE
n < @ATo
)
SELECT n AS Number from Numbers
)
GO
答案 1 :(得分：11)
declare @date int
WITH CTE_DatesTable
AS
(
SELECT CAST('20000101' as date) AS [date]
UNION ALL
SELECT DATEADD(dd, 1, [date])
FROM CTE_DatesTable
WHERE DATEADD(dd, 1, [date]) <= '21001231'
)
SELECT [DWDateKey]=[date],[DayDate]=datepart(dd,[date]),[DayOfWeekName]=datename(dw,[date]),[WeekNumber]=DATEPART( WEEK , [date]),[MonthNumber]=DATEPART( MONTH , [date]),[MonthName]=DATENAME( MONTH , [date]),[MonthShortName]=substring(LTRIM( DATENAME(MONTH,[date])),0, 4),[Year]=DATEPART(YY,[date]),[QuarterNumber]=DATENAME(quarter, [date]),[QuarterName]=DATENAME(quarter, [date]) into DimDate FROM CTE_DatesTable
OPTION (MAXRECURSION 0);
答案 2 :(得分：8)
这样可以快速创建结果。
select top 100000 identity (int ,1,1) as Sequence into Tally from sysobjects , sys.all_columns
select dateadd(dd,sequence,-1) Dates into CalenderTable from tally
delete from CalenderTable where dates < -- mention the mindate you need
delete from CalenderTable where dates > -- mention the max date you need
步骤1：创建序列表
步骤2：使用序列表生成所需的日期
第3步：删除不需要的日期
答案 3 :(得分：3)
由于这只是标记sql（并不表示任何特定的DBMS），这里是Postgres的解决方案：
select d::date
from generate_series(date '1990-01-01', date '1990-01-01' + interval '100' year, interval '1' day) as t(d);
如果你需要那么多，那么将它存储在一个表格中会更有效率（例如可以将其编入索引）：
create table calendar
as
select d::date as the_date
from generate_series(date '1990-01-01', date '1990-01-01' + interval '100' year, interval '1' day) as t(d);
答案 4 :(得分：2)
此SQL Server用户定义函数可以有效地解决问题。没有递归，没有复杂的循环。生成需要很短的时间。
ALTER FUNCTION [GA].[udf_GenerateCalendar]
(
@StartDate DATE -- StartDate
, @EndDate DATE -- EndDate
)
RETURNS @Results TABLE
(
Date DATE
)
AS
/**********************************************************
Purpose: Generate a sequence of dates based on StartDate and EndDate
***********************************************************/
BEGIN
DECLARE @counter INTEGER = 1
DECLARE @days table(
day INTEGER NOT NULL
)
DECLARE @months table(
month INTEGER NOT NULL
)
DECLARE @years table(
year INTEGER NOT NULL
)
DECLARE @calendar table(
Date DATE NOT NULL
)
-- Populate generic days
SET @counter = 1
WHILE @counter <= 31
BEGIN
INSERT INTO @days
SELECT @counter dia
SELECT @counter = @counter + 1
END
-- Populate generic months
SET @counter = 1
WHILE @counter <= 12
BEGIN
INSERT INTO @months
SELECT @counter month
SELECT @counter = @counter + 1
END
-- Populate generic years
SET @counter = YEAR(@StartDate)
WHILE @counter <= YEAR(@EndDate)
BEGIN
INSERT INTO @years
SELECT @counter year
SELECT @counter = @counter + 1
END
INSERT @calendar (Date)
SELECT Date
FROM (
SELECT
CONVERT(Date, [Date], 102) AS Date
FROM (
SELECT
CAST(
y.year * 10000
+ m.month * 100
+ d.day
AS VARCHAR(8)) AS Date
FROM @days d, @months m, @years y
WHERE
ISDATE(CAST(
y.year * 10000
+ m.month * 100
+ d.day
AS VARCHAR(8))
) = 1
) A
) A
INSERT @Results (Date)
SELECT Date
FROM @calendar
WHERE Date BETWEEN @StartDate AND @EndDate
RETURN
/*
DECLARE @StartDate DATE = '2015-08-01'
DECLARE @EndDate DATE = '2015-08-31'
select * from [GA].[udf_GenerateCalendar](@StartDate, @EndDate)
*/
END