我想在SQL Server 2000中生成datetime
范围。我想要与2000,2005和2008兼容的东西。如果它与2012兼容,那也会很棒。
所以,我在这里找到答案 - Generate Dates between date ranges。但是,我想要两列而不是1.我想要一个开始日期列和一个结束日期列。如何修改下面的查询才能获得它?
-- Declarations
DECLARE @dates TABLE(dt datetime)
DECLARE @dateFrom datetime
DECLARE @dateTo datetime
SET @dateFrom = '2001/01/01'
SET @dateTo = '2001/01/12'
-- Query:
WHILE(@dateFrom < @dateTo)
BEGIN
SELECT @dateFrom = DATEADD(day, 1,@dateFrom)
INSERT INTO @dates
SELECT @dateFrom
END
-- Output
SELECT * FROM @dates
我的代码 -
DECLARE @dates TABLE(StartDate datetime, EndDate datetime)
DECLARE @dateFrom datetime
DECLARE @dateTo datetime
DECLARE @StartDate datetime
DECLARE @EndDate datetime
SET @dateFrom = '2001/01/01'
SET @dateTo = '2001/01/12'
WHILE(@dateFrom < @dateTo)
BEGIN
SELECT @dateFrom, @dateFrom = DATEADD(day, 1,@dateFrom)
INSERT INTO @dates
SELECT @dateFrom
END
SELECT * FROM @dates
错误 - 为变量赋值的SELECT语句不能与数据检索操作结合使用。
答案 0 :(得分:5)
这是一个基于集合的解决方案,使用spt_values
,一个从2000年开始在所有SQL Server版本中找到的系统表 - &gt;在2000年,这将限制在256天的日期范围内。在更现代的版本中,这个上限将是2048。
通常,基于集合的解决方案比任何类型的WHILE
循环都更好。
DECLARE @start DATETIME, @end DATETIME;
SELECT @start = '20010101', @end = '20010112';
-- INSERT @dates(StartDate, EndDate)
SELECT
StartDate = DATEADD(DAY, number, @start),
EndDate = DATEADD(DAY, number+1, @start)
FROM master..spt_values
WHERE [type] = 'p'
AND number < DATEDIFF(DAY, @start, @end);
那就是说,特别是考虑到256天的限制,你应该考虑自己的数字和/或日历表来解决这个限制:
http://www.sqlperformance.com/generate-a-set-1
http://www.sqlperformance.com/generate-a-set-2
http://www.sqlperformance.com/generate-a-set-3
所以你可以使用以下代码创建一个代码(假设100,000天是你足够大的范围):
SELECT TOP 100000 Number = IDENTITY(INT,1,1)
INTO dbo.Numbers
FROM sysobjects AS s1
CROSS JOIN sysobjects AS s2
ORDER BY 1;
CREATE UNIQUE CLUSTERED INDEX n ON dbo.Numbers(Number);
以上在我的2000系统上花费了4.2秒。创建该表后,查询变得更加容易(并且限制性更小):
DECLARE @start DATETIME, @end DATETIME;
SELECT @start = '20010101', @end = '20010112';
-- INSERT @Dates(StartDate, EndDate)
SELECT
StartDate = DATEADD(DAY, Number-1, @start),
EndDate = DATEADD(DAY, Number, @start)
FROM dbo.Numbers
WHERE Number <= DATEDIFF(DAY, @start, @end);
答案 1 :(得分:0)
这使用数字表:
DECLARE @dateFrom datetime, @dateTo datetime
SELECT @dateFrom = '2001/01/01'
,@dateTo = '2001/01/12'
select
@dateFrom+Number-1
From Numbers
WHERE Number<=datediff(day,@dateFrom,@dateTo+1)
输出:
-----------------------
2001-01-01 00:00:00.000
2001-01-02 00:00:00.000
2001-01-03 00:00:00.000
2001-01-04 00:00:00.000
2001-01-05 00:00:00.000
2001-01-06 00:00:00.000
2001-01-07 00:00:00.000
2001-01-08 00:00:00.000
2001-01-09 00:00:00.000
2001-01-10 00:00:00.000
2001-01-11 00:00:00.000
2001-01-12 00:00:00.000
(12 row(s) affected)
我不确定OP想要什么“2”列,但在查询中很容易:
select
@dateFrom+Number-1,@dateFrom+Number
From Numbers
WHERE Number<=datediff(day,@dateFrom,@dateTo)
输出:
----------------------- -----------------------
2001-01-01 00:00:00.000 2001-01-02 00:00:00.000
2001-01-02 00:00:00.000 2001-01-03 00:00:00.000
2001-01-03 00:00:00.000 2001-01-04 00:00:00.000
2001-01-04 00:00:00.000 2001-01-05 00:00:00.000
2001-01-05 00:00:00.000 2001-01-06 00:00:00.000
2001-01-06 00:00:00.000 2001-01-07 00:00:00.000
2001-01-07 00:00:00.000 2001-01-08 00:00:00.000
2001-01-08 00:00:00.000 2001-01-09 00:00:00.000
2001-01-09 00:00:00.000 2001-01-10 00:00:00.000
2001-01-10 00:00:00.000 2001-01-11 00:00:00.000
2001-01-11 00:00:00.000 2001-01-12 00:00:00.000
如果您需要创建数字表,请参阅:What is the best way to create and populate a numbers table?
答案 2 :(得分:-4)
-- Declarations
DECLARE @dates TABLE(dt datetime, dt2 datetime)
DECLARE @dateFrom datetime
DECLARE @dateTo datetime
DECLARE @endDate datetime
SET @dateFrom = '2001/01/01'
SET @dateTo = '2001/01/12'
-- Query:
WHILE(@dateFrom < @dateTo)
BEGIN
SELECT @dateFrom = DATEADD(day, 1,@dateFrom)
SELECT @endDate = DATEADD(day, 1,@dateFrom)
INSERT INTO @dates
SELECT @dateFrom, @endDate
END
-- Output
SELECT * FROM @dates