t-sql获取2个日期之间的所有日期

Question

  

可能重复：
  Getting Dates between a range of dates

假设我有2个日期（仅限日期部分，没有时间），我希望获得这两个日期之间的所有日期，并将其插入表格中。是否有一种简单的方法可以使用SQL语句（即没有循环）？

Ex:
Date1: 2010-12-01
Date2: 2010-12-04

Table should have following dates:
2010-12-01, 2010-12-02, 2010-12-03, 2010-12-04

Answer 1

假设SQL Server 2005+，使用递归查询：

WITH sample AS (
  SELECT CAST('2010-12-01' AS DATETIME) AS dt
  UNION ALL
  SELECT DATEADD(dd, 1, dt)
    FROM sample s
   WHERE DATEADD(dd, 1, dt) <= CAST('2010-12-04' AS DATETIME))
SELECT * 
  FROM sample

返回：

 dt
 ---------
 2010-12-01 00:00:00.000
 2010-12-02 00:00:00.000
 2010-12-03 00:00:00.000
 2010-12-04 00:00:00.000

使用CAST/CONVERT to format as you like。

使用参数启动＆amp;端：

INSERT INTO dbo.YOUR_TABLE
  (datetime_column)
WITH sample AS (
    SELECT @start_date AS dt
    UNION ALL
    SELECT DATEADD(dd, 1, dt)
      FROM sample s
     WHERE DATEADD(dd, 1, dt) <= @end_date)
SELECT s.dt
  FROM sample s

Answer 2

你需要一个数字表。如果你没有永久的那个，那么a more efficient生成一个方法比使用递归CTE。只要从缓冲区缓存中读取永久文件就会更有效率。

DECLARE @D1 DATE = '2010-12-01'
DECLARE @D2 DATE = '2010-12-04'

;WITH 
L0 AS (SELECT 1 AS c UNION ALL SELECT 1),
L1 AS (SELECT 1 AS c FROM L0 A CROSS JOIN L0 B),
L2 AS (SELECT 1 AS c FROM L1 A CROSS JOIN L1 B),
L3 AS (SELECT 1 AS c FROM L2 A CROSS JOIN L2 B),
L4 AS (SELECT 1 AS c FROM L3 A CROSS JOIN L3 B),
Nums AS (SELECT ROW_NUMBER() OVER (ORDER BY (SELECT 0)) AS i FROM L4)
SELECT DATEADD(day,i-1,@D1)
 FROM Nums where i <= 1+DATEDIFF(day,@D1,@D2)

Answer 3

我刚刚做了这样的事情：

declare @dt datetime = '2010-12-01'
declare @dtEnd datetime = '2010-12-04'

WHILE (@dt < @dtEnd) BEGIN
    insert into table(datefield)
        values(@dt)
    SET @dt = DATEADD(day, 1, @dt)
END

Answer 4

重复提问

Getting Dates between a range of dates

DECLARE @DateFrom smalldatetime, @DateTo smalldatetime;
SET @DateFrom='20000101';
SET @DateTo='20081231';
-------------------------------
WITH T(date)
AS
( 
SELECT @DateFrom 
UNION ALL
SELECT DateAdd(day,1,T.date) FROM T WHERE T.date < @DateTo
)
SELECT date FROM T OPTION (MAXRECURSION 32767);