如何在sql中选择连续日期

Question

是否有任何功能可以检查连续日期。我在处理以下问题时遇到问题：

我的表有一个 datetime 列，其中包含以下数据：

----------
2015-03-11
2015-03-12
2015-03-13
2015-03-16

给定开始日期为2015-3-11，结束日期为2015-3-17。我希望结果为：

----------
2015-03-11
2015-03-12
2015-03-13

任何人都可以提出任何建议吗？

Answer 1

我认为这在某种程度上是 Grouping Islands of Contiguous Dates 问题的变体。这可以使用ROW_NUMBER()：

SQL Fiddle

CREATE TABLE Test(
    tDate   DATETIME
)
INSERT INTO Test VALUES
('20150311'), ('20150312'), ('20150313'), ('20150316');

DECLARE @startDate  DATE = '20150311'
DECLARE @endDate    DATE = '20150317'

;WITH Cte AS(
    SELECT
        *,
        RN = DATEADD(DD, - (ROW_NUMBER() OVER(ORDER BY tDATE) - 1), tDate)
    FROM Test
    WHERE 
        tDate >= @startDate
        AND tDate < DATEADD(DAY, 1, @endDate)
)
SELECT CAST(tDate AS DATE)
FROM CTE
WHERE RN = @startDate

结果

|------------|
| 2015-03-11 |
| 2015-03-12 |
| 2015-03-13 |

---

这是 SQL Server 2005 版本：

SQL Fiddle

DECLARE @startDate  DATETIME
DECLARE @endDate    DATETIME

SET @startDate  = '20150311'
SET @endDate    = '20150317'

;WITH Cte AS(
    SELECT
        *,
        RN = DATEADD(DD, -(ROW_NUMBER() OVER(ORDER BY tDATE)-1), tDate)
    FROM Test
    WHERE 
        tDate >= @startDate
        AND tDate < DATEADD(DAY, 1, @endDate)
)
SELECT CONVERT(VARCHAR(10), tDate, 121)
FROM CTE
WHERE RN = @startDate

Answer 2

对于MSSQL 2012。这将返回 MAX 个连续组：

DECLARE @t TABLE(d DATE)

INSERT INTO @t VALUES
('20150311'),
('20150312'),
('20150313'),
('20150316')


;WITH
c1 AS(SELECT d, IIF(DATEDIFF(dd,LAG(d, 1, DATEADD(dd, -1, d)) OVER(ORDER BY d), d) = 1, 0, 1) AS n FROM @t),
c2 AS(SELECT d, SUM(n) OVER(ORDER BY d) AS n FROM c1) 

SELECT TOP 1 WITH TIES MIN(d) AS StartDate, MAX(d) AS EndDate, COUNT(*) AS DayCount
FROM c2
GROUP BY n
ORDER BY DayCount desc

输出：

StartDate   EndDate     DayCount
2015-03-11  2015-03-13  3

对于

('20150311'),
('20150312'),
('20150313'),
('20150316'),
('20150317'),
('20150318'),
('20150319'),
('20150320')

输出：

StartDate   EndDate     DayCount
2015-03-16  2015-03-20  5

在c1 CTE中应用过滤：

c1 AS(SELECT d, IIF(DATEDIFF(dd,LAG(d, 1, DATEADD(dd, -1, d)) OVER(ORDER BY d), d) = 1, 0, 1) AS n FROM @t WHERE d BETWEEN '20150311' AND '20150320'),

对于MSSQL 2008：

;WITH
c1 AS(SELECT d, (SELECT MAX(d) FROM @t it WHERE it.d < ot.d) AS pd FROM @t ot),
c2 AS(SELECT d, CASE WHEN DATEDIFF(dd,ISNULL(pd, DATEADD(dd, -1, d)),  d) = 1 THEN  0 ELSE 1 END AS n FROM c1),
c3 AS(SELECT d, (SELECT SUM(n) FROM c2 ci WHERE ci.d <= co.d)  AS n FROM c2 co) 

SELECT TOP 1 WITH TIES MIN(d) AS StartDate, MAX(d) AS EndDate, COUNT(*) AS DayCount
FROM c3
GROUP BY n
ORDER BY DayCount desc

Answer 3

您不需要像其他答案所说的那样声明任何开始日期或结束日期，您需要一个 row_num 和 datediff 函数：

create table DateFragTest (cDate date);
insert into DateFragTest 
       values ('2015-3-11'),
              ('2015-3-12'),
              ('2015-3-13'),
              ('2015-3-16')

   with cte as 
      (select 
        cDate,
        row_number() over (order by cDate ) as rn
       from
        DateFragTest)
    select cDate 
    from cte t1  
    where datediff(day,
                   (select cDate from cte t2 where t2.rn=t1.rn+1),
                   t1.cDate)<>1

输出：

cDate
2015-03-11
2015-03-12
2015-03-13

SQLFIDDLE DEMO

Answer 4

对于sql server 2012-

WITH cte
AS
(
    SELECT [datex]
    ,      lead([datex]) OVER ( ORDER BY [datex]) lead_datex
    ,      datediff(dd,[datex],lead([datex]) OVER ( ORDER BY [datex]) ) AS diff
    FROM [dbo].[datex]
)
SELECT c.[datex]
FROM [cte] AS c
WHERE diff >=1

Answer 5

使用BETWEEN

查询将如下所示：

SELECT * 
FROM your_table_name
WHERE your_date_column_name BETWEEN '2015-3-11' AND '2015-3-13'

Answer 6

(dt between x and y) 或只是(dt >= x and dt <= y)。