日期范围的简单SQL查询

Question

我需要一种方法来找出每组相同ID集的天数。例如，如果有两行ID1，第一个日期是第一个日期，第二个日期是第四个日期，则1st的计算将为9月4日 - 9月1日（因为4日是ID1的下一个日期），而第4天的计算方法是今天（9月28日） - 9月4日（因为第4天是ID1的最后日期，而ID1的第4天之后没有更多日期）。我在下面列出了每列的计算公式。

IDs Date   
ID1 09/01/2015  
ID1 09/04/2015 
ID2 09/04/2015  
ID2 09/09/2015 
ID2 09/15/2015 
ID3 09/09/2015 
ID4 09/15/2015 

要

IDs Date   Days  Formula... 
ID1 09/01/2015  3   Sept 4th - Sept 1st 
ID1 09/04/2015  22  Today - Sept 4th 
ID2 09/04/2015  5   Sept 9th - Sept 4th 
ID2 09/09/2015  6   Sept 15th - Sept 9th 
ID2 09/15/2015  13  Today - Sept 15th 
ID3 09/09/2015  19  Today - Sept 9th 
ID4 09/15/2015  13  Today - Sept 15th 

Answer 1

你基本上是寻找lead()功能。而且，你没有它，所以你必须即兴创作。下面是使用cross apply的方法：

select t.*, 
     datediff(day, t.date, coalesce(n.date, getdate()) 
from table t outer apply 
    (select top 1 t2.date 
     from table t2 
     where t2.id = t.id and t2.date > t.date 
     order by t2.date 
    ) n; 

这应该与(id, date)指数合理的性能。

Answer 2

可以使用ROW_NUMBER与LEFT JOIN一起做到这一点：

SQL Fiddle

WITH Cte AS(
    SELECT *, 
    rn = ROW_NUMBER() OVER(PARTITION BY ID ORDER BY [Date]) 
    FROM tbl 
) 
SELECT 
    t1.*, 
    Days = DATEDIFF(DAY, t1.Date, ISNULL(t2.Date, GETDATE())) 
FROM Cte t1 
LEFT JOIN Cte t2 
    ON t1.ID = t2.ID 
    AND t1.rn = t2.rn - 1 

Answer 3

使用公共表表达式，你可以安排你的表与一个开始日期和结束日期为每个ID记录一个临时视图，就像这样：http://sqlfiddle.com/#!3/af078/7

with t1 (id, dt, ndx) as (
    select id, dt, row_number() over (partition by id order by dt) 
    from id_dates 
) 
, 
t2 (id, startdt, enddt) as (
    select t1.id, t1.dt, coalesce(t2.dt, convert(date, getdate())) 
    from t1 
    left join t1 t2 
    on t2.id = t1.id 
    and t2.ndx = t1.ndx + 1 
) 
select * from t2 

确保使用的​​3210功能获取最后一行的当前日期（否则它将从左连接中为空）。

从那里，它是一个使用datediff()功能的一个简单的问题：http://sqlfiddle.com/#!3/af078/8

with t1 (id, dt, ndx) as (
    select id, dt, row_number() over (partition by id order by dt) 
    from id_dates 
) 
, 
t2 (id, startdt, enddt) as (
    select t1.id, t1.dt, coalesce(t2.dt, convert(date, getdate())) 
    from t1 
    left join t1 t2 
    on t2.id = t1.id 
    and t2.ndx = t1.ndx + 1 
) 
select id, startdt, datediff(day, startdt, enddt) as dtcount 
from t2 

Answer 4

这里是工作的例子，你可以这样做没有任何连接。 只需使用Lead窗口功能即可。

--Here is your data 
DECLARE @Data TABLE(ID NVARCHAR(8),Date DATE) 
INSERT INTO @Data(ID,Date) 
VALUES 
('ID1','09/01/2015'), 
('ID1','09/04/2015'), 
('ID2','09/04/2015'), 
('ID2','09/09/2015'), 
('ID2','09/15/2015'), 
('ID3','09/09/2015'), 
('ID4','09/15/2015') 

--This is all you need, Assuming your using a version of SQL that supports LEAD!!! 
SELECT ID,DATEDIFF(D,Date,ISNULL(LEAD(Date) OVER(PARTITION BY ID ORDER BY Date),GETDATE())) 
FROM  @Data