如何仅在一个 SELECT 语句中使用 DateDiff？答案

【问题标题】：How to use DateDiff into only one SELECT statement?如何仅在一个 SELECT 语句中使用 DateDiff？
【发布时间】：2016-12-26 16:13:21
【问题描述】：

我想对我的 SQL 查询中的 DATEDIFF 函数制作一个简短的版本。在我的代码中，我创建了两个临时表，然后在那里选择并使用 DATEDIFF 函数。

我希望简化这段代码，并且只使用一个 SELECT 语句来提供相同的结果。有可能吗？

这是我的结果：

这是我的 SQL 查询

DECLARE @Temp TABLE (ID int, Stamp datetime)

INSERT INTO @Temp (ID, Stamp) VALUES (1, '2016-08-17')
INSERT INTO @Temp (ID, Stamp) VALUES (1, GETDATE())
INSERT INTO @Temp (ID, Stamp) VALUES (1, GETDATE()+0.5)
INSERT INTO @Temp (ID, Stamp) VALUES (2, '2016-08-16')
INSERT INTO @Temp (ID, Stamp) VALUES (2, GETDATE())
INSERT INTO @Temp (ID, Stamp) VALUES (2, GETDATE()+3)

SELECT ROW_NUMBER() OVER (ORDER BY ID) as c, ID, Stamp INTO #Temp2 
FROM @Temp

SELECT ROW_NUMBER() OVER (ORDER BY ID) as d, ID, Stamp INTO #Temp3 
FROM @Temp

SELECT temp2.ID, temp2.Stamp, ISNULL(DATEDIFF(day, temp3.Stamp, temp2.Stamp),0) as DateDiff
FROM #Temp2 as temp2
LEFT JOIN #Temp3 as temp3 on temp2.ID = temp3.ID and temp2.c = temp3.d + 1

谢谢！

【问题讨论】：

如果您在 2012 年以上，请使用 Lead()/Lag()

标签： sql sql-server select datediff

【解决方案1】：

如果您使用的是 SQL Server 2012：

select * ,isnull(datediff(day,lag(stamp) over(partition by id order by stamp),stamp) ,0) 
from @temp t1

别的用这个..

;with cte
as
(select * ,row_number() over (partition by id order by stamp ) as rownum
from @temp t1
)
select c1.id,c1.stamp,isnull(datediff(day,c2.stamp,c1.stamp),0) as datee
 from cte c1
left join
cte c2
on c1.id=c2.id and c1.rownum=c2.rownum+1

【讨论】：

【解决方案2】：

试一试，

DECLARE @Temp TABLE (ID int, Stamp datetime)

INSERT INTO @Temp (ID, Stamp) VALUES (1, '2016-08-17')
INSERT INTO @Temp (ID, Stamp) VALUES (1, GETDATE())
INSERT INTO @Temp (ID, Stamp) VALUES (1, GETDATE()+0.5)
INSERT INTO @Temp (ID, Stamp) VALUES (2, '2016-08-16')
INSERT INTO @Temp (ID, Stamp) VALUES (2, GETDATE())
INSERT INTO @Temp (ID, Stamp) VALUES (2, GETDATE()+3)

;WITH CTE AS 
(
    SELECT ROW_NUMBER() OVER (ORDER BY ID) as RowNo, ID, Stamp 
    FROM @Temp
)

SELECT temp2.ID, temp2.Stamp, ISNULL(DATEDIFF(day, temp3.Stamp, temp2.Stamp),0) as DateDiff
FROM CTE as temp2
LEFT JOIN CTE as temp3 on temp2.ID = temp3.ID 
    AND temp2.RowNo = temp3.RowNo + 1

【讨论】：

【解决方案3】：

在 SQL Server 2012+ 中，您只需使用 lag()：

select t.*
       isnull(datediff(day, lag(stamp) over (partition by id order by stamp), stamp), 0)
from @temp t;

在早期版本中，我会使用outer apply：

select t.*,
       isnull(datediff(day, t2.stamp, t.stamp), 0)
from @temp t outer apply
     (select top 1 t2.*
      from @temp t2
      where t2.id = t.id and t2.stamp < t.stamp
      order by t2.stamp desc
     ) t2;

【讨论】：

【解决方案4】：

您可以删除插入临时表并在最终查询中使用子选择：

DECLARE @Temp TABLE (ID int, Stamp datetime)

INSERT INTO @Temp (ID, Stamp) VALUES (1, '2016-08-17')
INSERT INTO @Temp (ID, Stamp) VALUES (1, GETDATE())
INSERT INTO @Temp (ID, Stamp) VALUES (1, GETDATE()+0.5)
INSERT INTO @Temp (ID, Stamp) VALUES (2, '2016-08-16')
INSERT INTO @Temp (ID, Stamp) VALUES (2, GETDATE())
INSERT INTO @Temp (ID, Stamp) VALUES (2, GETDATE()+3)

SELECT temp2.ID, temp2.Stamp, ISNULL(DATEDIFF(day, temp3.Stamp, temp2.Stamp),0) as DateDiff
FROM (SELECT ROW_NUMBER() OVER (ORDER BY ID) as c, ID, Stamp FROM @Temp) as temp2
LEFT JOIN (SELECT ROW_NUMBER() OVER (ORDER BY ID) as d, ID, Stamp FROM @Temp) as temp3 
on temp2.ID = temp3.ID and temp2.c = temp3.d + 1

【讨论】：