[英]SQL query to get DateDiff of the last two records
I have a table called Event with eventNum as the primary key and date as a datetime2(7) in SQL Server 2008 R2. 我有一个名为Event的表,其中eventNum作为主键,日期作为SQL Server 2008 R2中的datetime2(7)。 I am trying to get the date of the last two rows in the table and get the difference in minutes. 我试图获取表中最后两行的日期,并在几分钟内得到差异。 This is what I currently have: 这就是我目前拥有的:
Select DATEDIFF(MI, e.date,(Select e2.date from Event e2 where eventNum = (Select MAX(e2.eventNum))))
From Event e
Where eventNum = (Select MAX(e.eventNum)-1 from e)
and I get this error: 我收到此错误:
Invalid column name 'Select eventNum from Event Where eventNum = Select MAX(eventNum) from Event'. 无效的列名称'从Event中选择eventNum,其中eventNum =从Event中选择MAX(eventNum)。
I've changed this 100 times and can't get it to work. 我已经改变了100次,无法让它发挥作用。 Any help? 有帮助吗?
You could use ROW_NUMBER
你可以使用ROW_NUMBER
WITH CTE AS
(
SELECT RN = ROW_NUMBER() OVER (ORDER BY eventNum DESC)
, date
FROM Event
)
SELECT Minutes = DATEDIFF(minute,
(SELECT date FROM CTE WHERE RN = 2),
(SELECT date FROM CTE WHERE RN = 1))
Fiddle: http://www.sqlfiddle.com/#!3/3e9c8/17/0 小提琴: http ://www.sqlfiddle.com/#!3 / 3e9c8/17 /0
This doesn't have to go through the table twice like Tim's answer. 这不必像蒂姆的回答那样经过两次。
select datediff(mi, min(x.date), max(x.date))
from (
select top(2) *
from Event e
order by eventNum desc
) x
Assuming you always have 2 records or more, and the time is monotonously increasing, then the above works. 假设你总是有2个或更多的记录,并且时间单调增加,那么上述工作。
eg 例如
select top(1) datediff(mi, x.date, y.date)
from event x
join event y on y.eventnum < x.eventnum
order by x.eventnum desc, y.eventnum desc
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