伙计们我有一个表有一个名为time的列.它捕获每个记录条目进入数据库的时间.我想查询并返回另一列,显示一个条目和它之前的条目之间的持续时间.例如,如果我今天中午12点为约翰存储记录,然后在下午1:10存储阿里,我想要另一列显示01:10:00(即HH:MM:SS).
据我所知,我可以按如下方式查询每个列号.
SELECT ROW_NUMBER() OVER (ORDER BY [followuptime]) from [dbo].[FollowUp] .
我想查询最大行号AS,但它失败并返回错误“windowed ….”
SELECT MAX(ROW_NUMBER() OVER (ORDER BY [followuptime])) from [dbo].[FollowUp] .
我想使用DATEDIFF(interval,start_time,end_time); sql的功能,但就像现在一样,我被卡住了.非常感谢您的帮助或任何其他选择.
最佳答案 由于SQL-Server 2008R2不支持LAG / LEAD,因此您需要使用row_number进行自联接以获取上一行的时间:
WITH OrderedResults AS
( SELECT [id],
[followuptime],
[remark],
RowNumber = ROW_NUMBER() OVER (ORDER BY [followuptime])
FROM [dbo].[FollowUp]
)
SELECT a.ID,
a.FollowUpTime,
a.Remark,
PreviousTime = b.FollowUpTime,
MinutesDifference = DATEDIFF(MINUTE, b.FollowUpTime, a.FollowUpTime)
FROM OrderedResults a
LEFT JOIN OrderedResults b
ON b.RowNumber = a.RowNumber - 1
ORDER BY a.FollowUpTime;