Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
269 views
in Technique[技术] by (71.8m points)

sql server - SQL Datediff - find datediff between rows

I would like to query a database using sql to show the difference in time between id 1,2,3 and so on. basically it will compare the row below it for all records. any help would be appreciated.

IDCODE  DATE TIME        DIFFERENCE (MINS)
1      02/03/2011 08:00        0
2      02/03/2011 08:10        10
3      02/03/2011 08:23        13
4       02/03/2011 08:25        2
5       02/03/2011 09:25        60
6       02/03/2011 10:20        55
7       02/03/2011 10:34        14

Thanks!

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

If using SQL Server, one way is to do:

DECLARE @Data TABLE (IDCode INTEGER PRIMARY KEY, DateVal DATETIME)
INSERT @Data VALUES (1, '2011-03-02 08:00')
INSERT @Data VALUES (2, '2011-03-02 08:10')
INSERT @Data VALUES (3, '2011-03-02 08:23')
INSERT @Data VALUES (4, '2011-03-02 08:25')
INSERT @Data VALUES (5, '2011-03-02 09:25')
INSERT @Data VALUES (6, '2011-03-02 10:20')
INSERT @Data VALUES (7, '2011-03-02 10:34')

SELECT t1.IDCode, t1.DateVal, ISNULL(DATEDIFF(mi, x.DateVal, t1.DateVal), 0) AS Mins
FROM @Data t1
    OUTER APPLY (
        SELECT TOP 1 DateVal FROM @Data t2 
        WHERE t2.IDCode < t1.IDCode ORDER BY t2.IDCode DESC) x

Another way is using a CTE and ROW_NUMBER(), like this:

;WITH CTE AS (SELECT ROW_NUMBER() OVER (ORDER BY IDCode) AS RowNo, IDCode, DateVal FROM @Data)

SELECT t1.IDCode, t1.DateVal, ISNULL(DATEDIFF(mi, t2.DateVal, t1.DateVal), 0) AS Mins
FROM CTE t1
    LEFT JOIN CTE t2 ON t1.RowNo = t2.RowNo + 1
ORDER BY t1.IDCode

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...