SQL Datediff - 查找行之间的 datediff

Question

我想使用sql查询数据库以显示id 1,2,3等之间的时间差异。基本上它会比较其下面的行的所有记录。任何帮助，将不胜感激。

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

谢谢!

Answer 1

如果使用 SQL Server，一种方法是:

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

另一种方法是使用 CTE 和 ROW_NUMBER()，如下所示:

;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