按组 SQL 获取上一条记录

Question

我有下表：

ID  GROUPID     ODATE       OTIME       OVALUE
1   A           2014-05-31  00:00:00    1207432.6
2   A           2014-05-31  01:00:00    1209064     
3   A           2014-05-31  02:00:00    1210698     
4   A           2014-05-31  03:00:00    1212333.3   
5   A           2014-05-31  04:00:00    1213967.7   
6   B           2014-05-31  00:00:00    2110016     
7   B           2014-05-31  01:00:00    2110016     
8   B           2014-05-31  02:00:00    2110016     
9   B           2014-05-31  03:00:00    2110016     
10  B           2014-05-31  04:00:00    2110016     
11  C           2014-05-31  00:00:00    2326592.6   
12  C           2014-05-31  01:00:00    2328088.8
13  C           2014-05-31  02:00:00    2329590.3   
14  C           2014-05-31  03:00:00    2331094.5   
15  C           2014-05-31  04:00:00    2332598

然后我运行这个语法：

SELECT 
A.ID, A.GroupID, A.oDate, A.oTime, 
A.oValue, MAX(B.oValue) AS Prev_oValue, A.oValue - MAX(B.oValue) AS oResult
FROM
Table1 AS A LEFT OUTER JOIN Table1 AS B ON B.GroupID = A.GroupID AND B.oValue < A.oValue
GROUP BY
A.ID, A.GroupID, A.oDate, A.oTime, A.oValue
ORDER BY A.GroupID, A.oDate, A.oTime

我想得到以下结果：

ID  GROUPID     ODATE       OTIME       OVALUE      PREV_OVALUE     ORESULT
1   A           2014-05-31  00:00:00    1207432.6   (null)          (null)
2   A           2014-05-31  01:00:00    1209064     1207432.6       1631.4
3   A           2014-05-31  02:00:00    1210698     1209064         1634
4   A           2014-05-31  03:00:00    1212333.3   1210698         1635.3
5   A           2014-05-31  04:00:00    1213967.7   1212333.3        1634.4
6   B           2014-05-31  00:00:00    2110016     (null)          (null)
7   B           2014-05-31  01:00:00    2110016     2110016         0
8   B           2014-05-31  02:00:00    2110016     2110016         0
9   B           2014-05-31  03:00:00    2110016     2110016         0
10  B           2014-05-31  04:00:00    2110016     2110016         0
11  C           2014-05-31  00:00:00    2326592.6   (null)          (null)
12  C           2014-05-31  01:00:00    2328088.8   2326592.6       1496.2
13  C           2014-05-31  02:00:00    2329590.3   2328088.8       1501.5
14  C           2014-05-31  03:00:00    2331094.5   2329590.3       1504.2
15  C           2014-05-31  04:00:00    2332598     2331094.5       1503.5

查看fiddle

我想要的是，根据 GroupID 列和 Order by Date and Time 列获取先前的值。在我得到之前的值之后，当前记录减去之前的值作为结果。但是出了点问题，结果很糟糕。一些记录获得先前的值，而一些记录则没有。我无法理解。

有谁知道如何做到这一点？

谢谢。

Answer 1

您可以使用带有 ROW_NUMBER() 的公用表表达式来按时间顺序对组内的每一行进行编号。获取前一个值就像将 cte 与自身连接起来一样简单，以获取具有相同 groupid 和行号少一的行。类似的东西；

WITH cte AS (
  SELECT groupid, odate, otime, ovalue,
    ROW_NUMBER() OVER (PARTITION BY groupid ORDER BY odate, otime) rn
  FROM table1
)
SELECT a.groupid, a.odate, a.otime, a.ovalue, b.ovalue Prev_oValue,
       a.ovalue-b.ovalue oResult
FROM cte a
LEFT JOIN cte b
  ON a.groupid = b.groupid
 AND a.rn = b.rn + 1
ORDER BY a.groupid, a.odate, a.otime

An SQLfiddle to test with.