具有相同ID的多行之间的SQL差异

Question

SQL 服务器 2012

原始数据

ID   VAL   Time
+---+----+---------------------+
| 2 |  1 | 2015-05-09 12:54:39 |
| 3 | 10 | 2015-05-09 12:54:39 |
| 2 |  1 | 2015-05-09 12:56:39 |
| 3 | 10 | 2015-05-09 12:56:39 |
| 2 |  5 | 2015-05-09 13:48:30 |
| 3 | 16 | 2015-05-09 13:48:30 |
| 2 |  7 | 2015-05-09 15:01:09 |
| 3 | 20 | 2015-05-09 15:01:09 |
+---+----+---------------------+ 

我有一张表，其中 VAL 会随着时间的推移而永远增加。我想操纵数据以显示随着时间的推移每个 ID 的 VAL 增加了多少。所以时间 2 的值 - 时间 1 的值

理想结果：

ID   VALI   Time
+---+----+---------------------+
| 2 |  0 | 2015-05-09 12:56:39 |
| 3 |  0 | 2015-05-09 12:56:39 |
| 2 |  4 | 2015-05-09 13:48:30 |
| 3 |  6 | 2015-05-09 13:48:30 |
| 2 |  2 | 2015-05-09 15:01:09 |
| 3 |  4 | 2015-05-09 15:01:09 |
+---+----+---------------------+ 

到目前为止的代码：

select
t1.Time,t1.[ID],t2.[VAL]-t1.[VAL] AS [ValI]
from #tempTable t1
inner join #tempTable t2 ON t1.[ID]=t2.[ID]
AND t1.[Time]<t2.[Time]

我需要计算当前时间戳与仅当前时间戳之前的时间之间的差异，而不是当前时间戳之前的所有时间戳。到目前为止，当 VAL 没有改变时，我得到了很多重复值。

Answer 1

你可以用这个。

DECLARE @MyTable TABLE (ID INT,   VAL INT,  [Time] DATETIME)
INSERT INTO @MyTable VALUES
(2,  1 ,'2015-05-09 12:54:39'),
(3, 10 ,'2015-05-09 12:54:39'),
(2,  1 ,'2015-05-09 12:56:39'),
(3, 10 ,'2015-05-09 12:56:39'),
(2,  5 ,'2015-05-09 13:48:30'),
(3, 16 ,'2015-05-09 13:48:30'),
(2,  7 ,'2015-05-09 15:01:09'),
(3, 20 ,'2015-05-09 15:01:09')


;WITH CTE AS (
    SELECT *, ROW_NUMBER() OVER(PARTITION BY ID ORDER BY [Time]) RN FROM @MyTable
)
SELECT T1.ID, T2.VAL - T1.VAL AS VALI, T2.Time FROM CTE T1
    INNER JOIN CTE T2 ON T1.ID = T2.ID AND T1.RN = T2.RN - 1
ORDER BY T1.[Time], T1.ID

结果：

ID          VALI        Time
----------- ----------- -----------------------
2           0           2015-05-09 12:56:39.000
3           0           2015-05-09 12:56:39.000
2           4           2015-05-09 13:48:30.000
3           6           2015-05-09 13:48:30.000
2           2           2015-05-09 15:01:09.000
3           4           2015-05-09 15:01:09.000

Answer 2

这应该可以工作：

 select id, time, val-prevval val1 from (
 select * , lag(val, 1, 0) over(partition by  id order by val, time) prevVal from #Temp)A
 order by time

Answer 3

您可以首先在您的#tempTable 上放置一个排名，按时间降序排列，并按 ID 分区。

那么你的加入变成这样：

select
    t1.Time,
    t1.[ID],
    t1.[VAL] - t2.[VAL] AS [ValI]
from #tempTable t1
inner join #tempTable t2 ON t1.[ID] = t2.[ID]
    AND t2.Rank = (t1.Rank + 1)

Answer 4

LAG() 在 SQL 2012 中可用。这允许您获取当前行的 val 并从前一行中减去 val，按 id 分组并按 Time 排序。这将返回前两行的NULL，因为它们没有以前的记录可供比较。您可以通过将查询放在子选择中然后应用WHERE valDiff IS NULL 来排除它们，或者您可以使用LAG() > LAG(Val,1,0) 的第三个参数默认valDiff 将前两行默认为0 .

SQL Fiddle

MS SQL Server 2017 架构设置：

CREATE TABLE t1 ( ID int, VAL int, [Time] datetime) ;

INSERT INTO t1 ( ID, Val, [Time] )
VALUES 
    ( 2, 1 , '2015-05-09 12:54:39')
  , ( 3, 10, '2015-05-09 12:54:39')
  , ( 2, 1 , '2015-05-09 12:56:39')
  , ( 3, 10, '2015-05-09 12:56:39')
  , ( 2, 5 , '2015-05-09 13:48:30')
  , ( 3, 16, '2015-05-09 13:48:30')
  , ( 2, 7 , '2015-05-09 15:01:09')
  , ( 3, 20, '2015-05-09 15:01:09')
;

查询 1：

SELECT s1.ID
  , s1.ValDiff
  , FORMAT(s1.[Time], 'yyyy-MM-dd hh:mm:ss') AS fTime
FROM (
  SELECT ID
    , Val - LAG(Val,1) OVER ( PARTITION BY ID ORDER BY [Time],ID ) AS ValDiff
    , [Time]
  FROM t1
) s1
WHERE s1.valDiff IS NOT NULL
ORDER BY s1.[Time],s1.ID

Results：

| ID |    ValI |               fTime |
|----|---------|---------------------|
|  2 |       0 | 2015-05-09 12:56:39 |
|  3 |       0 | 2015-05-09 12:56:39 |
|  2 |       4 | 2015-05-09 01:48:30 |
|  3 |       6 | 2015-05-09 01:48:30 |
|  2 |       2 | 2015-05-09 03:01:09 |
|  3 |       4 | 2015-05-09 03:01:09 |

Answer 5

如果你有LAG

DEMO

   SELECT
        id
      , val - LAG(val, 1) OVER (PARTITION BY id ORDER BY time ASC) AS VALI
      , time
    FROM #TempTable
    ORDER BY time ASC, ID ASC