SQL server执行类似分组任务的操作

我有一个带有SQL server的表，如下所示

Date        Value
---------------------------------------------------
08-01-2016    1
08-02-2016    1
08-03-2016    1
08-04-2016    1
08-05-2016    1
08-06-2016    2
08-07-2016    2
08-08-2016    2
08-09-2016    2.5
08-10-2016    1
08-11-2016    1

由于原始表太大，即使我使用了“Results to file”，它仍然会引发异常“System.OutOfMemoryException”。这就是为什么我想把桌子组织成这样

但我没有一个好的逻辑来处理。因此，我想将表更改为如下所示的类型

Date_from      Date_to      Value
-------------------------------------------------
08-01-2016     08-05-2016   1
08-06-2016     08-08-2016   2
08-09-2016     08-09-2016   2.5
08-10-2016     08-11-2016   1

我很欣赏你的想法

通常称为群岛问题。这里有一个技巧可以做到这一点

;WITH data
    AS (SELECT *,Lag(Value, 1)OVER(ORDER BY Dates) [pVal]
        FROM   (VALUES ('08-01-2016',1 ),
                    ('08-02-2016',1 ),
                    ('08-03-2016',1 ),
                    ('08-04-2016',1 ),
                    ('08-05-2016',1 ),
                    ('08-06-2016',2 ),
                    ('08-07-2016',2 ),
                    ('08-08-2016',2 ),
                    ('08-09-2016',2.5 ),
                    ('08-10-2016',1 ),
                    ('08-11-2016',1 )) tc (Dates, Value)),
     intr
     AS (SELECT Dates,
                Value,
                Sum(Iif(pVal = Value, 0, 1)) OVER(ORDER BY Dates) AS [Counter]
         FROM   data)
SELECT Min(Dates) AS Dates_from,
       Max(Dates) AS Dates_to,
       Value
FROM   intr
GROUP  BY [Counter],
          Value 

---

累积和/滞后法是一种方法。在这种情况下，更简单的方法是：

select min(date) as date_from, max(date) as date_to, value
from (select t.*,
             dateadd(day, - row_number() over (partition by value order by date),date) as grp
      from t
     ) t
group by value, grp;

这使用了日期连续且没有间隔的观察结果。因此，从日期中减去一个序列将产生一个常数——当

值相同时。
以下是一个示例：

DECLARE @T TABLE (
    [Date] DATE,
    [Value] DECIMAL(9,2)
)

INSERT @T VALUES
( '08-01-2016', 1 ),
( '08-02-2016', 1 ),
( '08-03-2016', 1 ),
( '08-04-2016', 1 ),
( '08-05-2016', 1 ),
( '08-06-2016', 2 ),
( '08-07-2016', 2 ),
( '08-08-2016', 2 ),
( '08-09-2016', 2.5 ),
( '08-10-2016', 1 ),
( '08-11-2016', 1 )

SELECT * FROM @T

SELECT A.[Date] StartDate, B.[Date] EndDate, A.[Value] FROM (
    SELECT A.*, ROW_NUMBER() OVER (ORDER BY A.[Date], A.[Value]) O FROM @T A
    LEFT JOIN @T B ON B.[Value] = A.[Value] AND B.[Date] = DATEADD(d, -1, A.[Date])
    WHERE B.[Date] IS NULL
) A
JOIN (
    SELECT A.*, ROW_NUMBER() OVER (ORDER BY A.[Date], A.[Value]) O FROM @T A
    LEFT JOIN @T B ON B.[Value] = A.[Value] AND B.[Date] = DATEADD(d, 1, A.[Date])
    WHERE B.[Date] IS NULL
) B ON B.O = A.O

Prdp的解决方案非常好，但如果有人仍在使用SQL Server 2008，而在SQL Server 2008中，LAG（）和并行数据仓库（PDW）功能不可用，那么这里有一个替代方案：

样本数据：

IF OBJECT_ID('tempdb..#Temp') IS NOT NULL
    DROP TABLE #Temp;

CREATE TABLE #Temp([Dates] DATE
              , [Value] FLOAT);

INSERT INTO      #Temp([Dates]
                 , [Value])
VALUES
      ('08-01-2016'
     , 1),
      ('08-02-2016'
     , 1),
      ('08-03-2016'
     , 1),
      ('08-04-2016'
     , 1),
      ('08-05-2016'
     , 1),
      ('08-06-2016'
     , 2),
      ('08-07-2016'
     , 2),
      ('08-08-2016'
     , 2),
      ('08-09-2016'
     , 2.5),
      ('08-10-2016'
     , 1),
      ('08-11-2016'
     , 1); 

查询：

;WITH Seq
    AS (SELECT SeqNo = ROW_NUMBER() OVER(ORDER BY [Dates]
                                        , [Value])
            , t.Dates
            , t.[Value]
        FROM   #Temp t)
    SELECT StartDate = MIN([Dates])
        , EndDate = MAX([Dates])
        , [Value]
    FROM
          (SELECT [Value]
               , [Dates]
               , SeqNo
               , rn = SeqNo - ROW_NUMBER() OVER(PARTITION BY [Value] ORDER BY SeqNo)
           FROM   Seq s) a
    GROUP BY [Value]
          , rn
    ORDER BY StartDate;

结果:

你能解释一下你如何分组数据的逻辑吗？你使用的是哪个版本的sql server？@Prdp sql server 2014管理研究你在应用程序（VB或C）中使用这个吗？在代码中处理此问题可以正常工作，并且不会引发“内存不足”异常。否则，您将需要一个存储过程，在该过程中，您将遍历所有记录以生成结果。@DForck42您是指我如何得到这种结果的？我不能，所以我向你们征求意见~谢谢~@Gordon Linoff-对不起，我声明值数据类型为INT，现在我声明值数据类型为decimal，然后进行检查。它工作正常。@Mansoor。对我来说，这似乎是最简单的解决办法。