我有一个表,我想获得每组的最新条目。下面是表格:

DocumentStatusLogs表

|ID| DocumentID | Status | DateCreated |
| 2| 1          | S1     | 7/29/2011   |
| 3| 1          | S2     | 7/30/2011   |
| 6| 1          | S1     | 8/02/2011   |
| 1| 2          | S1     | 7/28/2011   |
| 4| 2          | S2     | 7/30/2011   |
| 5| 2          | S3     | 8/01/2011   |
| 6| 3          | S1     | 8/02/2011   |

该表将按documententid分组,并按DateCreated降序排序。对于每个documententid,我希望获得最新的状态。

我的首选输出:

| DocumentID | Status | DateCreated |
| 1          | S1     | 8/02/2011   |
| 2          | S3     | 8/01/2011   |
| 3          | S1     | 8/02/2011   |

Is there any aggregate function to get only the top from each group? See pseudo-code GetOnlyTheTop below: SELECT DocumentID, GetOnlyTheTop(Status), GetOnlyTheTop(DateCreated) FROM DocumentStatusLogs GROUP BY DocumentID ORDER BY DateCreated DESC If such function doesn't exist, is there any way I can achieve the output I want? Or at the first place, could this be caused by unnormalized database? I'm thinking, since what I'm looking for is just one row, should that status also be located in the parent table?

更多信息请参见父表:

当前文档表

| DocumentID | Title  | Content  | DateCreated |
| 1          | TitleA | ...      | ...         |
| 2          | TitleB | ...      | ...         |
| 3          | TitleC | ...      | ...         |

父表应该是这样的,以便我可以轻松地访问它的状态吗?

| DocumentID | Title  | Content  | DateCreated | CurrentStatus |
| 1          | TitleA | ...      | ...         | s1            |
| 2          | TitleB | ...      | ...         | s3            |
| 3          | TitleC | ...      | ...         | s1            |

更新 我刚刚学会了如何使用“apply”,它可以更容易地解决这类问题。


当前回答

这是这个话题中最容易找到的问题之一,所以我想给出一个现代的答案(既供我参考,也帮助其他人)。通过使用first_value和over,你可以简化上面的查询:

Select distinct DocumentID
  , first_value(status) over (partition by DocumentID order by DateCreated Desc) as Status
  , first_value(DateCreated) over (partition by DocumentID order by DateCreated Desc) as DateCreated
From DocumentStatusLogs

这应该工作在Sql Server 2008及更高。First_value可以被认为是在使用over子句时实现Select Top 1的一种方法。Over允许在选择列表中分组,而不是编写嵌套的子查询(像许多现有的答案那样),这以一种更可读的方式进行。希望这能有所帮助。

其他回答

我刚学会如何使用交叉应用。下面是如何在这种情况下使用它:

 select d.DocumentID, ds.Status, ds.DateCreated 
 from Documents as d 
 cross apply 
     (select top 1 Status, DateCreated
      from DocumentStatusLogs 
      where DocumentID = d.DocumentId
      order by DateCreated desc) as ds

CROSS APPLY是我在解决方案中使用的方法,因为它对我和客户的需求都有效。从我所读到的,应该提供最好的整体性能,如果他们的数据库大幅增长。

我相信这是可以做到的。这可能需要一些调整,但你可以从组中选择最大值。

这些答案太夸张了。

SELECT
  d.DocumentID,
  MAX(d.Status),
  MAX(d1.DateCreated)
FROM DocumentStatusLogs d, DocumentStatusLogs d1
USING DocumentID
GROUP BY 1
ORDER BY 3 DESC

从上面验证克林特的正确答案:

下面两个查询之间的性能非常有趣。52%是最高的。48%是第二个。使用DISTINCT而不是ORDER BY提高了4%的性能。但是ORDER BY具有按多列排序的优势。

IF (OBJECT_ID('tempdb..#DocumentStatusLogs') IS NOT NULL) BEGIN DROP TABLE #DocumentStatusLogs END

CREATE TABLE #DocumentStatusLogs (
    [ID] int NOT NULL,
    [DocumentID] int NOT NULL,
    [Status] varchar(20),
    [DateCreated] datetime
)

INSERT INTO #DocumentStatusLogs([ID], [DocumentID], [Status], [DateCreated]) VALUES (2, 1, 'S1', '7/29/2011 1:00:00')
INSERT INTO #DocumentStatusLogs([ID], [DocumentID], [Status], [DateCreated]) VALUES (3, 1, 'S2', '7/30/2011 2:00:00')
INSERT INTO #DocumentStatusLogs([ID], [DocumentID], [Status], [DateCreated]) VALUES (6, 1, 'S1', '8/02/2011 3:00:00')
INSERT INTO #DocumentStatusLogs([ID], [DocumentID], [Status], [DateCreated]) VALUES (1, 2, 'S1', '7/28/2011 4:00:00')
INSERT INTO #DocumentStatusLogs([ID], [DocumentID], [Status], [DateCreated]) VALUES (4, 2, 'S2', '7/30/2011 5:00:00')
INSERT INTO #DocumentStatusLogs([ID], [DocumentID], [Status], [DateCreated]) VALUES (5, 2, 'S3', '8/01/2011 6:00:00')
INSERT INTO #DocumentStatusLogs([ID], [DocumentID], [Status], [DateCreated]) VALUES (6, 3, 'S1', '8/02/2011 7:00:00')

选项1:

    SELECT
    [Extent1].[ID], 
    [Extent1].[DocumentID],
    [Extent1].[Status], 
    [Extent1].[DateCreated]
FROM #DocumentStatusLogs AS [Extent1]
    OUTER APPLY (
        SELECT TOP 1
            [Extent2].[ID], 
            [Extent2].[DocumentID],
            [Extent2].[Status], 
            [Extent2].[DateCreated]
        FROM #DocumentStatusLogs AS [Extent2]
        WHERE [Extent1].[DocumentID] = [Extent2].[DocumentID]
        ORDER BY [Extent2].[DateCreated] DESC, [Extent2].[ID] DESC
    ) AS [Project2]
WHERE ([Project2].[ID] IS NULL OR [Project2].[ID] = [Extent1].[ID])

选项2:

SELECT 
    [Limit1].[DocumentID] AS [ID], 
    [Limit1].[DocumentID] AS [DocumentID], 
    [Limit1].[Status] AS [Status], 
    [Limit1].[DateCreated] AS [DateCreated]
FROM (
    SELECT DISTINCT [Extent1].[DocumentID] AS [DocumentID] FROM #DocumentStatusLogs AS [Extent1]
) AS [Distinct1]
    OUTER APPLY  (
        SELECT TOP (1) [Project2].[ID] AS [ID], [Project2].[DocumentID] AS [DocumentID], [Project2].[Status] AS [Status], [Project2].[DateCreated] AS [DateCreated]
        FROM (
            SELECT 
                [Extent2].[ID] AS [ID], 
                [Extent2].[DocumentID] AS [DocumentID], 
                [Extent2].[Status] AS [Status], 
                [Extent2].[DateCreated] AS [DateCreated]
            FROM #DocumentStatusLogs AS [Extent2]
            WHERE [Distinct1].[DocumentID] = [Extent2].[DocumentID]
        )  AS [Project2]
        ORDER BY [Project2].[ID] DESC
    ) AS [Limit1]

在Microsoft SQL Server Management Studio中:在突出显示并运行第一个块后,突出显示选项1和选项2,右键单击->[显示估计的执行计划]。然后运行整个程序以查看结果。

选项1结果:

ID  DocumentID  Status  DateCreated
6   1   S1  8/2/11 3:00
5   2   S3  8/1/11 6:00
6   3   S1  8/2/11 7:00

选项2结果:

ID  DocumentID  Status  DateCreated
6   1   S1  8/2/11 3:00
5   2   S3  8/1/11 6:00
6   3   S1  8/2/11 7:00

注意:

当我想要一个连接为1到-(多个连接中的1个)时,我倾向于使用APPLY。 如果我想要连接为1对多或多对多,则使用JOIN。 我使用ROW_NUMBER()避免CTE,除非我需要做一些高级的事情,并且可以接受窗口性能损失。

我还避免在WHERE或ON子句中使用EXISTS / IN子查询,因为我曾经经历过这样会导致一些糟糕的执行计划。但具体里程各不相同。在需要的时候随时随地检查执行计划和概要文件的性能!

SELECT o.*
FROM `DocumentStatusLogs` o                   
  LEFT JOIN `DocumentStatusLogs` b                   
  ON o.DocumentID = b.DocumentID AND o.DateCreated < b.DateCreated
 WHERE b.DocumentID is NULL ;

如果您只想按DateCreated返回最近的文档顺序,它将只按documententid返回前1个文档