得到每组的前1行

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

;WITH cte AS
(
   SELECT *,
         ROW_NUMBER() OVER (PARTITION BY DocumentID ORDER BY DateCreated DESC) AS rn
   FROM DocumentStatusLogs
)
SELECT *
FROM cte
WHERE rn = 1

如果您希望每天有2个条目，那么这将任意选择一个。要获得一天的两个条目，请使用DENSE_RANK代替

至于是否正常化，这取决于你是否想:

在两个地方保持状态保存状态历史 .．.

目前，您保留了状态历史。如果你也想在父表中保持最新状态(这是非规范化的)，你需要一个触发器来维持父表中的“状态”。或者删除这个状态历史表。

2011-07-27 08:44:10

其他回答

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

2016-12-19 15:10:28

如果你担心性能问题，你也可以用MAX()这样做:

SELECT *
FROM DocumentStatusLogs D
WHERE DateCreated = (SELECT MAX(DateCreated) FROM DocumentStatusLogs WHERE ID = D.ID)

ROW_NUMBER()要求对SELECT语句中的所有行进行排序，而MAX则不需要。应该会大大加快你的查询速度。

2013-01-15 20:57:21

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

2012-08-30 06:10:55

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。

2020-02-12 16:54:50

此解决方案可用于获取每个分区的TOP N最近行(在本例中，WHERE语句中的N为1,partition为doc_id):

SELECT T.doc_id, T.status, T.date_created FROM 
(
    SELECT a.*, ROW_NUMBER() OVER (PARTITION BY doc_id ORDER BY date_created DESC) AS rnk FROM doc a
) T
WHERE T.rnk = 1;

2018-10-16 08:40:51

得到每组的前1行

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