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

我的代码从每组中选择top 1

select a.* from #DocumentStatusLogs a where 
 datecreated in( select top 1 datecreated from #DocumentStatusLogs b
where 
a.documentid = b.documentid
order by datecreated desc
)

其他回答

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。

;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。

SELECT * FROM
DocumentStatusLogs JOIN (
  SELECT DocumentID, MAX(DateCreated) DateCreated
  FROM DocumentStatusLogs
  GROUP BY DocumentID
  ) max_date USING (DocumentID, DateCreated)

什么数据库服务器?这段代码并不是对所有的都有效。

关于你问题的后半部分,我认为应该在一栏中写上“地位”。您可以将DocumentStatusLogs保留为日志,但仍然将最新信息存储在主表中。

顺便说一句,如果你已经在文档表中有DateCreated列,你可以使用它来加入DocumentStatusLogs(只要DateCreated在DocumentStatusLogs中是唯一的)。

编辑:MsSQL不支持USING,因此将其更改为:

ON DocumentStatusLogs.DocumentID = max_date.DocumentID AND DocumentStatusLogs.DateCreated = max_date.DateCreated
SELECT documentid, 
       status, 
       datecreated 
FROM   documentstatuslogs dlogs 
WHERE  status = (SELECT status 
                 FROM   documentstatuslogs 
                 WHERE  documentid = dlogs.documentid 
                 ORDER  BY datecreated DESC 
                 LIMIT  1) 
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