[英]SQL Server - SELECT one row with the MAX() value on a column
这是我的表表结构
id | Name | Date | VersionID
----+---------------+---------------+-------------------------------------
1 | Item 1 | 10/15/2012 | F8883CA4-3603-476D-BA33-1BBB6B87A79F
1 | Item 1 | 11/06/2012 | AA22DA27-87D1-42EC-96F6-A4846A45DF6B
1 | Item 1 | 11/06/2018 | AA22DA27-87D1-42EC-96F6-A4846A45DF6B
2 | Item 2 | 11/06/2015 | F8883CA4-3603-476D-BA33-1BBB6B87A79F
2 | Item 2 | 12/15/2012 | AA22DA27-87D1-42EC-96F6-A4846A45DF6B
2 | Item 2 | 1/19/2013 | F8883CA4-3603-476D-BA33-1BBB6B87A79F
从此表中,对于每个版本,我希望获得每个项目的最大日期。
例
id | Name | Date | VersionID
----+---------------+---------------+-------------------------------------
1 | Item 1 | 10/15/2012 | F8883CA4-3603-476D-BA33-1BBB6B87A79F
1 | Item 1 | 11/06/2018 | AA22DA27-87D1-42EC-96F6-A4846A45DF6B
2 | Item 2 | 11/06/2015 | F8883CA4-3603-476D-BA33-1BBB6B87A79F
2 | Item 2 | 12/15/2012 | AA22DA27-87D1-42EC-96F6-A4846A45DF6B
我尝试了rank
和dense_rank
函数,但是尝试的逻辑未返回预期结果。 有什么想法吗?
正常方法是使用row_number
:
select id, name, date, versionid
from (select s.*,
row_number() over (partition by id, version_id order by date desc) as seqnum
from structure s
) s
where seqnum = 1;
我们也可以通过使用MAX Condition和关联子查询来获得结果
DECLARE @Table1 TABLE
(id int, Name varchar(6), Date datetime, VersionID varchar(36))
;
INSERT INTO @Table1
(id, Name, Date, VersionID)
VALUES
(1, 'Item 1', '2012-10-15 05:30:00', 'F8883CA4-3603-476D-BA33-1BBB6B87A79F'),
(1, 'Item 1', '2012-11-06 05:30:00', 'AA22DA27-87D1-42EC-96F6-A4846A45DF6B'),
(1, 'Item 1', '2018-11-06 05:30:00', 'AA22DA27-87D1-42EC-96F6-A4846A45DF6B'),
(2, 'Item 2', '2015-11-06 05:30:00', 'F8883CA4-3603-476D-BA33-1BBB6B87A79F'),
(2, 'Item 2', '2012-12-15 05:30:00', 'AA22DA27-87D1-42EC-96F6-A4846A45DF6B'),
(2, 'Item 2', '2013-01-19 05:30:00', 'F8883CA4-3603-476D-BA33-1BBB6B87A79F')
;
Select T.id,T.Name,TT.Dt,T.VersionID from @Table1 T
INNER JOIN (
select id,Name,MAX(Date)Dt,VersionID from @Table1
GROUP BY id,Name,VersionID)TT
ON T.id = TT.id AND T.Date = TT.Dt
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