如何從sql表中獲取最大值與group by？

Question

我有一個表

Id | Version | DateFrom | DateTo  | 
_______________________________________ 
1 | 1  | 2015-09-15 | 2015-09-18 | 
1 | 2  | 2015-09-15 | 2015-09-18 | 
1 | 3  | 2015-09-15 | 2015-09-20 | --different date 
1 | 4  | 2015-09-15 | 2015-09-18 | 
2 | 1  | 2015-09-15 | 2015-09-18 | 
2 | 2  | 2015-09-15 | 2015-09-18 | 

而我試圖讓與最新版本的獨立於其他列則返回記錄的視圖。 因此，例如我想到：

Id | Version | DateFrom | DateTo  | 
_______________________________________ 
1 | 4  | 2015-09-15 | 2015-09-18 | 
2 | 2  | 2015-09-15 | 2015-09-18 | 

這是我已經做了：

Select 
    Id, 
    MAX([Version]) AS Version, 
    DateFrom, 
    DateTo 
FROM 
    dbo.Table_1 
Group By 
    Id, 
    DateFrom, 
    DateTo 

但結果是：

Id | Version | DateFrom | DateTo  | 
_______________________________________ 
1 | 4  | 2015-09-15 | 2015-09-18 | 
1 | 3  | 2015-09-15 | 2015-09-20 | 
2 | 2  | 2015-09-15 | 2015-09-18 | 

Answer 1

也可以用左做加盟：

SELECT t.* 
FROM YourTable t 
LEFT JOIN YourTable s 
ON(t.id = s.id AND t.version < s.version) 
WHERE s.version is null 

Answer 2

有一個子查詢，返回每個ID的最大版。與結果加入：

select t1.* 
from tablename t1 
join (select id, max(version) as maxversion 
     from tablename 
     group by id) t2 
    on t1.id = t2.id and t1.version = t2.maxversion 

Answer 3

您可以運行查詢類型：

Select ID, version, dateFrom, dateTo 
from table_1 a 
where not exists (
select 1 
from table_1 b 
where a.id = b.id and a.version < b.version)