MSSQL-选择表中多行的最新匹配项

所以，这一次我的头撞到了墙上。。。 我有一个表（几百万行），看起来像这样：

ID  Time            Area    Name
1   2017-01-01 01:42    21  Test
1   2017-01-02 02:13    21  Test
1   2017-02-01 12:45    30  N/A
1   2017-03-12 02:23    41  N/A
2   2017-02-15 04:25    20  N/A

ID  Time            Area    Name
1   2017-01-02 02:13    21  Test
1   2017-02-01 12:45    30  N/A
1   2017-03-12 02:23    41  N/A

select max(Time),
"of every area"
from TABLE
where ID = 1

我想运行一个查询，选择某个ID的最新出现的区域； 结果是这样的：

ID  Time            Area    Name
1   2017-01-01 01:42    21  Test
1   2017-01-02 02:13    21  Test
1   2017-02-01 12:45    30  N/A
1   2017-03-12 02:23    41  N/A
2   2017-02-15 04:25    20  N/A

ID  Time            Area    Name
1   2017-01-02 02:13    21  Test
1   2017-02-01 12:45    30  N/A
1   2017-03-12 02:23    41  N/A

select max(Time),
"of every area"
from TABLE
where ID = 1

---

如果有人能给我一个线索，那就太棒了

您可以使用

行号

：

SELECT * FROM (
  SELECT *, ROW_NUMBER() OVER(PARTITION BY Area ORDER BY Time DESC) AS rn
  FROM table) sub
WHERE rn = 1

还有一个选项（sans子查询和/或额外字段）

示例

 Select Top 1 with Ties *
  From  YourTable
  Where ID =1
  Order By Row_Number() over (Partition By Area Order By [Time] Desc)

返回

ID  Time                     Area   Name
1   2017-01-02 02:13:00.000  21     Test
1   2017-02-01 12:45:00.000  30     N/A
1   2017-03-12 02:23:00.000  41     N/A

这是正确的

要更具体地满足您对特定ID的要求，请使用

SELECT * FROM (
   SELECT *, ROW_NUMBER() OVER(PARTITION BY Area ORDER BY Time DESC) AS rn
   FROM table WHERE ID=1) sub
WHERE rn = 1

使用CTE，您可以使用与相同的

;WITH cte AS (
  SELECT [ID], [Time], [Area], [Name],
     row_number() OVER(PARTITION BY [Area] ORDER BY [Time] desc) AS [rn]
  FROM [AreaTable] WHERE ID=1
)
SELECT [ID], [Time], [Area], [Name] FROM cte WHERE [rn] = 1

---

或者更老派的方法是在派生表中使用GROUPBY

SELECT Test.* FROM dbo.testTable Test
INNER JOIN (
    SELECT Id, MAX(Time) Time, Area 
    FROM dbo.testTable
    GROUP BY Id, Area 
) TestMax ON TestMax.Id = Test.Id AND TestMax.Time = Test.Time AND 
TestMax.Area = Test.Area
ORDER BY Test.Id, Test.Area;