sql-server-2008 - 从表中选择列中具有唯一值的行

Question

我想选择按日期排序的前 3 条记录，以便这些记录没有相同的 ID，对于两个（或更多）相同 ID 的记录，请选择最新的记录。表是这样的：

ID  | Date                 | Value
1   | 2013-04-29T16:27:20  | value1
2   | 2013-04-29T19:46:25  | value2
2   | 2013-04-30T16:50:30  | value3
3   | 2013-04-30T16:40:30  | value4
1   | 2013-04-30T17:50:30  | value5

我希望结果是这样的：

ID  | Date                 | Value
1   | 2013-04-30T17:50:30  | value5
2   | 2013-04-30T16:50:30  | value3
3   | 2013-04-30T16:40:30  | value4

Answer 1

SELECT t.ID, t.Date, t.Value
FROM (
    SELECT ID, Date, Value, ROW_NUMBER() OVER (PARTITION BY ID ORDER BY Date DESC) AS RowNumber
    FROM table_name
) t
WHERE t.RowNumber = 1

Answer 2

由于您使用的是 SQL Server 2005+，因此您可以实施row_number()以获得结果：

select id, date, value
from 
(
  select id, date, value,
    row_number() over(partition by id order by date desc) seq
  from yourtable
) d
where seq = 1;

请参阅SQL Fiddle with Demo。

或者，您可以使用带有聚合函数的子查询来返回id最新日期：

select t1.id, t1.date, t1.value
from yourtable t1
inner join
(
  select id, max(date) date
  from yourtable
  group by id
) t2
  on t1.id = t2.id
  and t1.date = t2.date

请参阅带有演示的 SQL Fiddle