Question

我有一张桌子：

id|val|updated
1 | 31|2012-01-12
1 | 34|2012-01-15
2 | 41|2012-01-15
3 | 44|2012-01-15
4 | 66|2012-01-01

我需要选择具有最大更新值的id和val。所以对于这个表，结果应该是：

1| 34| 2012-01-15
2| 41| 2012-01-15
3| 44| 2012-01-15
4| 66| 2012-01-01

Answer 1

这需要在针对主表连接的子查询中使用MAX()聚合。

SELECT
  tbl.id,
  tbl.val,
  tbl.updated
FROM tbl JOIN (
  /* Subquery gets MAX(updated) per id  to JOIN against */
  SELECT 
    id,
    MAX(updated) as updated
  FROM tbl
  GROUP BY id
) t2 ON tbl.id = t2.id AND tbl.updated = t2.updated

Answer 2

其他几种“便携式”方法。在我对SQL SERVER 2008的快速测试中，Michael的查询表现最佳

SELECT T1.*
FROM yourTable AS T1
WHERE NOT EXISTS(
    SELECT *
    FROM yourTable AS T2
    WHERE T2.id = T1.id
      AND T2.updated > T1.updated
)

SELECT T1.*
FROM yourTable AS T1
LEFT JOIN yourTable AS T2 ON
(
         T2.id = T1.id 
    AND  T2.updated > T1.updated
)
WHERE T2.id IS NULL

Answer 3

选择的答案可以正常工作，但这是一个更快的解决方案。它不使用连接是可以减慢查询速度的组。这个使用比组和连接快得多的分区。看看：

select id, val, Maxdate as updated from
(
     select id, val,
            Max(updated) over(partition by id order by updated desc) as MaxDate, 
            Row_Number() over(partition by id order by updated desc) as RowNumber,
     from table1
)
where RowNumber = 1

选择具有最大时间戳的值

3 个答案: