我知道你要求使用Oracle，但是在SQL 2005中我们现在使用这个:

-- Single Value
;WITH ByDate
AS (
SELECT UserId, Value, ROW_NUMBER() OVER (PARTITION BY UserId ORDER BY Date DESC) RowNum
FROM UserDates
)
SELECT UserId, Value
FROM ByDate
WHERE RowNum = 1

-- Multiple values where dates match
;WITH ByDate
AS (
SELECT UserId, Value, RANK() OVER (PARTITION BY UserId ORDER BY Date DESC) Rnk
FROM UserDates
)
SELECT UserId, Value
FROM ByDate
WHERE Rnk = 1

检查这个链接，如果你的问题似乎类似的页面，然后我会建议你下面的查询，将给出该链接的解决方案

选择不同的sno,item_name,max(start_date) over(sno分区)，max(end_date) over(sno分区)，max(creation_date) over(sno分区)， Max (last_modified_date) over(sno分区) 从uniq_select_records 按sno,item_name asc排序;

是否会给出与该链接相关的准确结果

答案是Oracle。这里有一个更复杂的SQL回答:

谁的整体作业成绩最好(作业点数最多)?

SELECT FIRST, LAST, SUM(POINTS) AS TOTAL
FROM STUDENTS S, RESULTS R
WHERE S.SID = R.SID AND R.CAT = 'H'
GROUP BY S.SID, FIRST, LAST
HAVING SUM(POINTS) >= ALL (SELECT SUM (POINTS)
FROM RESULTS
WHERE CAT = 'H'
GROUP BY SID)

还有一个更难的例子，需要一些解释，我没有时间了

给出2008年最受欢迎的书(ISBN和书名)，即2008年最常被借阅的书。

SELECT X.ISBN, X.title, X.loans
FROM (SELECT Book.ISBN, Book.title, count(Loan.dateTimeOut) AS loans
FROM CatalogEntry Book
LEFT JOIN BookOnShelf Copy
ON Book.bookId = Copy.bookId
LEFT JOIN (SELECT * FROM Loan WHERE YEAR(Loan.dateTimeOut) = 2008) Loan 
ON Copy.copyId = Loan.copyId
GROUP BY Book.title) X
HAVING loans >= ALL (SELECT count(Loan.dateTimeOut) AS loans
FROM CatalogEntry Book
LEFT JOIN BookOnShelf Copy
ON Book.bookId = Copy.bookId
LEFT JOIN (SELECT * FROM Loan WHERE YEAR(Loan.dateTimeOut) = 2008) Loan 
ON Copy.copyId = Loan.copyId
GROUP BY Book.title);

希望这能对(任何人)有所帮助。：）

问候 古斯

Select  
   UserID,  
   Value,  
   Date  
From  
   Table,  
   (  
      Select  
          UserID,  
          Max(Date) as MDate  
      From  
          Table  
      Group by  
          UserID  
    ) as subQuery  
Where  
   Table.UserID = subQuery.UserID and  
   Table.Date = subQuery.mDate  

难道一个qualified子句不是既简单又最好吗?

select userid, my_date, ...
from users
qualify rank() over (partition by userid order by my_date desc) = 1

对于上下文，在Teradata这里一个像样的大小测试运行在17秒与这个合格版本和在23秒与“内联视图”/Aldridge解决方案#1。

这将检索my_date列值等于该用户id的my_date最大值的所有行。这可能会为最大日期位于多行上的用户id检索多行。

select userid,
       my_date,
       ...
from
(
select userid,
       my_date,
       ...
       max(my_date) over (partition by userid) max_my_date
from   users
)
where my_date = max_my_date

“解析函数摇滚”

编辑:关于第一条评论……

“使用分析查询和自连接违背了分析查询的目的”

这段代码中没有自连接。相反，在包含分析函数的内联视图的结果上放置了一个谓词——这是非常不同的事情，完全是标准的实践。

Oracle的默认窗口是从分区的第一行到当前一行

加窗条款仅适用于存在按顺序条款的情况。如果没有按子句排序，默认情况下就不会应用任何窗口子句，也不能显式地指定任何窗口子句。

代码可以工作。