select   UserId,max(Date) over (partition by UserId) value from users;

只是需要在工作中写一个“活”的例子:)

它支持在同一日期为UserId设置多个值。

列: 用户id，值，日期

SELECT
   DISTINCT UserId,
   MAX(Date) OVER (PARTITION BY UserId ORDER BY Date DESC),
   MAX(Values) OVER (PARTITION BY UserId ORDER BY Date DESC)
FROM
(
   SELECT UserId, Date, SUM(Value) As Values
   FROM <<table_name>>
   GROUP BY UserId, Date
)

您可以使用FIRST_VALUE而不是MAX，并在解释计划中查找它。我没有时间玩它。

当然，如果搜索巨大的表，在查询中使用FULL提示可能会更好。

我认为你应该对之前的查询进行修改:

SELECT UserId, Value FROM Users U1 WHERE 
Date = ( SELECT MAX(Date)    FROM Users where UserId = U1.UserId)

SELECT userid, MAX(value) KEEP (DENSE_RANK FIRST ORDER BY date DESC)
  FROM table
  GROUP BY userid

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

在PostgreSQL 8.4或更高版本中，你可以这样使用:

select user_id, user_value_1, user_value_2
  from (select user_id, user_value_1, user_value_2, row_number()
          over (partition by user_id order by user_date desc) 
        from users) as r
  where r.row_number=1