我已经很晚了，但下面的黑客将超越相关子查询和任何分析功能，但有一个限制:值必须转换为字符串。所以它适用于日期，数字和其他字符串。代码看起来不太好，但执行配置文件很棒。

select
    userid,
    to_number(substr(max(to_char(date,'yyyymmdd') || to_char(value)), 9)) as value,
    max(date) as date
from 
    users
group by
    userid

这段代码运行良好的原因是它只需要扫描表一次。它不需要任何索引，最重要的是，它不需要像大多数分析函数那样对表进行排序。如果您需要为单个用户id过滤结果，索引将有所帮助。

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)

我想是这样的。(请原谅我的语法错误;在这一点上，我习惯使用HQL !)

编辑:也误解了问题!修正了查询…

SELECT UserId, Value
FROM Users AS user
WHERE Date = (
    SELECT MAX(Date)
    FROM Users AS maxtest
    WHERE maxtest.UserId = user.UserId
)

这应该非常简单:

SELECT UserId, Value
FROM Users u
WHERE Date = (SELECT MAX(Date) FROM Users WHERE UserID = u.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

使用代码:

select T.UserId,T.dt from (select UserId,max(dt) 
over (partition by UserId) as dt from t_users)T where T.dt=dt;

这将检索结果，而不考虑UserId的重复值。 如果你的UserId是唯一的，它变得更简单:

select UserId,max(dt) from t_users group by UserId;