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

select VALUE from TABLE1 where TIME = 
   (select max(TIME) from TABLE1 where DATE= 
   (select max(DATE) from TABLE1 where CRITERIA=CRITERIA))

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 a.userid,a.values1,b.mm 
FROM table_name a,(SELECT userid,Max(date1)AS mm FROM table_name GROUP BY userid) b
WHERE a.userid=b.userid AND a.DATE1=b.mm;

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