表:

UserId, Value, Date.

我想获得UserId,为每个UserId的最大值(日期)的值。也就是说,具有最新日期的每个UserId的值。有没有一种方法可以在SQL中简单地做到这一点?(最好是Oracle)

更新:为任何歧义道歉:我需要得到所有的用户id。但是对于每个UserId,只有该用户拥有最新日期的行。


当前回答

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

其他回答

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

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

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。

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;