views:

50364

answers:

25

Table: UserId, Value, Date.

I want to get the UserId, Value for the max(Date) for each UserId. That is, the Value for each UserId that has the latest date. Is there a way to do this simply in SQL? (Preferably Oracle)

Thank in advance!

[Update:] Apologies for any ambiguity: I need to get ALL the UserIds. But for each UserId, only that row where that user has the latest date.

A: 

(T-SQL)

select top 1 UserId, Value
from theTable
order by [date] DESC

---EDIT---- I just re-read your question and realized that it's not this simple. Whoops!

Does this work?

select userId, value
from theTable
group by userId
having MAX([date])
Ben Scheirman
Nope, it doesn't work in Oracle
David Aldridge
(t-sql) value is not valid in the select clause because it's not used in the group clause and it's not an aggregate.
David B
Nope. Also, "having MAX(date)" will just take the global max date and pick out users having that.
Umang
A: 

I think something like this. (Forgive me for any syntax mistakes; I'm used to using HQL at this point!)

EDIT: Also misread the question! Corrected the query...

SELECT UserId, Value
FROM Users AS user
WHERE Date = (
    SELECT MAX(Date)
    FROM Users AS maxtest
    WHERE maxtest.UserId = user.UserId
)
jdmichal
Doesn't meet the "for each UserId" condition
David Aldridge
Where would it fail? For every UserID in Users, it will be guaranteed that at least one row containing that UserID will be returned. Or am I missing a special case somewhere?
jdmichal
+14  A: 

I don't know your exact columns names, but it would be something like this:

    select userid, value
      from users u1
     where date = (select max(date)
                     from users u2
                    where u1.userid = u2.userid)
Steve K
Probably not very efficent, Steve.
David Aldridge
You are probably underestimating the Oracle query optimizer.
Rafał Dowgird
Not at all. This will almost certainly be implemented as a full scan with a nested loop join to get the dates. You're talking about logical io's in the order of 4 times the number of rows in the table and be dreadful for non-trivial amounts of data.
David Aldridge
Not efficent, but works. Mine as well:http://stackoverflow.com/questions/121387/sql-fetch-the-row-which-has-the-max-value-for-a-column#121556
Zsolt Botykai
What about using analytical sql extensions for Oracle?
Mike McAllister
My analytic solution got voted down for some reason. No idea why -- it's almost the gold standard of how to tackle these problems in Oracle now.
David Aldridge
FYI, "Not efficient, but works" is the same as "Works, but is not efficient". When did we give up on efficient as a design goal?
David Aldridge
I voted your analytic query solution down because it was wrong. While efficient is a design goal, it comes after correctness. See my analytic solution instead.
Your comments on my analytic method are incorrect. See my edit.
David Aldridge
A: 

(T-SQL) First get all the users and their maxdate. Join with the table to find the corresponding values for the users on the maxdates.

create table users (userid int , value int , date datetime)
insert into users values (1, 1, '20010101')
insert into users values (1, 2, '20020101')
insert into users values (2, 1, '20010101')
insert into users values (2, 3, '20030101')

select T1.userid, T1.value, T1.date 
    from users T1,
    (select max(date) as maxdate, userid from users group by userid) T2    
    where T1.userid= T2.userid and T1.date = T2.maxdate

results:

userid      value       date                                    
----------- ----------- -------------------------- 
2           3           2003-01-01 00:00:00.000
1           2           2002-01-01 00:00:00.000
Frans
A: 

Your question is a little ambiguous - is there one max(date) or a max(date) for each user id?

Either way, embedded queries - either in the condition, in the select or in one of the joins should get you on the right track.

Unsliced
Sorry if it's ambiguous. Will just update the question.
Umang
+29  A: 

This will retrieve all rows for which the my___date column value is equal to the maximum value of my_date for that userid. This may retrieve multiple rows for the userid where the maximum date is on multiple rows.

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

"Analytic functions rock"

Edit: With regard to the first comment ...

"using analytic queries and a self-join defeats the purpose of analytic queries"

There is no self-join in this code. There is instead a predicate placed on the result of the inline view that contains the analytic function -- a very different matter, and completely standard practice.

"The default window in Oracle is from the first row in the partition to the current one"

The windowing clause is only applicable in the presence of the order by clause. With no order by clause, no windowing clause is applied by default and none can be explicitly specified.

The code works.

David Aldridge
Sorry, but I don't think this is right. The default window in Oracle is from the first row in the partition to the current one. This may or may not include the maximum date. Secondly using analytic queries and a self-join defeats the purpose of analytic queries.
Huh, I just double-checked the documentation and you are right. I've been bitten by the default window in the presence of an ORDER BY enough that I didn't realize that the default is different with no ORDER BY. I switched my vote.
"I've been bitten by the default window in the presence of an ORDER BY ..."Me too :)
David Aldridge
Works great! Especially when one has a 100 million rows to run through!Thanks. :)
Umang
A: 

If (UserID, Date) is unique, i.e. no date appears twice for the same user then:

select TheTable.UserID, TheTable.Value
from TheTable inner join (select UserID, max([Date]) MaxDate
                          from TheTable
                          group by UserID) UserMaxDate
     on TheTable.UserID = UserMaxDate.UserID
        TheTable.[Date] = UserMaxDate.MaxDate;
finnw
I believe that you need to join by the UserID as well
Tom H.
You're right. Fixed.
finnw
A: 

Hi, i thing you shuold make this variant to previous query:

SELECT UserId, Value FROM Users U1 WHERE 
Date = ( SELECT MAX(Date)    FROM Users where UserId = U1.UserId)
stefano m
A: 

Assuming Date is unique for a given UserID, here's some TSQL:

SELECT UserTest.UserID, UserTest.Value FROM UserTest INNER JOIN ( SELECT UserID, MAX(Date) MaxDate FROM UserTest GROUP BY UserID ) Dates ON UserTest.UserID = Dates.UserID AND UserTest.Date = Dates.MaxDate

marc
+2  A: 
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
Aheho
A: 
select userid, value, date
  from thetable t1 ,
       ( select t2.userid, max(t2.date) date2 
           from thetable t2 
          group by t2.userid ) t3
 where t3.userid t1.userid and
       t3.date2 = t1.date

IMHO this works. HTH

Zsolt Botykai
A: 

I think this should work?

Select
T1.UserId,
(Select Top 1 T2.Value From Table T2 Where T2.UserId = T1.UserId Order By Date Desc) As 'Value'
From
Table T1
Group By
T1.UserId
Order By
T1.UserId
GateKiller
+1  A: 

This should be as simple as:

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

Valerion
A: 

First try I misread the question, following the top answer, here is a complete example with correct results:

CREATE TABLE table_name (id int, the_value varchar(2), the_date datetime);

INSERT INTO table_name (id,the_value,the_date) VALUES(1 ,'a','1/1/2000');
INSERT INTO table_name (id,the_value,the_date) VALUES(1 ,'b','2/2/2002');
INSERT INTO table_name (id,the_value,the_date) VALUES(2 ,'c','1/1/2000');
INSERT INTO table_name (id,the_value,the_date) VALUES(2 ,'d','3/3/2003');
INSERT INTO table_name (id,the_value,the_date) VALUES(2 ,'e','3/3/2003');

--

  select id, the_value
      from table_name u1
      where the_date = (select max(the_date)
                     from table_name u2
                     where u1.id = u2.id)

--

id          the_value
----------- ---------
2           d
2           e
1           b

(3 row(s) affected)
KyleLanser
+1  A: 
SELECT userid, MAX(value) KEEP (DENSE_RANK FIRST ORDER BY date DESC)
  FROM table
  GROUP BY userid
Dave Costa
+2  A: 

I know you asked for Oracle, but in SQL 2005 we now use this:


-- 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
mancaus
A: 

Since this is tagged with oracle, analytic functions should be ok, and the best answer performance-wise: see David Aldridge's answer above

David's Aldridge's answer is second-best. See Dave Costa's solution for the best answer performance-wise.
Rob van Wijk
+3  A: 

I don't have Oracle to test it, but the most efficient solution is to use analytic queries. It should look something like this:

SELECT DISTINCT
    UserId
  , MaxValue
FROM (
    SELECT UserId
      , FIRST (Value) Over (
          PARTITION BY UserId
          ORDER BY Date DESC
        ) MaxValue
    FROM SomeTable
  )

I suspect that you can get rid of the outer query and put distinct on the inner, but I'm not sure. In the meantime I know this one works.

If you want to learn about analytic queries, I'd suggest reading http://www.orafaq.com/node/55 and http://www.akadia.com/services/ora_analytic_functions.html. Here is the short summary.

Under the hood analytic queries sort the whole dataset, then process it sequentially. As you process it you partition the dataset according to certain criteria, and then for each row looks at some window (defaults to the first value in the partition to the current row - that default is also the most efficient) and can compute values using a number of analytic functions (the list of which is very similar to the aggregate functions).

In this case here is what the inner query does. The whole dataset is sorted by UserId then Date DESC. Then it processes it in one pass. For each row you return the UserId and the first Date seen for that UserId (since dates are sorted DESC, that's the max date). This gives you your answer with duplicated rows. Then the outer DISTINCT squashes duplicates.

This is not a particularly spectacular example of analytic queries. For a much bigger win consider taking a table of financial receipts and calculating for each user and receipt, a running total of what they paid. Analytic queries solve that efficiently. Other solutions are less efficient. Which is why they are part of the 2003 SQL standard. (Unfortunately Postgres doesn't have them yet. Grrr...)

You also need to return the date value to answer the question completely. If that means another first_value clause then I'd suggest that the solution is more complex than it ought to be, and the analytic method based on max(date) reads better.
David Aldridge
The question statement says nothing about returning the date. You can do that either by adding another FIRST(Date) or else just by querying the Date and changing the outer query to a GROUP BY. I'd use the first and expect the optimizer to calculate both in one pass.
"The question statement says nothing about returning the date" ... yes, you're right. Sorry.But adding more FIRST_VALUE clauses would become messy pretty quickly. It's a single window sort, but if you had 20 columns to return for that row then you've written a lot of code to wade through.
David Aldridge
It also occurs to me that this solution is non-deterministic for data where a single userid has multiple rows that have the maximum date and different VALUEs. More a fault in the question than the answer though.
David Aldridge
I agree it is painfully verbose. However isn't that generally the case with SQL? And you're right that the solution is non-deterministic. There are multiple ways to deal with ties, and sometimes each is what you want.
+14  A: 

I see many people use subqueries or else vendor-specific features to do this, but I often do this kind of query without subqueries in the following way. It uses plain, standard SQL so it should work in any brand of RDBMS.

SELECT t1.*
FROM mytable AS t1
  LEFT OUTER JOIN mytable AS t2
    ON (t1.UserId = t2.UserId AND t1."Date" < t2."Date")
WHERE t2.UserId IS NULL;

In other words: fetch the row from t1 where no other row exists with the same UserId and a greater Date.

(I put the identifier "Date" in delimiters because it's an SQL reserved word.)

Bill Karwin
As as a developer who does not think in SQL very often, I find this very clever!
Michael Easter
+4  A: 

Not being at work, I don't have Oracle to hand, but I seem to recall that Oracle allows multiple columns to be matched in an IN clause, which should at least avoid the options that use a correlated subquery, which is seldom a good idea.

Something like this, perhaps (can't remember if the column list should be parenthesised or not):

SELECT * 
FROM MyTable
WHERE (User, Date) IN
  ( SELECT User, MAX(Date) FROM MyTable GROUP BY User)

EDIT: Just tried it for real:

SQL> create table MyTable (usr char(1), dt date);
SQL> insert into mytable values ('A','01-JAN-2009');
SQL> insert into mytable values ('B','01-JAN-2009');
SQL> insert into mytable values ('A', '31-DEC-2008');
SQL> insert into mytable values ('B', '31-DEC-2008');
SQL> select usr, dt from mytable
  2  where (usr, dt) in 
  3  ( select usr, max(dt) from mytable group by usr)
  4  /

U DT
- ---------
A 01-JAN-09
B 01-JAN-09

So it works, although some of the new-fangly stuff mentioned elsewhere may be more performant.

Mike Woodhouse
This works nicely on PostgreSQL too. And I like the simplicity and generality of it -- the subquery says "Here's my criteria", the outer query says "And here's the details I want to see". +1.
j_random_hacker
A: 

This will also take care of duplicates (return one row for each user_id):

SELECT *
FROM (
  SELECT u.*, FIRST_VALUE(u.rowid) OVER(PARTITION BY u.user_id ORDER BY u.date DESC) AS last_rowid
  FROM users u
) u2
WHERE u2.rowid = u2.last_rowid
na43251
A: 

The answer here is Oracle only. Here's a bit more sophisticated answer in all SQL:

Who has the best overall homework result (maximum sum of homework points)?

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)

And a more difficult example, which need some explanation, for which I don't have time atm:

Give the book (ISBN and title) that is most popular in 2008, i.e., which is borrowed most often in 2008.

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);

Hope this helps (anyone).. :)

Regards, Guus

Guus
A: 

thanks a lot guys, helped me with a simular issue I was struggling with - much obliged!

Steve

Steve
A: 

Just tested this and it seems to work on a logging table

select ColumnNames, max(DateColumn) from log  group by ColumnNames order by 1 desc
Mauro
+1  A: 

Just had to write a "live" example at work :)

This one supports multiple values for UserId on the same date.

Columns: UserId, Value, Date

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
)

You can use FIRST_VALUE instead of MAX and look it up in the explain plan. I didn't have the time to play with it.

Of course, if searching through huge tables, it's probably better if you use FULL hints in your query.

Truper