Fetch the rows which have the Max value for a column for each distinct value of another column

I see many people use subqueries or else window functions 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 t1 LEFT OUTER JOIN mytable 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.)

In case if t1."Date" = t2."Date", doubling appears. Usually tables has auto_inc(seq) key, e.g. id. To avoid doubling can be used follows:

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

Re comment from @Farhan:

Here's a more detailed explanation:

An outer join attempts to join t1 with t2. By default, all results of t1 are returned, and if there is a match in t2, it is also returned. If there is no match in t2 for a given row of t1, then the query still returns the row of t1, and uses NULL as a placeholder for all of t2's columns. That's just how outer joins work in general.

The trick in this query is to design the join's matching condition such that t2 must match the same userid, and a greater date. The idea being if a row exists in t2 that has a greater date, then the row in t1 it's compared against can't be the greatest date for that userid. But if there is no match -- i.e. if no row exists in t2 with a greater date than the row in t1 -- we know that the row in t1 was the row with the greatest date for the given userid.

In those cases (when there's no match), the columns of t2 will be NULL -- even the columns specified in the join condition. So that's why we use WHERE t2.UserId IS NULL, because we're searching for the cases where no row was found with a greater date for the given userid.

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

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 ) 

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.

I know you asked for Oracle, but in SQL 2005 we can now use ROW_NUMBER() and RANK() within a partition.

 -- 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 

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...)

Wouldn't a QUALIFY clause be both simplest and best?

select userid, my_date, ... from users qualify rank() over (partition by userid order by my_date desc) = 1 

For context, on Teradata here a decent size test of this runs in 17s with this QUALIFY version and in 23s with the 'inline view'/Aldridge solution #1.

In Oracle 12c+, you can use Top n queries along with analytic function rank to achieve this very concisely without subqueries:

select * from your_table order by rank() over (partition by user_id order by my_date desc) fetch first 1 row with ties; 

The above returns all the rows with max my_date per user.

If you want only one row with max date, then replace the rank with row_number:

select * from your_table order by row_number() over (partition by user_id order by my_date desc) fetch first 1 row with ties; 

With PostgreSQL 8.4 or later, you can use this:

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 

I'm quite late to the party but the following hack will outperform both correlated subqueries and any analytics function but has one restriction: values must convert to strings. So it works for dates, numbers and other strings. The code does not look good but the execution profile is great.

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 

The reason why this code works so well is that it only needs to scan the table once. It does not require any indexes and most importantly it does not need to sort the table, which most analytics functions do. Indexes will help though if you need to filter the result for a single userid.

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.

If you're using Postgres, you can use array_agg like

SELECT userid,MAX(adate),(array_agg(value ORDER BY adate DESC))[1] as value FROM YOURTABLE GROUP BY userid 

I'm not familiar with Oracle. This is what I came up with

SELECT userid, MAX(adate), SUBSTR( (LISTAGG(value, ',') WITHIN GROUP (ORDER BY adate DESC)), 0, INSTR((LISTAGG(value, ',') WITHIN GROUP (ORDER BY adate DESC)), ',')-1 ) as value FROM YOURTABLE GROUP BY userid 

Both queries return the same results as the accepted answer. See SQLFiddles:

1. Accepted answer
2. My solution with Postgres
3. My solution with Oracle

Use ROW_NUMBER() to assign a unique ranking on descending Date for each UserId, then filter to the first row for each UserId (i.e., ROW_NUMBER = 1).

SELECT UserId, Value, Date FROM (SELECT UserId, Value, Date, ROW_NUMBER() OVER (PARTITION BY UserId ORDER BY Date DESC) rn FROM users) u WHERE rn = 1; 

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 ) 

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) 

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 

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

(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 

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

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 

Solution for MySQL which doesn't have concepts of partition KEEP, DENSE_RANK.

select userid, my_date, ... from ( select @sno:= case when @pid<>userid then 0 else @sno+1 end as serialnumber, @pid:=userid, my_Date, ... from users order by userid, my_date ) a where a.serialnumber=0 

Reference: http://benincampus.blogspot.com/2013/08/select-rows-which-have-maxmin-value-in.html

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

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 

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) 

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 

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 

This should be as simple as:

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

Oracle Database 23ai added a new way to solve this - adding the partition byclause tofetch first:

create table t ( userid int, value int, dt date ); insert into t values ( 1, 1, date'2024-01-01' ), ( 1, 2, date'2024-02-02' ), ( 1, 3, date'2024-03-03' ), ( 2, 1, date'2024-01-01' ), ( 2, 2, date'2024-02-02' ), ( 2, 3, date'2024-03-03' ); select * from t order by userid, dt desc fetch first 9999999 partition by userid, 1 row only; USERID VALUE DT ---------- ---------- ----------- 1 3 03-MAR-2024 2 3 03-MAR-2024 

Note you have to specify a number before partition by; this states how many userids you want to fetch. Assuming you want them all, set this to a value (much) larger than the number of userids in the table.

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;