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

Question

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.

How do I do this in SQL? (Preferably Oracle.)

I need to get ALL the UserIds. But for each UserId, only that row where that user has the latest date.

Answer 1

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.

Answer 2

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.

Answer 3

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

Answer 4

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
)

Answer 5

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.

Answer 6

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

Answer 7

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...)

Answer 8

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; 

Answer 9

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.

Answer 10

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

Answer 11

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.

Answer 12

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.

Answer 13

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;

Answer 14

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

Answer 15

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
)

Answer 16

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)

Answer 17

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  

Answer 18

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

Answer 19

(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

Answer 20

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 

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

This should be as simple as:

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

Answer 26

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)

Answer 27

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

Answer 28

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

Answer 29

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;

Answer 30

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

Answer 31

check this link if your questions seems similar to that page then i would suggest you the following query which will give the solution for that link

select distinct sno,item_name,max(start_date) over(partition by sno),max(end_date) over(partition by sno),max(creation_date) over(partition by sno), max(last_modified_date) over(partition by sno) from uniq_select_records order by sno,item_name asc;

will given accurate results related to that link

Answer 32

Use the code:

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

This will retrieve the results, irrespective of duplicate values for UserId. If your UserId is unique, well it becomes more simple:

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

Answer 33

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;

Answer 34

Below query can work :

SELECT user_id, value, date , row_number() OVER (PARTITION BY user_id ORDER BY date desc) AS rn
FROM table_name
WHERE rn= 1

Answer 35

SELECT a.* 
FROM user a INNER JOIN (SELECT userid,Max(date) AS date12 FROM user1 GROUP BY userid) b  
ON a.date=b.date12 AND a.userid=b.userid ORDER BY a.userid;