Related
I was asked this question in an interview. A table, trips, contains the following columns( customer_id, start_from, end_at, start_at_time, end_at_time), with data structured so that each trip is stored as a separate row and a part of the table looks like this: How would you find the list of all the customers who started yesterday from point A and ended yesterday at point P?
I provided solution using windowing functions that identified the list of all customers that started their day at A and then did an inner join of a list of these customers with the customers who ended their day at P( using the same windowing functions).
The solution I gave was this:
SELECT a.customer_id
FROM
(SELECT a.customer_id
FROM
(SELECT customer_id,
start_from,
row_number() OVER (PARTITION BY customer_id
ORDER BY start_at_time ASC) AS rnk
FROM trips
WHERE to_date(start_at_time)= date_sub(CURRENT_DATE, 1) ) AS a
WHERE a.rnk=1
AND a.start_from='A' ) AS a
INNER JOIN
(SELECT a.customer_id
FROM
(SELECT customer_id,
end_at,
row_number() OVER (PARTITION BY customer_id
ORDER BY end_at_time DESC) AS rnk
FROM trips
WHERE to_date(end_at_time)= date_sub(CURRENT_DATE, 1) ) AS a
WHERE a.rnk=1
AND a.end_at='P' ) AS b ON a.customer_id=b.customer_id
My interviewer said my solution was correct but there is a more efficient way to solve this problem. I've searching and trying to find a more efficient way but I could not find one so far. Can you suggest a more efficient way to solve this problem?
I might use first_value() for this:
select t.customer_id
from (select t.*,
first_value(start_from) over (partition by customer_id order by start_at_time) as first_start,
first_value(end_at) over (partition by customer_id order by start_at_time desc) as last_end
from t
where start_at_time >= date_sub(CURRENT_DATE, 1) and
start_at_time < CURRENT_DATE
) t
where first_start = start_from and -- just some filtering so select distinct is not needed
first_start = 'A' and
last_end = 'P';
I should add that many databases support an equivalent function for aggregation, and I would use that instead.
This assumes that starts are not repeated. To be safe, you can add select distinct, but there is a performance hit for that.
A generalized version of what I would probably have done:
SELECT fandl.a
FROM (
SELECT a, MIN(start) AS t0, MAX(start) AS tN
FROM someTable
WHERE start >= DATE_SUB(CURRENT_DATE, 1) AND start < CURRENT_DATE
GROUP BY a
) AS fandl
INNER JOIN someTable AS st0 ON fandl.a = st0.a AND fandl.t0 = st0.start
INNER JOIN someTable AS stN ON fandl.a = stN.a AND fandl.tN = stN.start
WHERE st0.b1 = 'A' AND stN.b2 = 'P'
;
Using the date function you did, since you did not specify sql dialect.
Note that, in many RDBMS, if there is an (a, start) index, the subquery and joins can be done with the index alone; actual table access would only be required for the final WHERE evaluation.
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)
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.
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.
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.
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 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
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
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.
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.
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;
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:
Accepted answer
My solution with Postgres
My solution with Oracle
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)
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;
select UserId,max(Date) over (partition by UserId) value from users;
I am using a sql to select data from a table for a particular range of records.
I am using rownum to implement greater and less than logic to arrive at the required set of data.
With the use of below sql I can fetch the records from 21 to 40 from my table. In total for this condition table contains 100 rows.
Through this sql I want to fetch an indicator(value) which tells that there are more records for this condition.
I could not find any solution in google.
Sql -
select * from ( select rownum rnum, a.* from(SELECT TO_CHAR(D.DATE,'YYYYMMDD'),D.TYPE,
TO_CHAR(D.VDATE,'YYYYMMDD'),D.AMT,D.PARTICULAR,D.NUM,D.ID,
D.CODE,D.INFO FROM MySCHEMA
.MYTABLE D WHERE D.DATE >= TO_CHAR(TO_DATE('20160701','YYYYMMDD'),'DD-MON-RRRR')
AND D.DATE <= TO_CHAR(TO_DATE('20161105','YYYYMMDD'),'DD-MON-RRRR') AND D.XDATE >= TO_CHAR(TO_DATE('20160701','YYYYMMDD'),'DD-MON-RRRR')
AND D.XDATE <= TO_CHAR(TO_DATE('20161105','YYYYMMDD'),'DD-MON-RRRR')
AND D.FLG='Y' AND D.TYPE IN('D','C')
AND
D.ACI = 'CO6'
ORDER BY D.DATE DESC
)
a where rownum <= 40 ) where rnum >= 21;
You can add a count(*) over () total_rows to your inner select. That will tell you how many rows the inner query would return without the rownum predicates. That is going to mean, however, that every time you ask for a page of results, Oracle has to execute the inner query completely and then discard all the rows you aren't fetching. That's going to be more expensive than what you are currently doing
select *
from ( select rownum rnum, a.*
from(SELECT count(1) over () total_rows,
<<the rest of your query>>
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)
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.
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.
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.
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 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
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
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.
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.
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;
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:
Accepted answer
My solution with Postgres
My solution with Oracle
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)
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;
select UserId,max(Date) over (partition by UserId) value from users;
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)
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.
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.
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.
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 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
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
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.
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.
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;
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:
Accepted answer
My solution with Postgres
My solution with Oracle
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)
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;
select UserId,max(Date) over (partition by UserId) value from users;