how do I query sql for a latest record date for each user

I have a table that is a collection entries as to when a user was logged on.

username, date,      value
--------------------------
brad,     1/2/2010,  1.1
fred,     1/3/2010,  1.0
bob,      8/4/2009,  1.5
brad,     2/2/2010,  1.2
fred,     12/2/2009, 1.3

etc..

How do I create a query that would give me the latest date for each user?

Update: I forgot that I needed to have a value that goes along with the latest date.

select t.username, t.date, t.value
from MyTable t
inner join (
    select username, max(date) as MaxDate
    from MyTable
    group by username
) tm on t.username = tm.username and t.date = tm.MaxDate

Using window functions (works in Oracle, Postgres 8.4, SQL Server 2005, DB2, Sybase, Firebird 3.0, MariaDB 10.3)

select * from (
	select
		username,
		date,
		value,
		row_number() over(partition by username order by date desc) as rn
	from
		yourtable
) t
where t.rn = 1

I see most of the developers use an inline query without considering its impact on huge data.

Simply, you can achieve this by:

SELECT a.username, a.date, a.value
FROM myTable a
LEFT OUTER JOIN myTable b
ON a.username = b.username 
AND a.date < b.date
WHERE b.username IS NULL
ORDER BY a.date desc;

From my experience the fastest way is to take each row for which there is no newer row in the table.

Another advantage is that the syntax used is very simple, and that the meaning of the query is rather easy to grasp (take all rows such that no newer row exists for the username being considered).

NOT EXISTS

SELECT username, value
FROM t
WHERE NOT EXISTS (
  SELECT *
  FROM t AS witness
  WHERE witness.username = t.username AND witness.date > t.date
);

ROW_NUMBER

SELECT username, value
FROM (
  SELECT username, value, row_number() OVER (PARTITION BY username ORDER BY date DESC) AS rn
  FROM t
) t2
WHERE rn = 1

INNER JOIN

SELECT t.username, t.value
FROM t
INNER JOIN (
  SELECT username, MAX(date) AS date
  FROM t
  GROUP BY username
) tm ON t.username = tm.username AND t.date = tm.date;

LEFT OUTER JOIN

SELECT username, value
FROM t
LEFT OUTER JOIN t AS w ON t.username = w.username AND t.date < w.date
WHERE w.username IS NULL

To get the whole row containing the max date for the user:

select username, date, value
from tablename where (username, date) in (
    select username, max(date) as date
    from tablename
    group by username
)

SELECT *     
FROM MyTable T1    
WHERE date = (
   SELECT max(date)
   FROM MyTable T2
   WHERE T1.username=T2.username
)

This one should give you the correct result for your edited question.

The sub-query makes sure to find only rows of the latest date, and the outer GROUP BY will take care of ties. When there are two entries for the same date for the same user, it will return the one with the highest value.

SELECT t.username, t.date, MAX( t.value ) value
FROM your_table t
JOIN (
       SELECT username, MAX( date ) date
       FROM your_table
       GROUP BY username
) x ON ( x.username = t.username AND x.date = t.date )
GROUP BY t.username, t.date

SELECT DISTINCT Username, Dates,value 
FROM TableName
WHERE  Dates IN (SELECT  MAX(Dates) FROM TableName GROUP BY Username)


Username	Dates	    value
bob       	2010-02-02	1.2       
brad      	2010-01-02	1.1       
fred      	2010-01-03	1.0       

This is similar to one of the answers above, but in my opinion it is a lot simpler and tidier. Also, shows a good use for the cross apply statement. For SQL Server 2005 and above...

select
    a.username,
    a.date,
    a.value,
from yourtable a
cross apply (select max(date) 'maxdate' from yourtable a1 where a.username=a1.username) b
where a.date=b.maxdate

You could also use analytical Rank Function

	with temp as 
(
select username, date, RANK() over (partition by username order by date desc) as rnk from t
)
select username, rnk from t where rnk = 1

SELECT MAX(DATE) AS dates 
FROM assignment  
JOIN paper_submission_detail ON  assignment.PAPER_SUB_ID = 
     paper_submission_detail.PAPER_SUB_ID 

If your database syntax supports it, then TOP 1 WITH TIES can be a lifesafer in combination with ROWNUMER.

With the example data you provided, use this query:

SELECT TOP 1 WITH TIES
  username, date, value
FROM user_log_in_attempts
ORDER BY ROW_NUMBER() OVER (PARTITION BY username ORDER BY date DESC)

It yields:

username | date      | value
-----------------------------
bob      | 8/4/2009  | 1.5
brad     | 2/2/2010  | 1.2
fred     | 12/2/2009 | 1.3

Demo

How it works:

• ROWNUMBER() OVER (PARTITION BY... ORDER BY...) For each username a list of rows is calculated from the youngest (rownumber=1) to the oldest (rownumber=high)
• ORDER BY ROWNUMBER... sorts the youngest rows of each user to the top, followed by the second-youngest rows of each user, and so on
• TOP 1 WITH TIES Because each user has a youngest row, those youngest rows are equal in the sense of the sorting criteria (all have rownumber=1). All those youngest rows will be returned.

Tested with SQL-Server.

SELECT Username, date, value
 from MyTable mt
 inner join (select username, max(date) date
              from MyTable
              group by username) sub
  on sub.username = mt.username
   and sub.date = mt.date

Would address the updated problem. It might not work so well on large tables, even with good indexing.

SELECT *
FROM ReportStatus c
inner join ( SELECT 
  MAX(Date) AS MaxDate
  FROM ReportStatus ) m
on  c.date = m.maxdate

SELECT t1.username, t1.date, value
FROM MyTable as t1
INNER JOIN (SELECT username, MAX(date)
            FROM MyTable
            GROUP BY username) as t2 ON  t2.username = t1.username AND t2.date = t1.date

For Oracle sorts the result set in descending order and takes the first record, so you will get the latest record:

select * from mytable
where rownum = 1
order by date desc

Select * from table1 where lastest_date=(select Max(latest_date) from table1 where user=yourUserName)

Inner Query will return the latest date for the current user, Outer query will pull all the data according to the inner query result.

I used this way to take the last record for each user that I have on my table. It was a query to get last location for salesman as per recent time detected on PDA devices.

CREATE FUNCTION dbo.UsersLocation()
RETURNS TABLE
AS
RETURN
Select GS.UserID, MAX(GS.UTCDateTime) 'LastDate'
From USERGPS GS
where year(GS.UTCDateTime) = YEAR(GETDATE()) 
Group By GS.UserID
GO
select  gs.UserID, sl.LastDate, gs.Latitude , gs.Longitude
		from USERGPS gs
		inner join USER s on gs.SalesManNo = s.SalesmanNo 
		inner join dbo.UsersLocation() sl on gs.UserID= sl.UserID and gs.UTCDateTime = sl.LastDate 
		order by LastDate desc

SELECT * FROM TABEL1 WHERE DATE= (SELECT MAX(CREATED_DATE) FROM TABEL1)

My small compilation

• self join better than nested select
• but group by doesn't give you primary key which is preferable for join
• this key can be given by partition by in conjunction with first_value (docs)

So, here is a query:

select
t.*
from
Table t inner join (
select distinct first_value(ID) over(partition by GroupColumn order by DateColumn desc) as ID
from Table
where FilterColumn = 'value'
) j on t.ID = j.ID

Pros:

• Filter data with where statement using any column
• select any columns from filtered rows

Cons:

• Need MS SQL Server starting with 2012.

I did somewhat for my application as it:

Below is the query:

select distinct i.userId,i.statusCheck, l.userName from internetstatus 
as i inner join login as l on i.userID=l.userID 
where nowtime in((select max(nowtime) from InternetStatus group by userID));	

This should also work in order to get all the latest entries for users.

SELECT username, MAX(date) as Date, value
FROM MyTable
GROUP BY username, value

You would use aggregate function MAX and GROUP BY

SELECT username, MAX(date), value FROM tablename GROUP BY username, value