SQL - SELECT MAX() and accompanying field

Question

What I have is basically a problem which is easily solved with multiple tables, but I have only a single table to do it.

Consider the following database table

<

Answer 1

I think I have a solution that's different from the ones already proposed:

select *
from foo
where id = (
  select id
  from foo F
  where F.bar = foo.bar
  order by F.baz
  limit 1
)

This gives you all the foo records that have the greatest baz compared to other foo records with the same bar.

Answer 2

If you're on SQL Server 2005 or higher,

SELECT  UserID, UserName, EmailAddress, Source
FROM    (SELECT UserID, UserName, EmailAddress, Source,
                ROW_NUMBER() OVER (PARTITION BY UserID
                                   ORDER BY EmailAddress DESC) 
                    AS RowNumber
         FROM   MyTable) AS a
WHERE   a.RowNumber = 1

Of course there are ways to do the same task without the (SQL-Standard) ranking functions such as ROW_NUMBER, which SQL Server implemented only since 2005 -- including nested dependent queries and self left joins with an ON including a '>' and a WHERE ... IS NULL trick -- but the ranking functions make for code that's readable and (in theory) well optimizable by the SQL Server Engine.

Edit: this article is a nice tutorial on ranking, but it uses RANK in the examples instead of ROW_NUMBER (or the other ranking function, DENSE_RANK) -- the distinction matters when there are "ties" among grouped rows in the same partition according to the ordering criteria. this post does a good job explaining the difference.

Answer 3

select distinct * from table t1
where EmailAddress = 
(select max(EmailAddress) from table t2
where t1.userId = t2.userId)

Answer 4

select distinct
     *  
from    
   SomeTable a
inner join (
  select max(emailAddress), userId
  from
     SomeTable 
  group by 
     userId
) b on a.emailAddress = b.emailAddress and a.userId = b.userId