Question

In PostgreSQL i want to fetch every users at once and order them by date.

This is my query:

SELECT id, useridx, isread, message, date
  FROM messages
 WHERE isread = 1
 GROUP BY useridx
 ORDER BY date DESC

This is a sample data:

------------------------------------------------------
+  id  |  useridx |  isread  |  messsage |  date     +
------------------------------------------------------
   1   |  1       |  0        | Hello    |  2012-01-01    
   2   |  2       |  1        | Hi       |  2012-01-02    
   3   |  3       |  1        | Test     |  2012-01-03    
   4   |  3       |  0        | My Msg   |  2012-01-04    
   5   |  4       |  1        | sadasd   |  2012-01-05    
   6   |  4       |  1        | sdfsdfd  |  2012-01-06    
   7   |  4       |  0        | sdfsdfsd |  2012-01-07    
   8   |  5       |  0        | 5345634  |  2012-01-08
   9   |  6       |  0        | sdfdfsd  |  2012-01-09
   10  |  7       |  0        | sdfsdfsf |  2012-01-10
------------------------------------------------------

Now, what i want to do is fetch this table by grouping them via useridx and order by date.

Expected Result:

------------------------------------------------------
+  id  |  useridx |  isread  |  messsage |  date     +
------------------------------------------------------  
   6   |  4       |  1        | sdfsdfd  |  2012-01-06 
   3   |  3       |  1        | Test     |  2012-01-03  
   2   |  2       |  1        | Hi       |  2012-01-02    
------------------------------------------------------

Actual Result

ERROR:  column "messages.date" must appear in the GROUP BY clause or be used in an aggregate function

I do not want to group date either. I just want to group with useridx and sort them by date DESC.

Any help/idea is appreciated!

Note: I also tried Distinct. Not fit my needs or i did wrongly.

I am very confused and stuck between DISTINCT ON and rank() methods.

Conclusion:
For who get the same problem here can read this as an answer. Both @kgrittn’s and @mu is too short’s answers are correct. I will continue to use both answers and schemas on my project and in time i can understand which one is the best -i guess-. So, pick one of them and continue to your work. You will be just fine.

Last Update: Sometimes, Distinct On excludes some ids from result. Lets say i have a id column and i have 6 rows which is same. So, distinct on exlude it from the result BUT rank() just result it. So, use rank()!

Answer 1

You want to use the rank() window function to order the results within each useridx group and then peel off the first one by wrapping the ranked results in a derived table:

select id, useridx, isread, message, date
from (
    select id, useridx, isread, message, date,
           rank() over (partition by useridx order by date desc) as r
    from messages
    where isread = 1
) as dt
where r = 1

That will give your the rows with id 2, 3, and 6 from your sample. You might want to add a secondary sort key in the over to consistently make a choice when you have multiple messages per useridx on the same date.

You’ll need at least PostgreSQL 8.4 (AFAIK) to have window functions.