PostgreSQL does not allow me to group a column with order

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.

PostgreSQL, unlike MySQL, does not show random data for columns which are not aggregated in an aggregated query.

The solution is in the error message

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

Which means you must GROUP BY the "messages.date" column or use an aggregate function like MIN() or MAX() when selection this column

Example:

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

Another option is to use SELECT DISTINCT ON (which is very different from a simple SELECT DISTINCT):

SELECT *
  FROM (SELECT DISTINCT ON (useridx)
            id, useridx, isread, message, date
          FROM messages
          WHERE isread = 1
          ORDER BY useridx, date DESC) x
  ORDER BY date DESC;

In some cases this can scale better than the other approaches.

Years later, but can't you just order in the FROM subquery:

SELECT m.id, m.useridx, m.isread, m.message, m.date
FROM (
   SELECT m2.id, m2.useridx, m2.isread, m2.message, m2.date 
   FROM message m2 
   ORDER BY m2.id ASC, m2.date DESC
) m
WHERE isread = 1
GROUP BY useridx

This works for me in PostgreSQL 9.2