Re: Need SQL Help Finding Current Status of members

On Thu, Dec 15, 2005 at 08:31:09PM -0500, Michael Avila wrote:
> What I want to do is find the latest status for each member. Actually I want
> to find all those with an status of "A". But it must be the current (latest)
> status. How do I find the most current date for each member in a pile of
> many records for many members with many status settings with one SQL
> statement?

Suppose you have this table:

SELECT * FROM memberstatus;
member_id | status_code | status_date 
-----------+-------------+-------------        1 | a           | 2005-01-01        2 | x           | 2005-01-01
3| x           | 2005-01-01        4 | x           | 2005-01-01        1 | x           | 2005-12-15        2 | a
  | 2005-12-15        3 | y           | 2005-12-15        4 | a           | 2005-12-15
 
(8 rows)

Let's order the data so all of a member's records are shown together,
with the latest one first:

SELECT * FROM memberstatus
ORDER BY member_id, status_date DESC;
member_id | status_code | status_date 
-----------+-------------+-------------        1 | x           | 2005-12-15        1 | a           | 2005-01-01
2| a           | 2005-12-15        2 | x           | 2005-01-01        3 | y           | 2005-12-15        3 | x
  | 2005-01-01        4 | a           | 2005-12-15        4 | x           | 2005-01-01
 
(8 rows)

One way to get only the first record for each member is to use
PostgreSQL's nonstandard DISTINCT ON construct:

SELECT DISTINCT ON (member_id) * FROM memberstatus
ORDER BY member_id, status_date DESC;
member_id | status_code | status_date 
-----------+-------------+-------------        1 | x           | 2005-12-15        2 | a           | 2005-12-15
3| y           | 2005-12-15        4 | a           | 2005-12-15
 
(4 rows)

We could put the above in a subquery and restrict the output to the
records we want:

SELECT * FROM ( SELECT DISTINCT ON (member_id) * FROM memberstatus ORDER BY member_id, status_date DESC
) AS s
WHERE status_code = 'a'
ORDER BY member_id;
member_id | status_code | status_date 
-----------+-------------+-------------        2 | a           | 2005-12-15        4 | a           | 2005-12-15
(2 rows)

This isn't the only way; search the archives for alternatives.

-- 
Michael Fuhr