Return the data from the rows with the most recent date of each distinct candidate_id

Question

I am attempting to return the data from the rows with the most recent date of each distinct candidate_id. It is correctly returning the most recent date (the created_unix co

Answer 1

SELECT m1.*
FROM messages as m1
  LEFT OUTER JOIN messages AS m2
    ON    (m1.candidate_id = m2.candidate_id
      AND (m1.created_unix < m2.created_unix)
WHERE m2.created_unix is NULL
AND employer_id='$employerid' AND last='company'

This joins messages to itself on candidate_id and makes rows with picks all dates in m2 that are greater than each date in m1, substituting NULL if none are greater. So you get NULL precisely when there is no greater date within that candidate_id.

Answer 2

You must group by everything not using an aggregate function:

SELECT candidate_id, message, max(created_unix), jobpost_id, staffuserid 
    FROM messages 
       WHERE employer_id='$employerid' AND last='company' 
          GROUP BY candidate_id, message, jobpost_id, staffuserid 

If your message is different per row and you want to group by candidate_id, then you must not be using message. In that case, simply remove it from your select list and you won't need it in your group by list. The same goes for any other field you aren't using.

Remember, when using aggregate functions, you must contain each field in either an aggregate function or the group by. Otherwise, SQL won't know from which row to pull the data for the row returned.

Update:

After seeing what you're looking for, this will do the trick:

SELECT candidate_id, message, max(created_unix), jobpost_id, staffuserid 
    FROM messages 
       WHERE employer_id='$employerid' AND last='company' AND
       created_unix = (
           SELECT max(subm.created_unix)
           FROM messages subm
           WHERE subm.candidate_id = messages.candidate_id
       )