Select Earliest Date and Time from List of Distinct User Sessions

Question

I have a table of user access sessions which records website visitor activity:

accessid, userid, date, time, url

I\'m trying to retrieve all dis

Answer 1

This is a variation of the "greatest-n-per-group" problem that comes up on StackOverflow several times per week.

SELECT 
        a1.accessid, 
        a1.date, 
        a1.time 
FROM 
        accesslog a1
LEFT OUTER JOIN
        accesslog a2
  ON (a1.accessid = a2.accessid AND a1.userid = a2.userid
    AND (a1.date > a2.date OR a1.date = a2.date AND a1.time > a2.time))
WHERE a1.userid = '1234'
  AND a2.accessid IS NULL;

The way this works is that we try to find a row (a2) that has the same accessid and userid, and an earlier date or time than the row a1. When we can't find an earlier row, then a1 must be the earliest row.

---

Re your comment, I just tried it with the sample data you provided. Here's what I get:

+----------+------------+----------+
| accessid | date       | time     |
+----------+------------+----------+
|        1 | 2009-08-15 | 01:01:01 | 
|        2 | 2009-09-01 | 14:01:01 | 
+----------+------------+----------+

I'm using MySQL 5.0.75 on Mac OS X.

Answer 2

Try this

SELECT 
    accessid, 
    date, 
    time 
FROM 
    accesslog 
WHERE userid = '1234' 
GROUP BY accessid
HAVING MIN(date)

It will return all unique accesses with minimum time for each for userid = '1234'.