Question

I should have paid more attention in relational databases class, so I need some help.

I have a table structured like so (there’s more, but I’m only posting relevant info):

+------------------+-------------+------+-----+-------------------+-------+ | Field            | Type        | Null | Key | Default           | Extra | +------------------+-------------+------+-----+-------------------+-------+ | session_id       | char(32)    | NO   |     | NULL              |       | | file             | varchar(30) | NO   |     | NULL              |       | | access_time      | timestamp   | NO   |     | CURRENT_TIMESTAMP |       | +------------------+-------------+------+-----+-------------------+-------+` 

I want to select the file field of the record with the greatest access_time for each session_id. So pretty much the last page visited in each session.

I’ve tried countless combinations of DISTINCT and the GROUP BY clause, but I can’t seem to wrap my head around it for some reason. I’m using MySQL. Help please.

Answer 1

I know this will work in SQL Server, and it should port to MySQL.

select session_id, timestamp, file  from table t1 join (select session_id, max(timestamp) as timestamp)          from table group by session_id) as t2    on t1.session_id = t2.session_id      and t1.timestamp = t2.timestamp