MySQL query, MAX() + GROUP BY

Question

Daft SQL question. I have a table like so (\'pid\' is auto-increment primary col)

CREATE TABLE theTable (
    `pid` INT UNSIGNED PRIMARY KEY AUTO_INCREMENT,
          


        

Answer 1

(Tested in PostgreSQL 9.something)

Identify the rid and timestamp.

select rid, max(timestamp) as ts
from test
group by rid;

1   2011-04-14 18:46:00
2   2011-04-14 14:59:00

Join to it.

select test.pid, test.cost, test.timestamp, test.rid
from test
inner join 
    (select rid, max(timestamp) as ts
    from test
    group by rid) maxt
on (test.rid = maxt.rid and test.timestamp = maxt.ts)

Answer 2

select *
from (
    select `pid`, `timestamp`, `cost`, `rid`
    from theTable 
    order by `timestamp` desc
) as mynewtable
group by mynewtable.`rid`
order by mynewtable.`timestamp`

Hope I helped !

Answer 3

If you want to avoid a JOIN, you can use:

SELECT pid, rid FROM theTable t1 WHERE t1.pid IN ( SELECT MAX(t2.pid) FROM theTable t2 GROUP BY t2.rid);

Answer 4

You could also have subqueries like that:

SELECT ( SELECT MIN(t2.pid)
         FROM test t2
         WHERE t2.rid = t.rid
           AND t2.timestamp = maxtimestamp
       ) AS pid 
     , MAX(t.timestamp) AS maxtimestamp
     , t.rid
FROM test t
GROUP BY t.rid

But this way, you'll need one more subquery if you want cost included in the shown columns, etc.

So, the group by and join is better solution.

Answer 5

SELECT t.pid, t.cost, to.timestamp, t.rid
FROM test as t
JOIN (
    SELECT rid, max(tempstamp) AS maxtimestamp
    FROM test GROUP BY rid
) AS tmax
    ON t.pid = tmax.pid and t.timestamp = tmax.maxtimestamp

Answer 6

Try:

select pid,cost, timestamp, rid from theTable order by timestamp DESC limit 2;