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MySQL query, MAX() + GROUP BY

开发者 https://www.devze.com 2023-02-24 06:37 出处:网络
Daft SQL question. I have a table like so (\'pid\' is auto-increment primary col) CREATE TABLE theTable (

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,
    `timestamp` TIMESTAMP DEFAULT CURRENT_TIMESTAMP,
    `cost` INT UNSIGNED NOT NULL,
    `rid` INT NOT NULL,
) Engine=InnoDB;

Actual table data:

INSERT INTO theTable (`pid`, `timestamp`, `cost`, `rid`)
VALUES
  (1, '2011-04-14 01:05:07', 1122, 1),
  (2, '2011-04-14 00:05:07', 2233, 1),
  (3, '2011-04-14 01:05:41', 4455, 2),
  (4, '2011-04-14 01:01:11', 5566, 2),
  (5, '2011-04-14 01:06:06', 345, 1),
  (6, '2011-04-13 22:06:06', 543, 2),
  (7, '2011-04-14 01:14:14', 5435, 3),
  (8, '2011-04-14 01:10:13', 6767, 3)
;

I want to get the PID of the latest row for each rid (1 result per unique RID). For the sample data, I'd like:

pid | MAX(timestamp)      | rid
-----------------------------------
5   | 2011-04-14 01:06:06 | 1
3   | 2011-04-14 01:05:41 | 2
7   | 2011-04-14 01:14:14 | 3

I've tried running the following query:

SELECT MAX(timestamp),rid,pid FROM theTable GROUP BY rid

and I get:

max(timestamp)     ; rid; pid
------开发者_运维问答----------------------
2011-04-14 01:06:06; 1  ; 1
2011-04-14 01:05:41; 2  ; 3
2011-04-14 01:14:14; 3  ; 7

The PID returned is always the first occurence of PID for an RID (row / pid 1 is frst time rid 1 is used, row / pid 3 the first time RID 2 is used, row / pid 7 is first time rid 3 is used). Though returning the max timestamp for each rid, the pids are not the pids for the timestamps from the original table. What query would give me the results I'm looking for?


(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)


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 !


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


I created an index on rid and timestamp.

SELECT test.pid, test.cost, test.timestamp, test.rid
FROM theTable AS test
LEFT JOIN theTable maxt 
ON maxt.rid = test.rid
AND maxt.timestamp > test.timestamp
WHERE maxt.rid IS NULL 

Showing rows 0 - 2 (3 total, Query took 0.0104 sec)

This method will select all the desired values from theTable (test), left joining itself (maxt) on all timestamps higher than the one on test with the same rid. When the timestamp is already the highest one on test there are no matches on maxt - which is what we are looking for - values on maxt become NULL. Now we use the WHERE clause maxt.rid IS NULL or any other column on maxt.


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.


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);


Try:

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