MySQL查询，MAX（）+ GROUP BY

Question

Daft SQL问题。我有一个像这样的表（“PID”是自动增量主要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; 

实际的表数据：

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) 
; 

我想要得到的最新行的PID为每个RID（1每个唯一RID的结果）。对于样本数据，我想：

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 

我试着运行下面的查询：

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

，我也得到：

max(timestamp)  ; rid; pid 
---------------------------- 
2011-04-14 01:06:06; 1 ; 1 
2011-04-14 01:05:41; 2 ; 3 
2011-04-14 01:14:14; 3 ; 7 

的PID返回总是在第一次发生PID的RID（行/ pid 1是第一次使用rid 1，row/pid 3是第一次使用RID 2，row/pid 7是第一次使用rid 3）。虽然返回每个rid的最大时间戳，但pid不是来自原始表的时间戳的pid。什么查询会给我我要找的结果？

Answer 1

（PostgreSQL中9.something测试）

确定RID和时间戳。

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 

加入到它。

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

尝试：

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

Answer 3

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 4

你也可以有这样的子查询：

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 

但这样一来，就需要多一个子查询，如果你想包含在显示的列cost等

所以，group by和join是更好的解决方案。

Answer 5

我在rid和timestamp上创建了一个索引。

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 

显示行0 - 2（3总计，查询花费0.0104秒）

此方法将从theTable（测试）选择所有的需要的值，留在所有的时间标记加入本身（MAXT）高于在同一个摆脱测试的人。当时间戳已经是测试中的最高时间时，maxt上没有匹配 - 这是我们正在寻找的 - maxt上的值变为NULL。现在我们使用WHERE子句maxt.rid IS NULL或maxt上的任何其他列。

Answer 6

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

希望我帮了忙！

Answer 7

如果你想避免JOIN，你可以使用：

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