每个键选择两行最新时间

Question

我在 MySQL 表中有数据 tbl_test:

create table `tbl_test` (
  `id` INT(10) NOT NULL UNIQUE,
  `value` char(30),
  `time` timestamp,
  `id1` int(10)
  );

INSERT INTO
`tbl_test` (`id`,`value`,`time`,`id1`)
VALUES
  ('3638','value1','2014-11-16 02:01:48','1'),
  ('3639','value2','2014-11-14 13:00:45','1'),
  ('3642','value3','2014-11-14 13:00:40','1'),
  ('3769','value4','2014-11-15 22:21:50','2'),
  ('3770','value5','2014-11-15 22:21:55','2'),
  ('3789','value6','2014-11-14 16:08:20','2'),
  ('3870','value7','2014-11-16 02:01:49','1');

期望的结果（每个id1 2 行，每个order by time desc limit 2）：

+------+--------+---------------------+----+
| id   | value  | time                |id1 |
+------+--------+---------------------+----+
| 3769 | value4 | 2014-11-15 22:21:50 | 2  |
| 3770 | value5 | 2014-11-15 22:21:55 | 2  |
| 3638 | value1 | 2014-11-16 02:01:48 | 1  |
| 3870 | value7 | 2014-11-16 02:01:49 | 1  |
+------+--------+---------------------+----+

我的查询最接近：

select * from (
   select max(id) as id from tbl_test group by id1
   union all
   select max(id) from tbl_test tmp1
   where id not in (SELECT max(id) from tbl_test tmp2 where tmp1.id1 = tmp2.id1)
   group by id1
   ) as tmp
left join tbl_test USING(id);

+------+--------+---------------------+----+
| id   | value  | time                |id1 |
+------+--------+---------------------+----+
| 3770 | value5 | 2014-11-15 22:21:55 | 2  |
| 3789 | value6 | 2014-11-14 16:08:20 | 2  |
| 3642 | value3 | 2014-11-14 13:00:40 | 1  |
| 3870 | value7 | 2014-11-16 02:01:49 | 1  |
+------+--------+---------------------+----+

但我对max(id) 有疑问。表格是网格的，我需要时间 desc 限制 2。

http://sqlfiddle.com/#!2/f6f53/3

Answer 1

我认为这就是你要找的东西：

“对于每个id1，返回最新的time 的两行（最大的id 作为决胜局）”：

使用标准 SQL 窗口函数会很简单：

SELECT id, value, time, id1
FROM  (
   SELECT t.*
        , row_number() OVER (PARTITION BY id1 ORDER BY time DESC, id DESC) AS rn
   FROM   tbl_test t   
   ) sub
WHERE  rn < 3
ORDER  BY id1 DESC, time DESC, id DESC;

但是MySQL不提供窗口函数。您可以用用户定义的变量替换：

SELECT id, value, time, id1
FROM  (
   SELECT t.*
        , @rn := CASE WHEN @prev = id1 THEN @rn + 1 ELSE 1 END rn
        , @prev := id1
   FROM   tbl_test t
       , (SELECT @rn := 0) r
   ORDER BY id1 DESC, time DESC, id DESC
   ) sub
WHERE  rn < 3
ORDER  BY id1 DESC, time DESC, id DESC;

SQL Fiddle.

相关问题：

• In SQL, how to select the top 2 rows for each group

Answer 2

您似乎想要每个 id1 的最近时间的行。

select t.*
from tbl_test t  join
     (select id1, max(time) as maxtime
      from tbl_test
      group by id1
     ) tt
     on t.id1 = tt.id1 and t.time = tt.maxtime
order by time;

http://sqlfiddle.com/#!2/4e010/1/0