你想找出每个组的前n行。这个答案使用与OP不同的示例数据提供了一个通用的解决方案。

在MySQL 8或更高版本中，您可以根据top 5的确切定义使用ROW_NUMBER, RANK或DENSE_RANK函数。下面是这些函数根据值降序排序生成的数字。注意领带是如何处理的:

一旦你选择了函数，就像这样使用它:

SELECT *
FROM (
    SELECT *, ROW_NUMBER() OVER (PARTITION BY id ORDER BY value DESC) AS n
    FROM t
) AS x
WHERE n <= 5

DB < >小提琴

在MySQL 5。X，你可以使用穷人的排名超过分区，以达到预期的结果:外部连接表本身和每一行，计算它之前的行数(例如，前一行可以是一个较高的值)。

下面将产生类似RANK函数的结果:

SELECT t.pkid, t.catid, t.value, COUNT(b.value) + 1 AS rank
FROM t
LEFT JOIN t AS b ON b.catid = t.catid AND b.value > t.value
GROUP BY t.pkid, t.catid, t.value
HAVING COUNT(b.value) + 1 <= 5
ORDER BY t.catid, t.value DESC, t.pkid

进行以下更改以产生与DENSE_RANK函数类似的结果:

COUNT(DISTINCT b.value)

或进行以下更改，以产生类似于ROW_NUMBER函数的结果:

ON b.catid = t.catid AND (b.value > t.value OR b.value = t.value AND b.pkid < t.pkid)

DB < >小提琴

我刚刚为MYSQL创建了一个top操作。代码很简单。

drop table if exists h;
create table h(id varchar(5), year int, rate numeric(8,2), primary key(id,year));
insert into h(year, id, rate) values
(2006,'p01',8),
(2003,'p01',7.4),
(2008,'p01',6.8),
(2001,'p01',5.9),
(2007,'p01',5.3),
(2009,'p01',4.4),
(2002,'p01',3.9),
(2004,'p01',3.5),
(2005,'p01',2.1),
(2000,'p01',0.8),
(2001,'p02',12.5),
(2004,'p02',12.4),
(2002,'p02',12.2),
(2003,'p02',10.3),
(2000,'p02',8.7),
(2006,'p02',4.6),
(2007,'p02',3.3);

select id, year, rate
from 
(
    select id, year, rate, @last, if(@last=id,@top:=@top+1, @top:=0) as ztop, @last:=id update_last
    from h
    order by id, rate desc, year desc
) t2
where ztop<5

对我来说

SUBSTRING_INDEX(group_concat(col_name order by desired_col_order_name), ',', N) 

完美的工作。没有复杂的查询。

例如:每组取top 1

SELECT 
    *
FROM
    yourtable
WHERE
    id IN (SELECT 
            SUBSTRING_INDEX(GROUP_CONCAT(id
                            ORDER BY rate DESC),
                        ',',
                        1) id
        FROM
            yourtable
        GROUP BY year)
ORDER BY rate DESC;

请尝试下面的存储过程。我已经核实了。我得到正确的结果，但没有使用groupby。

CREATE DEFINER=`ks_root`@`%` PROCEDURE `first_five_record_per_id`()
BEGIN
DECLARE query_string text;
DECLARE datasource1 varchar(24);
DECLARE done INT DEFAULT 0;
DECLARE tenants varchar(50);
DECLARE cur1 CURSOR FOR SELECT rid FROM demo1;
DECLARE CONTINUE HANDLER FOR NOT FOUND SET done = 1;

    SET @query_string='';

      OPEN cur1;
      read_loop: LOOP

      FETCH cur1 INTO tenants ;

      IF done THEN
        LEAVE read_loop;
      END IF;

      SET @datasource1 = tenants;
      SET @query_string = concat(@query_string,'(select * from demo  where `id` = ''',@datasource1,''' order by rate desc LIMIT 5) UNION ALL ');

       END LOOP; 
      close cur1;

    SET @query_string  = TRIM(TRAILING 'UNION ALL' FROM TRIM(@query_string));  
  select @query_string;
PREPARE stmt FROM @query_string;
EXECUTE stmt;
DEALLOCATE PREPARE stmt;

END

花了一些工作，但我认为我的解决方案将是一些分享，因为它看起来很优雅，以及相当快。

SELECT h.year, h.id, h.rate 
  FROM (
    SELECT id, 
      SUBSTRING_INDEX(GROUP_CONCAT(CONCAT(id, '-', year) ORDER BY rate DESC), ',' , 5) AS l
      FROM h
      WHERE year BETWEEN 2000 AND 2009
      GROUP BY id
      ORDER BY id
  ) AS h_temp
    LEFT JOIN h ON h.id = h_temp.id 
      AND SUBSTRING_INDEX(h_temp.l, CONCAT(h.id, '-', h.year), 1) != h_temp.l

请注意，这个示例是为问题的目的而指定的，可以很容易地修改以用于其他类似的目的。

SELECT year, id, rate
FROM (SELECT
  year, id, rate, row_number() over (partition by id order by rate DESC)
  FROM h
  WHERE year BETWEEN 2000 AND 2009
  AND id IN (SELECT rid FROM table2)
  GROUP BY id, year
  ORDER BY id, rate DESC) as subquery
WHERE row_number <= 5

子查询与您的查询几乎相同。只有改变是增加

row_number() over (partition by id order by rate DESC)