查询:
SELECT
year, id, rate
FROM h
WHERE year BETWEEN 2000 AND 2009
AND id IN (SELECT rid FROM table2)
GROUP BY id, year
ORDER BY id, rate DESC
收益率:
year id rate
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
我想要的是每个id只有前5个结果:
2006 p01 8
2003 p01 7.4
2008 p01 6.8
2001 p01 5.9
2007 p01 5.3
2001 p02 12.5
2004 p02 12.4
2002 p02 12.2
2003 p02 10.3
2000 p02 8.7
是否有一种方法来做到这一点,使用一些限制之类的修饰符,在GROUP BY中工作?
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:
pkid |
catid |
value |
row_number |
rank |
dense_rank |
1 |
p01 |
100 |
*1 |
*1 |
*1 |
2 |
p01 |
90 |
*2 |
*2 |
*2 |
3 |
p01 |
90 |
*3 |
*2 |
*2 |
4 |
p01 |
80 |
*4 |
*4 |
*3 |
5 |
p01 |
80 |
*5 |
*4 |
*3 |
6 |
p01 |
80 |
6 |
*4 |
*3 |
7 |
p01 |
70 |
7 |
7 |
*4 |
8 |
p01 |
60 |
8 |
8 |
*5 |
9 |
p01 |
50 |
9 |
9 |
6 |
10 |
p01 |
40 |
10 |
10 |
7 |
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
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)
这需要一系列子查询对值进行排序、限制,然后在分组时执行求和
@Rnk:=0;
@N:=2;
select
c.id,
sum(c.val)
from (
select
b.id,
b.bal
from (
select
if(@last_id=id,@Rnk+1,1) as Rnk,
a.id,
a.val,
@last_id=id,
from (
select
id,
val
from list
order by id,val desc) as a) as b
where b.rnk < @N) as c
group by c.id;
对于那些像我一样有查询超时的人。我做了下面的限制和任何其他由特定的组。
DELIMITER $$
CREATE PROCEDURE count_limit200()
BEGIN
DECLARE a INT Default 0;
DECLARE stop_loop INT Default 0;
DECLARE domain_val VARCHAR(250);
DECLARE domain_list CURSOR FOR SELECT DISTINCT domain FROM db.one;
OPEN domain_list;
SELECT COUNT(DISTINCT(domain)) INTO stop_loop
FROM db.one;
-- BEGIN LOOP
loop_thru_domains: LOOP
FETCH domain_list INTO domain_val;
SET a=a+1;
INSERT INTO db.two(book,artist,title,title_count,last_updated)
SELECT * FROM
(
SELECT book,artist,title,COUNT(ObjectKey) AS titleCount, NOW()
FROM db.one
WHERE book = domain_val
GROUP BY artist,title
ORDER BY book,titleCount DESC
LIMIT 200
) a ON DUPLICATE KEY UPDATE title_count = titleCount, last_updated = NOW();
IF a = stop_loop THEN
LEAVE loop_thru_domain;
END IF;
END LOOP loop_thru_domain;
END $$
它循环遍历一个域列表,然后每个域只插入200个限制
试试这个:
SELECT h.year, h.id, h.rate
FROM (SELECT h.year, h.id, h.rate, IF(@lastid = (@lastid:=h.id), @index:=@index+1, @index:=0) indx
FROM (SELECT h.year, h.id, h.rate
FROM h
WHERE h.year BETWEEN 2000 AND 2009 AND id IN (SELECT rid FROM table2)
GROUP BY id, h.year
ORDER BY id, rate DESC
) h, (SELECT @lastid:='', @index:=0) AS a
) h
WHERE h.indx <= 5;
试试这个:
SET @num := 0, @type := '';
SELECT `year`, `id`, `rate`,
@num := if(@type = `id`, @num + 1, 1) AS `row_number`,
@type := `id` AS `dummy`
FROM (
SELECT *
FROM `h`
WHERE (
`year` BETWEEN '2000' AND '2009'
AND `id` IN (SELECT `rid` FROM `table2`) AS `temp_rid`
)
ORDER BY `id`
) AS `temph`
GROUP BY `year`, `id`, `rate`
HAVING `row_number`<='5'
ORDER BY `id`, `rate DESC;