我们有一个拥有 38 所小学的学区。 children 参加了考试。学校的平均水平分布广泛,但我想比较每所学校前 10 名学生的平均水平。
要求:仅使用临时表。
我以一种非常繁重、容易出错的方式完成了这项工作,如下所示。
(sch_code = 例如,9043;
-- schabbrev = 例如,“卡特”;
-- totpct_stu = 例如,61.3)
DROP TEMPORARY TABLE IF EXISTS avg_top10 ;
CREATE TEMPORARY TABLE avg_top10
( sch_code VARCHAR(4),
schabbrev VARCHAR(75),
totpct_stu DECIMAL(5,1)
);
INSERT
INTO avg_top10
SELECT sch_code
, schabbrev
, totpct_stu
FROM test_table
WHERE sch_code IN ('5489')
ORDER
BY totpct_stu DESC
LIMIT 10;
-- I do that last query for EVERY school, so the total
-- length of the code is well in excess of 300 lines.
-- Then, finally...
SELECT schabbrev, ROUND( AVG( totpct_stu ), 1 ) AS top10
FROM avg_top10
GROUP
BY schabbrev
ORDER
BY top10 ;
-- OUTPUT:
-----------------------------------
schabbrev avg_top10
---------- ---------
Goulding 75.4
Garth 77.7
Sperhead 81.4
Oak_P 83.7
Spring 84.9
-- etc...
问题:所以这是有效的,但没有更好的方法来做到这一点吗?
谢谢!
PS——看起来像家庭作业,但这是,嗯……真的。
最佳答案
使用 this technique 。
select sch_code,
schabbrev,
ROUND( AVG( totpct_stu ), 1 ) AS top10
from (select sch_code,
schabbrev,
totpct_stu,
@num := if(@group = sch_code, @num + 1, 1) as row_number,
@group := sch_code as dummy
from test_table
order by sch_code, totpct_stu desc) as x
where row_number <= 10
GROUP BY sch_code,
schabbrev
关于mysql - 获取每所学校前 10 名学生的平均值,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/4661373/