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/

10-11 15:45