mysql – 获得每所学校前10名学生的平均成绩

我们有一所拥有38所小学的学区.孩子们参加了考试.学校的平均分布广泛,但我想比较每所学校的10名学生的平均分.

要求：仅使用临时表.

我这样做的工作量非常大,容易出错,如下所示.
(sch_code =例如,9043;
  – schabbrev =例如,“Carter”;
  – 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：这样可行,但有没有更好的方法呢？

谢谢！

PS – 看起来像家庭作业,但这是……真实的.

解决方法:

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

标签：mysql,limit,temp-tables
来源： https://codeday.me/bug/20190610/1211926.html