“SQL 学生 课程 分数”的版本间的差异
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Jihongchang(讨论 | 贡献) |
Jihongchang(讨论 | 贡献) |
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第56行: | 第56行: | ||
|210 | |210 | ||
|} | |} | ||
− | + | 所以应该是:<syntaxhighlight lang="sql"> | |
+ | select distinct (student_id) | ||
+ | from score | ||
+ | group by student_id | ||
+ | having sum(score) = ( | ||
+ | select sum(score) | ||
+ | from score | ||
+ | group by student_id | ||
+ | order by sum(score) desc | ||
+ | limit 1 | ||
+ | ) | ||
+ | </syntaxhighlight> | ||
{| class="wikitable" | {| class="wikitable" | ||
|+ | |+ |
2024年7月7日 (日) 04:18的版本
建表
create table course (
id integer primary key,
name text not null
);
create table student (
id integer primary key,
name text not null
);
create table score (
id integer primary key,
course_id integer not null,
student_id integer not null,
score integer not null
);
准备数据
insert into course values (1, '语文'), (2, '数学'),(3,'外语');
insert into student values (1, '小张'), (2, '小王'), (3, '小马');
insert into score values (1, 1, 1, 80), (2,2,1,90), (3,3,1,70);
insert into score values (4,1,2,70),(5,2,2,90),(6,3,2,80);
insert into score values (7,1,3,80),(8,2,3,60),(9,3,3,70);
查询总分最高的学生分数是多少
先查最高分是多少
select student_id, sum(score)
from score
group by student_id
order by sum(score) desc
limit 1;
但其实可能存在多个学生的总分一样:
select student_id, sum(score)
from score
group by student_id
order by sum(score) desc;
student_id | sum(score) |
---|---|
1 | 240 |
2 | 240 |
3 | 210 |
所以应该是:
select distinct (student_id)
from score
group by student_id
having sum(score) = (
select sum(score)
from score
group by student_id
order by sum(score) desc
limit 1
)