“SQL 学生 课程 分数 2”的版本间的差异
跳到导航
跳到搜索
Jihongchang(讨论 | 贡献) |
Jihongchang(讨论 | 贡献) (→准备数据) |
||
第23行: | 第23行: | ||
insert into course values (1, '语文'), (2, '数学'),(3,'外语'); | 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'), (4, '小李'); | + | insert into student values (1, '小张'), (2, '小王'), (3, '小马'), (4, '小李'),(5,'小赵'),(6, '小孙'); |
#小张只考了语文 | #小张只考了语文 | ||
第33行: | 第33行: | ||
#小李考了语文、数学、外语,外语考了两次 | #小李考了语文、数学、外语,外语考了两次 | ||
insert into score values (10,1,4,80),(11,2,4,60),(12,3,4,70),(13,3,4,80); | insert into score values (10,1,4,80),(11,2,4,60),(12,3,4,70),(13,3,4,80); | ||
+ | #小赵考了两遍语文 | ||
+ | insert into score values (14,1,5,80),(15,1,5,60); | ||
+ | #小孙考了数学和外语 | ||
+ | insert into score values (16,2,6,80),(17,3,6,60); | ||
</syntaxhighlight> | </syntaxhighlight> |
2024年7月9日 (二) 07:31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'), (4, '小李'),(5,'小赵'),(6, '小孙');
#小张只考了语文
insert into score values (1, 1, 1, 80);
#小王考了语文和数学
insert into score values (4,1,2,70),(5,2,2,90);
#小马考了语文、数学、外语
insert into score values (7,1,3,80),(8,2,3,60),(9,3,3,70);
#小李考了语文、数学、外语,外语考了两次
insert into score values (10,1,4,80),(11,2,4,60),(12,3,4,70),(13,3,4,80);
#小赵考了两遍语文
insert into score values (14,1,5,80),(15,1,5,60);
#小孙考了数学和外语
insert into score values (16,2,6,80),(17,3,6,60);
查询考了语文数学的学生,显示姓名,要求两种实现,并且其中一种要用到group by
第一种
select s.id, s.name
from (
select distinct student_id, course_id
from score
where course_id in (1,2)
) sc left join student s
on sc.student_id = s.id
group by student_id having count(1) = 2;
第二种
select t1.student_id, s.name
from (
select distinct student_id
from score
where course_id = 1
) t1 inner join (
select distinct student_id
from score
where course_id = 2
) t2 on t1.student_id = t2.student_id
left join student s on t1.student_id = s.id;