“SQL 学生 课程 分数 2”的版本间的差异
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=== 查询没有考完所有科目的学生 === | === 查询没有考完所有科目的学生 === | ||
+ | <syntaxhighlight lang="sql"> | ||
+ | select s.id, s.name | ||
+ | from ( | ||
+ | select distinct student_id, course_id | ||
+ | from score | ||
+ | ) score_tmp | ||
+ | right join student s on score_tmp.student_id = s.id | ||
+ | group by s.id | ||
+ | having count(*) < ( | ||
+ | select count(*) | ||
+ | from course | ||
+ | ) | ||
+ | </syntaxhighlight> |
2024年7月9日 (二) 10:51的最新版本
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;
思路:
先用where条件筛选出考试过语文、数学的学生;
又因为同一个学生可能多次考试同一个科目,所以对子查询中的考试结果去重,得到的结果是一个学生对应一个科目就一条记录;
然后左连接student表后按student_id分组,就会产生一个学生对应多个科目的一组,然后筛选出分组结果中为2个科目的记录。
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;
查询只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 student_id
from (
select distinct student_id, course_id
from score
) t1
group by student_id having count(*) = 2
) t2 inner join (
select distinct student_id
from score
where course_id = 1
) t3 on t2.student_id = t3.student_id
inner join (
select distinct student_id
from score
where course_id = 2
) t4 on t2.student_id = t4.student_id
left join student s on t2.student_id = s.id;
第二种
查询没有考完所有科目的学生
select s.id, s.name
from (
select distinct student_id, course_id
from score
) score_tmp
right join student s on score_tmp.student_id = s.id
group by s.id
having count(*) < (
select count(*)
from course
)