“SQL 学生 课程 分数 2”的版本间的差异
		
		
		
		
		
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Jihongchang(讨论 | 贡献)  (→准备数据)  | 
				Jihongchang(讨论 | 贡献)   | 
				||
| (未显示同一用户的8个中间版本) | |||
| 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  | |
insert into score values (1, 1, 1, 80);  | 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);  | 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);  | 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 (  | + | 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>  | ||
| + | |||
| + | === 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 ====  | ||
| + | <syntaxhighlight lang="sql">  | ||
| + | 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;  | ||
| + | </syntaxhighlight>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 ====  | ||
| + | <syntaxhighlight lang="sql">  | ||
| + | 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;  | ||
</syntaxhighlight>  | </syntaxhighlight>  | ||
=== 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 ===  | === 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 ====  | ||
| + | <syntaxhighlight lang="sql">  | ||
| + | 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;  | ||
| + | </syntaxhighlight>  | ||
| + | |||
| + | ==== 第二种 ====  | ||
=== 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
)