一、题目
1、查询所有的课程的名称以及对应的任课老师姓名
2、查询学生表中男女生各有多少人
3、查询物理成绩等于100的学生的姓名
4、查询平均成绩大于八十分的同学的姓名和平均成绩
5、查询所有学生的学号,姓名,选课数,总成绩
6、 查询姓李老师的个数
7、 查询没有报李平老师课的学生姓名
8、 查询物理课程的分数比生物课程的分数高的学生的学号
9、 查询没有同时选修物理课程和体育课程的学生姓名
10、查询挂科超过两门(包括两门)的学生姓名和班级
11、查询选修了所有课程的学生姓名
12、查询李平老师教的课程的所有成绩记录
13、查询全部学生都选修了的课程号和课程名
14、查询每门课程被选修的次数
15、查询只选修了一门课程的学生学号和姓名
16、查询所有学生考出的成绩并按从高到低排序(成绩去重)
17、查询平均成绩大于85的学生姓名和平均成绩
18、查询生物成绩不及格的学生姓名和对应生物分数
19、查询在所有选修了李平老师课程的学生中,这些课程(李平老师的课程,不是所有课程)平均成绩最高的学生姓名
20、查询每门课程成绩最好的课程id、学生姓名和分数
21、查询不同课程但成绩相同的课程号、学生号、成绩
22、查询没学过“李平”老师课程的学生姓名以及选修的课程名称
23、查询所有选修了学号为2的同学选修过的一门或者多门课程的同学学号和姓名
24、任课最多的老师中学生单科成绩最高的课程id、学生姓名和分数
二、答案
1、查询所有的课程的名称以及对应的任课老师姓名
SELECT
course.cname,
teacher.tname
FROM
course
INNER JOIN teacher ON course.teacher_id = teacher.tid;
2、查询学生表中男女生各有多少人
SELECT
gender,
count(sid)
FROM
student
GROUP BY
gender;
3、查询物理成绩等于100的学生的姓名
SELECT
sname
FROM
student
WHERE
sid IN (
SELECT
score.student_id
FROM
score
LEFT JOIN course ON score.course_id = course.cid
WHERE
score.num = 100
AND course.cname = "物理"
);
4、查询平均成绩大于八十分的同学的姓名和平均成绩
SELECT
t1.sname,
t2.avg_num
FROM
student t1
INNER JOIN (
SELECT
student_id,
avg(num) avg_num
FROM
score
GROUP BY
student_id
HAVING
avg(num) > 80
) t2 ON t1.sid = t2.student_id;
5、查询所有学生的学号,姓名,选课数,总成绩(注意:对于那些没有选修任何课程的学生也算在内)
SELECT
sid,
sname,
t1.count_course,
t1.sum_num
FROM
student
LEFT JOIN (
SELECT
student_id,
count(course_id) count_course,
sum(num) sum_num
FROM
score
GROUP BY
student_id
) t1 ON student.sid = t1.student_id;
6、 查询姓李老师的个数
SELECT
count(tid)
FROM
teacher
WHERE
tname LIKE '李%';
7、 查询没有报李平老师课的学生姓名(找出报名李平老师课程的学生,然后取反就可以)
SELECT
sname
FROM
student
WHERE
sid NOT IN (
SELECT DISTINCT
student_id
FROM
score
WHERE
course_id IN (
SELECT
cid
FROM
course
WHERE
teacher_id = (
SELECT
tid
FROM
teacher
WHERE
tname = '李平'
)
)
);
8、 查询物理课程的分数比生物课程的分数高的学生的学号
SELECT
t1.student_id
FROM
(
SELECT
student_id,
num
FROM
score
LEFT JOIN course ON score.course_id = course.cid
WHERE
cname = '物理'
) t1
LEFT JOIN (
SELECT
student_id,
num
FROM
score
LEFT JOIN course ON score.course_id = course.cid
WHERE
cname = '生物'
) t2 ON t1.student_id = t2.student_id
WHERE
t1.num > t2.num;
9、 查询没有同时选修物理课程和体育课程的学生姓名(没有同时选修指的是选修了一门的,思路是得到物理+体育课程的学生信息表,然后基于学生分组,统计count(课程)=1)
SELECT
sname
FROM
student
WHERE
sid IN (
SELECT
student_id
FROM
score
WHERE
course_id IN (
SELECT
cid
FROM
course
WHERE
cname IN ('物理', '体育')
)
GROUP BY
student_id
HAVING
count(student_id) = 1
);
10、查询挂科超过两门(包括两门)的学生姓名和班级(求出<60的表,然后对学生进行分组,统计课程数目>=2)
# 方式一:
SELECT
t1.sname,
t2.caption
FROM
(
SELECT
class_id,
sname
FROM
student
WHERE
sid IN (
SELECT
student_id
FROM
score
WHERE
num < 60
GROUP BY
student_id
HAVING
count(course_id) > 1
)
) t1
INNER JOIN (SELECT cid, caption FROM class) t2 ON t1.class_id = t2.cid;
# 方式二:
SELECT
student.sname,
class.caption
FROM
student
INNER JOIN (
SELECT
student_id
FROM
score
WHERE
num < 60
GROUP BY
student_id
HAVING
count(course_id) >= 2
) AS t1
INNER JOIN class ON student.sid = t1.student_id
AND student.class_id = class.cid;
11、查询选修了所有课程的学生姓名(先从course表统计课程的总数,然后基于score表按照student_id分组,统计课程数据等于课程总数即可)
SELECT
sname
FROM
student
WHERE
sid IN (
SELECT
student_id
FROM
score
GROUP BY
student_id
HAVING
count(course_id) = (
SELECT
count(cid) AS course_total
FROM
course
)
);
12、查询李平老师教的课程的所有成绩记录
# 方式一:
SELECT
*
FROM
score
WHERE
course_id IN (
SELECT
cid
FROM
course
WHERE
teacher_id = (
SELECT
tid
FROM
teacher
WHERE
tname = '李平'
)
);
# 方式二:
SELECT
*
FROM
score
WHERE
course_id IN (
SELECT
cid
FROM
course
INNER JOIN teacher ON course.teacher_id = teacher.tid
WHERE
teacher.tname = '李平'
);
13、查询全部学生都选修了的课程号和课程名(先统计出有课程的学生总数,再以课程分组统计出有多少学生学,如果有课程跟学生总数相等,就说明这就是都选修的课程)
SELECT
cid,
cname
FROM
course
WHERE
cid IN (
SELECT
course_id
FROM
score
GROUP BY
course_id
HAVING
count(student_id) = (
SELECT
count(DISTINCT student_id)
FROM
score
)
)
14、查询每门课程被选修的次数
SELECT
course_id,
count(student_id)
FROM
score
GROUP BY
course_id;
15、查询只选修了一门课程的学生学号和姓名
SELECT
sid,
sname
FROM
student
WHERE
sid IN (
SELECT
student_id
FROM
score
GROUP BY
student_id
HAVING
count(course_id) = 1
);
16、查询所有学生考出的成绩并按从高到低排序(成绩去重)
SELECT DISTINCT
num
FROM
score
ORDER BY
num DESC;
17、查询平均成绩大于85的学生姓名和平均成绩
SELECT
sname,
t1.avg_num
FROM
student
INNER JOIN (
SELECT
student_id,
avg(num) AS avg_num
FROM
score
GROUP BY
student_id
HAVING
avg(num) > 85
) t1 ON student.sid = t1.student_id;
18、查询生物成绩不及格的学生姓名和对应生物分数
SELECT
sname,
t1.num
FROM
student
INNER JOIN (
SELECT
student_id,
num
FROM
score
WHERE
course_id = (
SELECT
cid
FROM
course
WHERE
cname = '生物'
)
HAVING
num < 60
) t1 ON student.sid = t1.student_id;
19、查询在所有选修了李平老师课程的学生中,这些课程(李平老师的课程,不是所有课程)平均成绩最高的学生姓名
SELECT
sname
FROM
student
WHERE
sid = (
SELECT
student_id
FROM
score
WHERE
course_id IN (
SELECT
course.cid
FROM
teacher
INNER JOIN course ON teacher.tid = course.teacher_id
WHERE
teacher.tname = '李平'
)
GROUP BY
student_id
ORDER BY
avg(num) DESC
LIMIT 1
);
20、查询每门课程成绩最好的课程id、学生姓名和分数
SELECT
course_id,
sname,
num
FROM
student
INNER JOIN (
SELECT
t1.course_id,
student_id,
num
FROM
score
INNER JOIN (
SELECT
course_id,
max(num) AS max_num
FROM
score
GROUP BY
course_id
) t1 ON score.course_id = t1.course_id
WHERE
score.num = t1.max_num
) t2 ON student.sid = t2.student_id;
21、查询不同课程但成绩相同的课程号、学生号、成绩
SELECT DISTINCT
s1.course_id,
s2.course_id,
s1.student_id,
s2.student_id,
s1.num,
s2.num
FROM
score AS s1,
score AS s2
WHERE
s1.num = s2.num
AND s1.course_id != s2.course_id;
22、查询没学过“李平”老师课程的学生姓名以及选修的课程名称
SELECT
student.sname,
course.cname
FROM
student
INNER JOIN (
SELECT
student_id,
course_id
FROM
score
WHERE
student_id IN (
SELECT
sid
FROM
student
WHERE
sid NOT IN (
SELECT
student_id
FROM
score
WHERE
course_id IN (
SELECT
cid
FROM
course
INNER JOIN teacher ON course.teacher_id = teacher.tid
WHERE
tname = '李平'
)
)
)
) t1 ON student.sid = t1.student_id
INNER JOIN course ON course.cid = t1.course_id;
23、查询所有选修了学号为2的同学选修过的一门或者多门课程的同学学号和姓名
SELECT
sid,
sname
FROM
student
WHERE
sid IN (
SELECT DISTINCT
student_id
FROM
score
WHERE
course_id IN (
SELECT
course_id
FROM
score
WHERE
student_id = 2
)
)
AND sid != 2;
24、任课最多的老师中学生单科成绩最高的课程id、学生姓名和分数
SELECT
t1.course_id,
student.sname,
num
FROM
score
INNER JOIN (
SELECT
course_id,
max(num) AS max_num
FROM
score
WHERE
course_id IN (
SELECT
cid
FROM
course
WHERE
teacher_id = (
SELECT
teacher_id
FROM
course
GROUP BY
teacher_id
ORDER BY
count(teacher_id) DESC
LIMIT 1
)
)
GROUP BY
course_id
) t1 ON score.course_id = t1.course_id
INNER JOIN student ON score.student_id = student.sid
WHERE
score.course_id = t1.course_id
AND num = t1.max_num;
参考答案