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、 查询物理课程比生物课程高的学生的学号
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
、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、查询之选修了一门课程的学生姓名和学号
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前两名学生姓名
21、查询不同课程但成绩相同的学号,课程号,成绩
22、查询没学过“叶平”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1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学生姓名
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 性别,
count(1) 人数
FROM
student
GROUP BY
gender;
#3、查询物理成绩等于100的学生的姓名
SELECT
student.sname
FROM
student
WHERE
sid IN (
SELECT
student_id
FROM
score
INNER JOIN course ON score.course_id = course.cid
WHERE
course.cname = '物理'
AND score.num = 100
);
#4、查询平均成绩大于八十分的同学的姓名和平均成绩
SELECT
student.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) > 80
) AS t1 ON student.sid = t1.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
student.sid,
student.sname,
t1.course_num,
t1.total_num
FROM
student
LEFT JOIN (
SELECT
student_id,
COUNT(course_id) course_num,
sum(num) total_num
FROM
score
GROUP BY
student_id
) AS 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
student.sname
FROM
student
WHERE
sid NOT IN (
SELECT DISTINCT
student_id
FROM
score
WHERE
course_id IN (
SELECT
course.cid
FROM
course
INNER JOIN teacher ON course.teacher_id = teacher.tid
WHERE
teacher.tname = '李平老师'
)
);
#8、 查询物理课程比生物课程高的学生的学号(分别得到物理成绩表与生物成绩表,然后连表即可)
SELECT
t1.student_id
FROM
(
SELECT
student_id,
num
FROM
score
WHERE
course_id = (
SELECT
cid
FROM
course
WHERE
cname = '物理'
)
) AS t1
INNER JOIN (
SELECT
student_id,
num
FROM
score
WHERE
course_id = (
SELECT
cid
FROM
course
WHERE
cname = '生物'
)
) AS 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
student.sname
FROM
student
WHERE
sid IN (
SELECT
student_id
FROM
score
WHERE
course_id IN (
SELECT
cid
FROM
course
WHERE
cname = '物理'
OR cname = '体育'
)
GROUP BY
student_id
HAVING
COUNT(course_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)
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
student.sname
FROM
student
WHERE
sid IN (
SELECT
student_id
FROM
score
GROUP BY
student_id
HAVING
COUNT(course_id) = (SELECT count(cid) FROM course)
);
#12、查询李平老师教的课程的所有成绩记录
SELECT
*
FROM
score
WHERE
course_id IN (
SELECT
cid
FROM
course
INNER JOIN teacher ON course.teacher_id = teacher.tid
WHERE
teacher.tname = '李平老师'
);
#13、查询全部学生都选修了的课程号和课程名(取所有学生数,然后基于score表的课程分组,找出count(student_id)等于学生数即可)
SELECT
cid,
cname
FROM
course
WHERE
cid IN (
SELECT
course_id
FROM
score
GROUP BY
course_id
HAVING
COUNT(student_id) = (
SELECT
COUNT(sid)
FROM
student
)
);
#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、查询之选修了一门课程的学生姓名和学号
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) 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 姓名,
num 生物成绩
FROM
score
LEFT JOIN course ON score.course_id = course.cid
LEFT JOIN student ON score.student_id = student.sid
WHERE
course.cname = '生物'
AND score.num < 60;
#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
course
INNER JOIN teacher ON course.teacher_id = teacher.tid
WHERE
teacher.tname = '李平老师'
)
GROUP BY
student_id
ORDER BY
AVG(num) DESC
LIMIT 1
);
#20、查询每门课程成绩最好的前两名学生姓名
#查看每门课程按照分数排序的信息,为下列查找正确与否提供依据
SELECT
*
FROM
score
ORDER BY
course_id,
num DESC;
#表1:求出每门课程的课程course_id,与最高分数first_num
SELECT
course_id,
max(num) first_num
FROM
score
GROUP BY
course_id;
#表2:去掉最高分,再按照课程分组,取得的最高分,就是第二高的分数second_num
SELECT
score.course_id,
max(num) second_num
FROM
score
INNER JOIN (
SELECT
course_id,
max(num) first_num
FROM
score
GROUP BY
course_id
) AS t ON score.course_id = t.course_id
WHERE
score.num < t.first_num
GROUP BY
course_id;
#将表1和表2联合到一起,得到一张表t3,包含课程course_id与该们课程的first_num与second_num
SELECT
t1.course_id,
t1.first_num,
t2.second_num
FROM
(
SELECT
course_id,
max(num) first_num
FROM
score
GROUP BY
course_id
) AS t1
INNER JOIN (
SELECT
score.course_id,
max(num) second_num
FROM
score
INNER JOIN (
SELECT
course_id,
max(num) first_num
FROM
score
GROUP BY
course_id
) AS t ON score.course_id = t.course_id
WHERE
score.num < t.first_num
GROUP BY
course_id
) AS t2 ON t1.course_id = t2.course_id;
#查询前两名的学生(有可能出现并列第一或者并列第二的情况)
SELECT
score.student_id,
t3.course_id,
t3.first_num,
t3.second_num
FROM
score
INNER JOIN (
SELECT
t1.course_id,
t1.first_num,
t2.second_num
FROM
(
SELECT
course_id,
max(num) first_num
FROM
score
GROUP BY
course_id
) AS t1
INNER JOIN (
SELECT
score.course_id,
max(num) second_num
FROM
score
INNER JOIN (
SELECT
course_id,
max(num) first_num
FROM
score
GROUP BY
course_id
) AS t ON score.course_id = t.course_id
WHERE
score.num < t.first_num
GROUP BY
course_id
) AS t2 ON t1.course_id = t2.course_id
) AS t3 ON score.course_id = t3.course_id
WHERE
score.num >= t3.second_num
AND score.num <= t3.first_num;
#排序后可以看的明显点
SELECT
score.student_id,
t3.course_id,
t3.first_num,
t3.second_num
FROM
score
INNER JOIN (
SELECT
t1.course_id,
t1.first_num,
t2.second_num
FROM
(
SELECT
course_id,
max(num) first_num
FROM
score
GROUP BY
course_id
) AS t1
INNER JOIN (
SELECT
score.course_id,
max(num) second_num
FROM
score
INNER JOIN (
SELECT
course_id,
max(num) first_num
FROM
score
GROUP BY
course_id
) AS t ON score.course_id = t.course_id
WHERE
score.num < t.first_num
GROUP BY
course_id
) AS t2 ON t1.course_id = t2.course_id
) AS t3 ON score.course_id = t3.course_id
WHERE
score.num >= t3.second_num
AND score.num <= t3.first_num
ORDER BY
course_id;
#可以用以下命令验证上述查询的正确性
SELECT
*
FROM
score
ORDER BY
course_id,
num DESC;
-- 21、查询不同课程但成绩相同的学号,课程号,成绩
-- 22、查询没学过“叶平”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1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学生姓名
更多练习以及参考答案:https://www.cnblogs.com/clschao/articles/9995768.html