CREATE DATABASE javacto;
USE javacto;
#班级表
CREATE TABLE class(
cid INT PRIMARY KEY,
caption VARCHAR(6)
);
#老师表:teacher
CREATE TABLE teacher(
tid INT PRIMARY KEY,
tname VARCHAR(8) NOT NULL
);
#学生表
CREATE TABLE student(
sid INT PRIMARY KEY,
sname VARCHAR(8) NOT NULL,
gender VARCHAR(2) NOT NULL,
class_id INT NOT NULL
);
#课程表 course
CREATE TABLE course(
cid INT PRIMARY KEY,
cname VARCHAR(8) NOT NULL,
teacher_id INT NOT NULL,
FOREIGN KEY(teacher_id) REFERENCES teacher(tid)
);
#成绩表
CREATE TABLE score(
sid INT PRIMARY KEY,
student_id INT NOT NULL,
scorse_id INT NOT NULL,
number INT NOT NULL,
FOREIGN KEY(student_id) REFERENCES student(sid),
FOREIGN KEY(scorse_id) REFERENCES course(cid)
);
#建库建表
1、自行创建测试数据
2、查询“生物”课程比“物理”课程成绩高的所有学生的学号;
3、查询平均成绩大于60分的同学的学号和平均成绩;
SELECT student_id,AVG(number) FROM score GROUP BY student_id HAVING AVG(number)>60;
问题:很明显是平均之后进行比较(用HAVING),出现HAVING之后必须在前有GROUP BY;
4、查询所有同学的学号、姓名、选课数、总成绩;
SELECT student.sid,student.sname,COUNT(scorse_id),SUM(number)FROM student LEFT JOIN score ON student.sid=score.student_id GROUP BY student.sid
5、查询姓“李”的老师的个数;
问题:LIKE模糊匹配不清楚
SELECT COUNT(tid) FROM teacher WHERE tname LIKE '李%';
6、查询没学过“叶平”老师课的同学的学号、姓名;
7、查询学过“001”并且也学过编号“002”课程的同学的学号、姓名;
8、查询学过“叶平”老师所教的所有课的同学的学号、姓名;
9、查询课程编号“002”的成绩比课程编号“001”课程低的所有同学的学号、姓名;
10、查询有课程成绩小于60分的同学的学号、姓名;
11、查询没有学全所有课的同学的学号、姓名;
12、查询至少有一门课与学号为“001”的同学所学相同的同学的学号和姓名;
13、查询至少学过学号为“001”同学所选课程中任意一门课的其他同学学号和姓名;
14、查询和“002”号的同学学习的课程完全相同的其他同学学号和姓名;
15、删除学习“叶平”老师课的SC表记录;
16、向SC表中插入一些记录,这些记录要求符合以下条件:①没有上过编号“002”课程的同学学号;②插入“002”号课程的平均成绩;
17、按平均成绩从低到高显示所有学生的“语文”、“数学”、“英语”三门的课程成绩,按如下形式显示: 学生ID,语文,数学,英语,有效课程数,有效平均分;
18、查询各科成绩最高和最低的分:以如下形式显示:课程ID,最高分,最低分;
19、按各科平均成绩从低到高和及格率的百分数从高到低顺序;
20、课程平均分从高到低显示(现实任课老师);
21、查询各科成绩前三名的记录:(不考虑成绩并列情况)
22、查询每门课程被选修的学生数;
23、查询出只选修了一门课程的全部学生的学号和姓名;
24、查询男生、女生的人数;
25、查询姓“张”的学生名单;
26、查询同名同姓学生名单,并统计同名人数;
27、查询每门课程的平均成绩,结果按平均成绩升序排列,平均成绩相同时,按课程号降序排列;
28、查询平均成绩大于85的所有学生的学号、姓名和平均成绩;
29、查询课程名称为“数学”,且分数低于60的学生姓名和分数;
30、查询课程编号为003且课程成绩在80分以上的学生的学号和姓名;
31、求选了课程的学生人数
32、查询选修“杨艳”老师所授课程的学生中,成绩最高的学生姓名及其成绩;
33、查询各个课程及相应的选修人数;
34、查询不同课程但成绩相同的学生的学号、课程号、学生成绩;
35、查询每门课程成绩最好的前两名;
36、检索至少选修两门课程的学生学号;
37、查询全部学生都选修的课程的课程号和课程名;
38、查询没学过“叶平”老师讲授的任一门课程的学生姓名;
39、查询两门以上不及格课程的同学的学号及其平均成绩;
40、检索“004”课程分数小于60,按分数降序排列的同学学号;
41、删除“002”同学的“001”课程的成绩;
#答案下面
2、查询“生物”课程比“物理”课程成绩高的所有学生的学号;
思路:
获取所有有生物课程的人(学号,成绩) - 临时表
获取所有有物理课程的人(学号,成绩) - 临时表
根据【学号】连接两个临时表:
学号 物理成绩 生物成绩
然后再进行筛选
SELECT A.student_id,sw,ty FROM
(SELECT student_id,num AS sw FROM score LEFT JOIN course ON score.course_id = course.cid WHERE course.cname = '生物') AS A
LEFT JOIN
(SELECT student_id,num AS ty FROM score LEFT JOIN course ON score.course_id = course.cid WHERE course.cname = '体育') AS B
ON A.student_id = B.student_id WHERE sw > IF(ISNULL(ty),0,ty);
3、查询平均成绩大于60分的同学的学号和平均成绩;
思路:
根据学生分组,使用AVG获取平均值,通过HAVING对AVG进行筛选
SELECT student_id,AVG(num) FROM score GROUP BY student_id HAVING AVG(num) > 60
4、查询所有同学的学号、姓名、选课数、总成绩;
SELECT score.student_id,SUM(score.num),COUNT(score.student_id),student.sname
FROM
score LEFT JOIN student ON score.student_id = student.sid
GROUP BY score.student_id
5、查询姓“李”的老师的个数;
SELECT COUNT(tid) FROM teacher WHERE tname LIKE '李%'
SELECT COUNT(1) FROM (SELECT tid FROM teacher WHERE tname LIKE '李%') AS B
6、查询没学过“叶平”老师课的同学的学号、姓名;
思路:
先查到“李平老师”老师教的所有课ID
获取选过课的所有学生ID
学生表中筛选
SELECT * FROM student WHERE sid NOT IN (
SELECT DISTINCT student_id FROM score WHERE score.course_id IN (
SELECT cid FROM course LEFT JOIN teacher ON course.teacher_id = teacher.tid WHERE tname = '李平老师'
)
)
7、查询学过“001”并且也学过编号“002”课程的同学的学号、姓名;
思路:
先查到既选择001又选择002课程的所有同学
根据学生进行分组,如果学生数量等于2表示,两门均已选择
SELECT student_id,sname FROM
(SELECT student_id,course_id FROM score WHERE course_id = 1 OR course_id = 2) AS B
LEFT JOIN student ON B.student_id = student.sid GROUP BY student_id HAVING COUNT(student_id) > 1
8、查询学过“叶平”老师所教的所有课的同学的学号、姓名;
同上,只不过将001和002变成 IN (叶平老师的所有课)
9、查询课程编号“002”的成绩比课程编号“001”课程低的所有同学的学号、姓名;
同第1题
10、查询有课程成绩小于60分的同学的学号、姓名;
SELECT sid,sname FROM student WHERE sid IN (
SELECT DISTINCT student_id FROM score WHERE num < 60
)
11、查询没有学全所有课的同学的学号、姓名;
思路:
在分数表中根据学生进行分组,获取每一个学生选课数量
如果数量 == 总课程数量,表示已经选择了所有课程
SELECT student_id,sname
FROM score LEFT JOIN student ON score.student_id = student.sid
GROUP BY student_id HAVING COUNT(course_id) = (SELECT COUNT(1) FROM course)
12、查询至少有一门课与学号为“001”的同学所学相同的同学的学号和姓名;
思路:
获取 001 同学选择的所有课程
获取课程在其中的所有人以及所有课程
根据学生筛选,获取所有学生信息
再与学生表连接,获取姓名
SELECT student_id,sname, COUNT(course_id)
FROM score LEFT JOIN student ON score.student_id = student.sid
WHERE student_id != 1 AND course_id IN (SELECT course_id FROM score WHERE student_id = 1) GROUP BY student_id
13、查询至少学过学号为“001”同学所有课的其他同学学号和姓名;
先找到和001的学过的所有人
然后个数 = 001所有学科 ==》 其他人可能选择的更多
SELECT student_id,sname, COUNT(course_id)
FROM score LEFT JOIN student ON score.student_id = student.sid
WHERE student_id != 1 AND course_id IN (SELECT course_id FROM score WHERE student_id = 1) GROUP BY student_id HAVING COUNT(course_id) = (SELECT COUNT(course_id) FROM score WHERE student_id = 1)
14、查询和“002”号的同学学习的课程完全相同的其他同学学号和姓名;
个数相同
002学过的也学过
SELECT student_id,sname FROM score LEFT JOIN student ON score.student_id = student.sid WHERE student_id IN (
SELECT student_id FROM score WHERE student_id != 1 GROUP BY student_id HAVING COUNT(course_id) = (SELECT COUNT(1) FROM score WHERE student_id = 1)
) AND course_id IN (SELECT course_id FROM score WHERE student_id = 1) GROUP BY student_id HAVING COUNT(course_id) = (SELECT COUNT(1) FROM score WHERE student_id = 1)
15、删除学习“叶平”老师课的score表记录;
DELETE FROM score WHERE course_id IN (
SELECT cid FROM course LEFT JOIN teacher ON course.teacher_id = teacher.tid WHERE teacher.name = '叶平'
)
16、向SC表中插入一些记录,这些记录要求符合以下条件:①没有上过编号“002”课程的同学学号;②插入“002”号课程的平均成绩;
思路:
由于INSERT 支持
inset INTO tb1(xx,xx) SELECT x1,x2 FROM tb2;
所有,获取所有没上过002课的所有人,获取002的平均成绩
INSERT INTO score(student_id, course_id, num) SELECT sid,2,(SELECT AVG(num) FROM score WHERE course_id = 2)
FROM student WHERE sid NOT IN (
SELECT student_id FROM score WHERE course_id = 2
)
17、按平均成绩从低到高 显示所有学生的“语文”、“数学”、“英语”三门的课程成绩,按如下形式显示: 学生ID,语文,数学,英语,有效课程数,有效平均分;
SELECT sc.student_id,
(SELECT num FROM score LEFT JOIN course ON score.course_id = course.cid WHERE course.cname = "生物" AND score.student_id=sc.student_id) AS sy,
(SELECT num FROM score LEFT JOIN course ON score.course_id = course.cid WHERE course.cname = "物理" AND score.student_id=sc.student_id) AS wl,
(SELECT num FROM score LEFT JOIN course ON score.course_id = course.cid WHERE course.cname = "体育" AND score.student_id=sc.student_id) AS ty,
COUNT(sc.course_id),
AVG(sc.num)
FROM score AS sc
GROUP BY student_id DESC
18、查询各科成绩最高和最低的分:以如下形式显示:课程ID,最高分,最低分;
SELECT course_id, MAX(num) AS max_num, MIN(num) AS min_num FROM score GROUP BY course_id;
19、按各科平均成绩从低到高和及格率的百分数从高到低顺序;
思路:CASE WHEN .. THEN
SELECT course_id, AVG(num) AS avgnum,SUM(CASE WHEN score.num > 60 THEN 1 ELSE 0 END)/COUNT(1)*100 AS percent FROM score GROUP BY course_id ORDER BY avgnum ASC,percent DESC;
20、课程平均分从高到低显示(现实任课老师);
SELECT AVG(IF(ISNULL(score.num),0,score.num)),teacher.tname FROM course
LEFT JOIN score ON course.cid = score.course_id
LEFT JOIN teacher ON course.teacher_id = teacher.tid
GROUP BY score.course_id
21、查询各科成绩前三名的记录:(不考虑成绩并列情况)
SELECT score.sid,score.course_id,score.num,T.first_num,T.second_num FROM score LEFT JOIN
(
SELECT
sid,
(SELECT num FROM score AS s2 WHERE s2.course_id = s1.course_id ORDER BY num DESC LIMIT 0,1) AS first_num,
(SELECT num FROM score AS s2 WHERE s2.course_id = s1.course_id ORDER BY num DESC LIMIT 3,1) AS second_num
FROM
score AS s1
) AS T
ON score.sid =T.sid
WHERE score.num <= T.first_num AND score.num >= T.second_num
22、查询每门课程被选修的学生数;
SELECT course_id, COUNT(1) FROM score GROUP BY course_id;
23、查询出只选修了一门课程的全部学生的学号和姓名;
SELECT student.sid, student.sname, COUNT(1) FROM score
LEFT JOIN student ON score.student_id = student.sid
GROUP BY course_id HAVING COUNT(1) = 1
24、查询男生、女生的人数;
SELECT * FROM
(SELECT COUNT(1) AS man FROM student WHERE gender='男') AS A ,
(SELECT COUNT(1) AS feman FROM student WHERE gender='女') AS B
25、查询姓“张”的学生名单;
SELECT sname FROM student WHERE sname LIKE '张%';
26、查询同名同姓学生名单,并统计同名人数;
SELECT sname,COUNT(1) AS COUNT FROM student GROUP BY sname;
27、查询每门课程的平均成绩,结果按平均成绩升序排列,平均成绩相同时,按课程号降序排列;
SELECT course_id,AVG(IF(ISNULL(num), 0 ,num)) AS AVG FROM score GROUP BY course_id ORDER BY AVG ASC,course_id DESC;
28、查询平均成绩大于85的所有学生的学号、姓名和平均成绩;
SELECT student_id,sname, AVG(IF(ISNULL(num), 0 ,num)) FROM score LEFT JOIN student ON score.student_id = student.sid GROUP BY student_id;
29、查询课程名称为“数学”,且分数低于60的学生姓名和分数;
SELECT student.sname,score.num FROM score
LEFT JOIN course ON score.course_id = course.cid
LEFT JOIN student ON score.student_id = student.sid
WHERE score.num < 60 AND course.cname = '生物'
30、查询课程编号为003且课程成绩在80分以上的学生的学号和姓名;
SELECT * FROM score WHERE score.student_id = 3 AND score.num > 80
31、求选了课程的学生人数
SELECT COUNT(DISTINCT student_id) FROM score
SELECT COUNT(c) FROM (
SELECT COUNT(student_id) AS c FROM score GROUP BY student_id) AS A
32、查询选修“杨艳”老师所授课程的学生中,成绩最高的学生姓名及其成绩;
SELECT sname,num FROM score
LEFT JOIN student ON score.student_id = student.sid
WHERE score.course_id IN (SELECT course.cid FROM course LEFT JOIN teacher ON course.teacher_id = teacher.tid WHERE tname='张磊老师') ORDER BY num DESC LIMIT 1;
33、查询各个课程及相应的选修人数;
SELECT course.cname,COUNT(1) FROM score
LEFT JOIN course ON score.course_id = course.cid
GROUP BY course_id;
34、查询不同课程但成绩相同的学生的学号、课程号、学生成绩;
SELECT DISTINCT s1.course_id,s2.course_id,s1.num,s2.num FROM score AS s1, score AS s2 WHERE s1.num = s2.num AND s1.course_id != s2.course_id;
35、查询每门课程成绩最好的前两名;
SELECT score.sid,score.course_id,score.num,T.first_num,T.second_num FROM score LEFT JOIN
(
SELECT
sid,
(SELECT num FROM score AS s2 WHERE s2.course_id = s1.course_id ORDER BY num DESC LIMIT 0,1) AS first_num,
(SELECT num FROM score AS s2 WHERE s2.course_id = s1.course_id ORDER BY num DESC LIMIT 1,1) AS second_num
FROM
score AS s1
) AS T
ON score.sid =T.sid
WHERE score.num <= T.first_num AND score.num >= T.second_num
36、检索至少选修两门课程的学生学号;
SELECT student_id FROM score GROUP BY student_id HAVING COUNT(student_id) > 1
37、查询全部学生都选修的课程的课程号和课程名;
SELECT course_id,COUNT(1) FROM score GROUP BY course_id HAVING COUNT(1) = (SELECT COUNT(1) FROM student);
38、查询没学过“叶平”老师讲授的任一门课程的学生姓名;
SELECT student_id,student.sname FROM score
LEFT JOIN student ON score.student_id = student.sid
WHERE score.course_id NOT IN (
SELECT cid FROM course LEFT JOIN teacher ON course.teacher_id = teacher.tid WHERE tname = '张磊老师'
)
GROUP BY student_id
39、查询两门以上不及格课程的同学的学号及其平均成绩;
SELECT student_id,COUNT(1) FROM score WHERE num < 60 GROUP BY student_id HAVING COUNT(1) > 2
40、检索“004”课程分数小于60,按分数降序排列的同学学号;
SELECT student_id FROM score WHERE num< 60 AND course_id = 4 ORDER BY num DESC;
41、删除“002”同学的“001”课程的成绩;
DELETE FROM score WHERE course_id = 1 AND student_id = 2
更多学习资料:javacto.taobao.com