【Python全棧-後端開發】MySQL數據庫-練習題
MySQL數據庫-練習題
一、表關系
請創建如下表,並創建相關約束
二、操作表
1、自行創建測試數據
2、查詢“生物”課程比“物理”課程成績高的所有學生的學號;
3、查詢平均成績大於60分的同學的學號和平均成績;
4、查詢所有同學的學號、姓名、選課數、總成績;
5、查詢姓“李”的老師的個數;
6、查詢沒學過“葉平”老師課的同學的學號、姓名;
7、查詢學過“001”並且也學過編號“002”課程的同學的學號、姓名;
8、查詢學過“葉平”老師所教的所有課的同學的學號、姓名;
9、查詢課程編號“002”的成績比課程編號“001”課程低的所有同學的學號、姓名;
10、查詢有課程成績小於60分的同學的學號、姓名;
11、查詢沒有學全所有課的同學的學號、姓名;
12、查詢至少有一門課與學號為“001”的同學所學相同的同學的學號和姓名;
13、查詢至少學過學號為“001”同學所有一門課的其他同學學號和姓名;
14、查詢和“002”號的同學學習的課程完全相同的其他同學學號和姓名;
15、刪除學習“葉平”老師課的SC表記錄;
16、向SC表中插入一些記錄,這些記錄要求符合以下條件:①沒有上過編號“002”課程的同學學號;②插入“002”號課程的平均成績;
17、按平均成績從低到高顯示所有學生的“語文”、“數學”、“英語”三門的課程成績,按如下形式顯示: 學生ID,語文,數學,英語,有效課程數,有效平均分;
18、查詢各科成績最高和最低的分:以如下形式顯示:課程ID,最高分,最低分;
19、按各科平均成績從低到高和及格率的百分數從高到低順序;
20、課程平均分從高到低顯示(現實任課老師);
21、查詢各科成績前三名的記錄:(不考慮成績並列情況)
22、查詢每門課程被選修的學生數;
23、查詢出只選修了一門課程的全部學生的學號和姓名;
24、查詢男生、女生的人數;
25、查詢姓“張”的學生名單;
26、查詢同名同姓學生名單,並統計同名人數;
27、查詢每門課程的平均成績,結果按平均成績升序排列,平均成績相同時,按課程號降序排列;
28、查詢平均成績大於85的所有學生的學號、姓名和平均成績;
29、查詢課程名稱為“數學”,且分數低於60的學生姓名和分數;
30、查詢課程編號為003且課程成績在80分以上的學生的學號和姓名;
31、求選了課程的學生人數
32、查詢選修“楊艷”老師所授課程的學生中,成績最高的學生姓名及其成績;
33、查詢各個課程及相應的選修人數;
34、查詢不同課程但成績相同的學生的學號、課程號、學生成績;
35、查詢每門課程成績最好的前兩名;
36、檢索至少選修兩門課程的學生學號;
37、查詢全部學生都選修的課程的課程號和課程名;
38、查詢沒學過“葉平”老師講授的任一門課程的學生姓名;
39、查詢兩門以上不及格課程的同學的學號及其平均成績;
40、檢索“004”課程分數小於60,按分數降序排列的同學學號;
41、刪除“002”同學的“001”課程的成績;
創建數據:
-- 創建班級表 create table Class( cid int primary key auto_increment , caption varchar(20) ); insert into Class (caption) values ("三年二班"), ("一年三班"), ("三年一班"); -- 創建老師表 create table teacher( tid int primary key auto_increment, tname varchar(20) ); insert into teacher (tname) values ("波多"), ("蒼空"), ("飯島"); -- 創建學生表 create table student( sid int primary key auto_increment, sname varchar (20), gender varchar(10), class_id TINYINT ); insert into student (sname,gender,class_id) values ("鋼蛋","男",1), ("翠花","女",1), ("鴨蛋","女",2); -- 創建課程表 create table course( cid int primary key auto_increment, cname varchar (20), teacher_id INT ); insert into course (cname,teacher_id) values ("生物",1), ("體育",1), ("物理",2); -- 創建成績表 create table score( sid int primary key auto_increment, student_id INT, teacher_id INT, number int ); insert into score (student_id,teacher_id,number) values (1,1,60), (1,2,59), (2,2,100);
參考答案:
2、查詢“生物”課程比“物理”課程成績高的所有學生的學號; 思路: 獲取所有有生物課程的人(學號,成績) - 臨時表 獲取所有有物理課程的人(學號,成績) - 臨時表 根據【學號】連接兩個臨時表: 學號 物理成績 生物成績 然後再進行篩選 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 = ‘體育‘) as B on A.student_id = B.student_id where sw > if(isnull(ty),0,ty); 3、查詢平均成績大於60分的同學的學號和平均成績; 思路: 根據學生分組,使用avg獲取平均值,通過having對avg進行篩選 select student_id,avg(num) from score group by student_id having avg(num) > 60 4、查詢所有同學的學號、姓名、選課數、總成績; 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、查詢姓“李”的老師的個數; select count(tid) from teacher where tname like ‘李%‘ select count(1) from (select tid from teacher where tname like ‘李%‘) as B 6、查詢沒學過“葉平”老師課的同學的學號、姓名; 思路: 先查到“李平老師”老師教的所有課ID 獲取選過課的所有學生ID 學生表中篩選 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 = ‘李平老師‘ ) ) 7、查詢學過“001”並且也學過編號“002”課程的同學的學號、姓名; 思路: 先查到既選擇001又選擇002課程的所有同學 根據學生進行分組,如果學生數量等於2表示,兩門均已選擇 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、查詢學過“葉平”老師所教的所有課的同學的學號、姓名; 同上,只不過將001和002變成 in (葉平老師的所有課) 9、查詢課程編號“002”的成績比課程編號“001”課程低的所有同學的學號、姓名; 同第1題 10、查詢有課程成績小於60分的同學的學號、姓名; select sid,sname from student where sid in ( select distinct student_id from score where num < 60 ) 11、查詢沒有學全所有課的同學的學號、姓名; 思路: 在分數表中根據學生進行分組,獲取每一個學生選課數量 如果數量 == 總課程數量,表示已經選擇了所有課程 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、查詢至少有一門課與學號為“001”的同學所學相同的同學的學號和姓名; 思路: 獲取 001 同學選擇的所有課程 獲取課程在其中的所有人以及所有課程 根據學生篩選,獲取所有學生信息 再與學生表連接,獲取姓名 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、查詢至少學過學號為“001”同學所有課的其他同學學號和姓名; 先找到和001的學過的所有人 然後個數 = 001所有學科 ==》 其他人可能選擇的更多 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、查詢和“002”號的同學學習的課程完全相同的其他同學學號和姓名; 個數相同 002學過的也學過 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、刪除學習“葉平”老師課的score表記錄; delete from score where course_id in ( select cid from course left join teacher on course.teacher_id = teacher.tid where teacher.name = ‘葉平‘ ) 16、向SC表中插入一些記錄,這些記錄要求符合以下條件:①沒有上過編號“002”課程的同學學號;②插入“002”號課程的平均成績; 思路: 由於insert 支持 inset into tb1(xx,xx) select x1,x2 from tb2; 所有,獲取所有沒上過002課的所有人,獲取002的平均成績 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、按平均成績從低到高 顯示所有學生的“語文”、“數學”、“英語”三門的課程成績,按如下形式顯示: 學生ID,語文,數學,英語,有效課程數,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 = "體育" 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、查詢各科成績最高和最低的分:以如下形式顯示:課程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、按各科平均成績從低到高和及格率的百分數從高到低順序; 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、課程平均分從高到低顯示(現實任課老師); 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、查詢各科成績前三名的記錄:(不考慮成績並列情況) 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、查詢每門課程被選修的學生數; select course_id, count(1) from score group by course_id; 23、查詢出只選修了一門課程的全部學生的學號和姓名; 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、查詢男生、女生的人數; 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、查詢姓“張”的學生名單; select sname from student where sname like ‘張%‘; 26、查詢同名同姓學生名單,並統計同名人數; select sname,count(1) as count from student group by sname; 27、查詢每門課程的平均成績,結果按平均成績升序排列,平均成績相同時,按課程號降序排列; 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、查詢平均成績大於85的所有學生的學號、姓名和平均成績; 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、查詢課程名稱為“數學”,且分數低於60的學生姓名和分數; 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、查詢課程編號為003且課程成績在80分以上的學生的學號和姓名; select * from score where score.student_id = 3 and score.num > 80 31、求選了課程的學生人數 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、查詢選修“楊艷”老師所授課程的學生中,成績最高的學生姓名及其成績; 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=‘張磊老師‘) order by num desc limit 1; 33、查詢各個課程及相應的選修人數; select course.cname,count(1) from score left join course on score.course_id = course.cid group by course_id; 34、查詢不同課程但成績相同的學生的學號、課程號、學生成績; 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、查詢每門課程成績最好的前兩名; 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、檢索至少選修兩門課程的學生學號; select student_id from score group by student_id having count(student_id) > 1 37、查詢全部學生都選修的課程的課程號和課程名; select course_id,count(1) from score group by course_id having count(1) = (select count(1) from student); 38、查詢沒學過“葉平”老師講授的任一門課程的學生姓名; 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 = ‘張磊老師‘ ) group by student_id 39、查詢兩門以上不及格課程的同學的學號及其平均成績; select student_id,count(1) from score where num < 60 group by student_id having count(1) > 2 40、檢索“004”課程分數小於60,按分數降序排列的同學學號; select student_id from score where num< 60 and course_id = 4 order by num desc; 41、刪除“002”同學的“001”課程的成績; delete from score where course_id = 1 and student_id = 2View Code
【Python全棧-後端開發】MySQL數據庫-練習題