Linux練習題-SQL語句

練習題

一、導入hellodb.sql生成數據庫

mysql -uroot < hellodb.sql，use hellodb

(1) 在students表中，查詢年齡大於25歲，且爲男性的同窗的名字和年齡

select Name,Age from students where Age>25;

(2) 以ClassID爲分組依據，顯示每組的平均年齡

select ClassID,avg(Age) from students group by ClassID

(3)顯示第2題中平均年齡大於30的分組及平均年齡

select ClassID,avg(Age) from students group by ClassID having avg(Age) > 30

(4) 顯示以L開頭的名字的同窗的信息

select * from students where name like 'L%';

(5) 顯示TeacherID非空的同窗的相關信息

select * from students where TeacherID is not null;

(6) 以年齡排序後，顯示年齡最大的前10位同窗的信息

select * from students order by age desc limit 10;

(7) 查詢年齡大於等於20歲，小於等於25歲的同窗的信息

select * from students where age between 20 and 25;

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二、導入hellodb.sql，如下操做在students表上執行

(1)以ClassID分組，顯示每班的同窗的人數

 select classID,count(*) as total_num from students group by ClassID;

(2)以Gender分組，顯示其年齡之和

select gender,sum(Age) from students group by gender;

(3)以ClassID分組，顯示其平均年齡大於25的班級

select ClassID,avg(Age)  from students group by ClassID having avg(Age) > 25;

(4)以Gender分組，顯示各組中年齡大於25的學員的年齡之和

select gender,sum(Age) from students where age > 25 group by gender;

(5)顯示前5位同窗的姓名、課程及成績

select s.stuid as stu_id,s.name as stu_name,sc.score,c.course from students as s inner join scores sc on s.stuid=sc.stuid inner join courses c on sc.courseid=c.courseid where s.stuid <= 5;

(6)顯示其成績高於80的同窗的名稱及課程；

select s.stuid as stu_id,s.name as stu_name,sc.score,c.course from students as s inner join scores sc on s.stuid=sc.stuid inner join courses c on sc.courseid=c.courseid where sc.score > 80;

(7)求前8位同窗每位同窗本身兩門課的平均成績，並按降序排列

select s.stuid,s.name,avg(sc.score) as avg_score from students as s inner join scores sc on s.stuid=sc.stuid group by stuid order by avg_score;

(8)取每位同窗各門課的平均成績，顯示成績前三名的同窗的姓名和平均成績

select s.name,avg(sc.score) from students as s inner join scores sc on s.stuid=sc.stuid group by s.name order by avg(sc.score) desc limit 3;

(9)顯示每門課程課程名稱及學習了這門課的同窗的個數

第一步：通過篩選，決定使用courses表、coc表和students表作查詢

第二步：經過聯繫將三個表聯繫在了一塊兒，select count(s.stuid),c.course from students as s inner join coc on s.classid=coc.classid inner join courses c on coc.courseid=c.courseid;

第三步，分組，獲得的這個表用哪一個分組合適？毫無疑問，用course字段合適，將該字段加入分組，select count(s.stuid) as Stu_Num,c.course from students as s inner join coc on s.classid=coc.classid inner join courses c on coc.courseid=c.courseid group by c.course;

(10)顯示其年齡大於平均年齡的同窗的名字

    有子查詢和內鏈接兩種寫法，雖然子查詢方式便於理解，但因爲數據庫性能不高在此推薦用內鏈接的寫法

    第一種，子查詢：select name,age from students where age > (select avg(Age) from students);

    第二種，內鏈接寫法：slect s.name,age from students as s inner join (select avg(age) as age from students) as ss on s.age > ss.age;

(11)顯示其學習的課程爲第一、2，4或第7門課的同窗的名字

select s.name,a.courseid from students as s inner join (select * from coc where courseid in ('1','2','4','7')) as a on s.classid=a.classid;

(12)顯示其成員數最少爲3個的班級的同窗中年齡大於同班同窗平均年齡的同窗

先查詢成員數量最少爲3個的班級，且每一個班的平均年齡，select classid,count(stuid),avg(age) from students group by classid having count(stuid)>=3;

再將學生表的name、age與生成的該表作對比，對比時候用where條件判斷每一個學生分別對應本身的classid進行比較

(13)統計各班級中年齡大於全校同窗平均年齡的同窗

select stuid,name,age,classid from students where age > (select avg(age) from students);