数据库习题参考答案

数据库练习题参考答案

]#mkdir /myload
]#cp /etc/passwd /myload
]#chown mysql /myload
]#vim /etc/my.cnf
[mysql]
secure_file_priv="/myload"
]#systemctl restart mysqld
mysql> load data infile “/myload/passwd” into table userdb.user fields terminated by “:” lines terminated by “\n” ;
2.alter table userdb.user add s_year int default 1990 after name;
3.alter table userdb.user add sex enum(“boy”,“girl”) default “boy” after name;
4.alter table userdb.user add age int unsigned default 21 after sex;
5.update userdb.user set sex=“girl” where uid between 10 and 50;
6.select count(*) from userdb.user where sex=“girl”;
7.select name from userdb.user where sex = “girl” order by uid desc limit 1;
8.insert into userdb.user(name,uid) values (“rtestd”,1000);
9.insert into userdb.user(name,uid) values (“rtest2d”,2000);
10.select name,uid from userdb.user where uid like “____”;

11.select name,uid from userdb.user where name regexp “^r.*d $"; 12.select name from user where name regexp "^a.*c$”;
13.update user set homedir = “/root” where gid between 100 and 500;
14.update user set shell = “/sbin/nologin” where name in (“root”,“bin”,“synv”);
15.select name,gid,shell from user where gid<10;
16.select name from user where name like “d%”;
或者 select name from user where name regexp “^d”;
17.select name,gid,shell from user order by gid desc limit 5;
18.select name,uid,homedir from user where homedir is null;

21.insert into userdb.user values(“lucy”,“girl”,22,1992,“x”,1000,1000,"","/home/lucy","/bin/bash");
22.alter table user drop info;
23. alter table user modify name char(20) not null;
24.alter table user drop homedir;
25.select name,shell,gid from user where gid > 500;
26.delete * from user where uid > 100;
27.select count(*) from user where uid between 10 and 30;
28.select name,shell,uid from user where uid < 100;
29.select * from user order by uid limit 10;
30.select * from user limit 9,6;

31.select * from user where name like “%a%” and uid < 50;
32.select * from user where name in (“root”,“bin”,“daemon”);
33.select * from user where name not in (“root”);
34.select count(*) from user where name is not null;
35.select * from user where name like “%c%”;
36.alter table user add pay float(7,2) default 15000;
37.update user set pay=10000 where sex=“girl”;
38.update user set pay = 30000 where name = “root”;
39.update user set pay = pay+500 where name = “adm”;
40.select name,pay from user;

41.select avg(pay) from user;
42.select name,pay from user where pay < (select avg(pay) from user );
43.select name,uid from user where sex=“girl” order by uid desc limit 1;
44.select sum((select uid from user where name=“bin”)+(select gid from user where name=“bin”)) from user where name = “bin”;