SQL_基础更改；创建表、索引：create，insert into，index，alter，update, replace , rename，delete

1、创建一个actor表，包含如下列信息
列表 类型 是否为NULL 含义
actor_id smallint(5) not null 主键id
first_name varchar(45) not null 名字
last_name varchar(45) not null 姓氏
last_update timestamp not null 最后更新时间，默认是系统的当前时间

答案：

create table actor
(
actor_id smallint(5) not null primary key,
first_name varchar(45) not null,
last_name varchar(45) not null,
last_update timestamp not null default(datetime('now','localtime'))-- ,
-- primary key(actor_id)
)

2、对于表actor批量插入如下数据
CREATE TABLE IF NOT EXISTS actor (
actor_id smallint(5) NOT NULL PRIMARY KEY,
first_name varchar(45) NOT NULL,
last_name varchar(45) NOT NULL,
last_update timestamp NOT NULL DEFAULT (datetime(‘now’,’localtime’)))

答案1：union select

insert into actor
select 1, 'PENELOPE', 'GUINESS', '2006-02-15 12:34:33'
union select 2,'NICK', 'WAHLBERG', '2006-02-15 12:34:33'

答案2：values( )

INSERT INTO actor
VALUES (1, 'PENELOPE', 'GUINESS', '2006-02-15 12:34:33'),
(2, 'NICK', 'WAHLBERG', '2006-02-15 12:34:33')

3、对于表actor批量插入如下数据,如果数据已经存在，请忽略，不使用replace操作
CREATE TABLE IF NOT EXISTS actor (
actor_id smallint(5) NOT NULL PRIMARY KEY,
first_name varchar(45) NOT NULL,
last_name varchar(45) NOT NULL,
last_update timestamp NOT NULL DEFAULT (datetime(‘now’,’localtime’)))

答案：insert or ignore into

insert or ignore into actor
values(3,'ED','CHASE','2006-02-15 12:34:33')

4、对于如下表actor，其对应的数据为:

actor_id	first_name	actor_id	first_name	last_update
1	PENELOPE	GUINESS	2006-02-15	12:34:33
2	NICK	WAHLBERG	2006-02-15	12:34:33

创建一个actor_name表，将actor表中的所有first_name以及last_name导入改表。 actor_name表结构如下：

列表	类型	是否为NULL	含义
first_name	varchar(45)	not null	名字
last_name	varchar(45)	not null	名字

答案：

create table actor_name
(
first_name varchar(45)not null,
last_name varchar(45)not null);
insert into actor_name select first_name,last_name from actor

5、针对如下表actor结构创建索引：
CREATE TABLE IF NOT EXISTS actor (
actor_id smallint(5) NOT NULL PRIMARY KEY,
first_name varchar(45) NOT NULL,
last_name varchar(45) NOT NULL,
last_update timestamp NOT NULL DEFAULT (datetime(‘now’,’localtime’)))
对first_name创建唯一索引uniq_idx_firstname，对last_name创建普通索引idx_lastname

答案：create (unique) index on

create unique index uniq_idx_firstname on actor(first_name);
create index idx_lastname on actor(last_name);

6、针对actor表创建视图actor_name_view，只包含first_name以及last_name两列，并对这两列重新命名，first_name为first_name_v，last_name修改为last_name_v：
CREATE TABLE IF NOT EXISTS actor (
actor_id smallint(5) NOT NULL PRIMARY KEY,
first_name varchar(45) NOT NULL,
last_name varchar(45) NOT NULL,
last_update timestamp NOT NULL DEFAULT (datetime(‘now’,’localtime’)))

答案：

create view actor_name_view (first_name_v,last_name_v) as
select first_name, last_name from actor

7、针对salaries表emp_no字段创建索引idx_emp_no，查询emp_no为10005, 使用强制索引。
CREATE TABLE salaries (
emp_no int(11) NOT NULL,
salary int(11) NOT NULL,
from_date date NOT NULL,
to_date date NOT NULL,
PRIMARY KEY (emp_no,from_date));
create index idx_emp_no on salaries(emp_no);

答案：

select * from salaries indexed by idx_emp_no where emp_no = 10005

8、存在actor表，包含如下列信息：
CREATE TABLE IF NOT EXISTS actor (
actor_id smallint(5) NOT NULL PRIMARY KEY,
first_name varchar(45) NOT NULL,
last_name varchar(45) NOT NULL,
last_update timestamp NOT NULL DEFAULT (datetime(‘now’,’localtime’)));
现在在last_update后面新增加一列名字为create_date, 类型为datetime, NOT NULL，默认值为’0000 00:00:00’

alter table … add

答案1：

alter table actor 
add column create_date datetime not null default '0000-00-00 00:00:00'

答案2：

alter table actor add create_date datetime DEFAULT '0000-00-00 00:00:00' NOT NULL ;

9、将所有to_date为9999-01-01的全部更新为NULL,且 from_date更新为2001-01-01。
CREATE TABLE IF NOT EXISTS titles_test (
id int(11) not null primary key,
emp_no int(11) NOT NULL,
title varchar(50) NOT NULL,
from_date date NOT NULL,
to_date date DEFAULT NULL);

insert into titles_test values (‘1’, ‘10001’, ‘Senior Engineer’, ‘1986-06-26’, ‘9999-01-01’),
(‘2’, ‘10002’, ‘Staff’, ‘1996-08-03’, ‘9999-01-01’),
(‘3’, ‘10003’, ‘Senior Engineer’, ‘1995-12-03’, ‘9999-01-01’),
(‘4’, ‘10004’, ‘Senior Engineer’, ‘1995-12-03’, ‘9999-01-01’),
(‘5’, ‘10001’, ‘Senior Engineer’, ‘1986-06-26’, ‘9999-01-01’),
(‘6’, ‘10002’, ‘Staff’, ‘1996-08-03’, ‘9999-01-01’),
(‘7’, ‘10003’, ‘Senior Engineer’, ‘1995-12-03’, ‘9999-01-01’);

答案：

update titles_test set to_date=null,from_date='2001-01-01'
where to_date='9999-01-01'

10、将id=5以及emp_no=10001的行数据替换成id=5以及emp_no=10005,其他数据保持不变，使用replace实现。
CREATE TABLE IF NOT EXISTS titles_test (
id int(11) not null primary key,
emp_no int(11) NOT NULL,
title varchar(50) NOT NULL,
from_date date NOT NULL,
to_date date DEFAULT NULL);

insert into titles_test values (‘1’, ‘10001’, ‘Senior Engineer’, ‘1986-06-26’, ‘9999-01-01’),
(‘2’, ‘10002’, ‘Staff’, ‘1996-08-03’, ‘9999-01-01’),
(‘3’, ‘10003’, ‘Senior Engineer’, ‘1995-12-03’, ‘9999-01-01’),
(‘4’, ‘10004’, ‘Senior Engineer’, ‘1995-12-03’, ‘9999-01-01’),
(‘5’, ‘10001’, ‘Senior Engineer’, ‘1986-06-26’, ‘9999-01-01’),
(‘6’, ‘10002’, ‘Staff’, ‘1996-08-03’, ‘9999-01-01’),
(‘7’, ‘10003’, ‘Senior Engineer’, ‘1995-12-03’, ‘9999-01-01’);

答案：replace into … values

replace into titles_test values (5, 10005, 'Senior Engineer', '1986-06-26', '9999-01-01')

11、将titles_test表名修改为titles_2017。
CREATE TABLE IF NOT EXISTS titles_test (
id int(11) not null primary key,
emp_no int(11) NOT NULL,
title varchar(50) NOT NULL,
from_date date NOT NULL,
to_date date DEFAULT NULL);

答案：alter … rename to …

alter table titles_test rename to titles_2017

12、将所有获取奖金的员工当前的薪水增加10%。
create table emp_bonus(
emp_no int not null,
recevied datetime not null,
btype smallint not null);
CREATE TABLE salaries (
emp_no int(11) NOT NULL,
salary int(11) NOT NULL,
from_date date NOT NULL,
to_date date NOT NULL, PRIMARY KEY (emp_no,from_date));

答案：
update … set

update salaries set salary=salary*1.1 where emp_no in 
(select s.emp_no from salaries s inner join emp_bonus b on s.emp_no=b.emp_no and s.to_date='9999-01-01')

13、删除emp_no重复的记录，只保留最小的id对应的记录。
CREATE TABLE IF NOT EXISTS titles_test (
id int(11) not null primary key,
emp_no int(11) NOT NULL,
title varchar(50) NOT NULL,
from_date date NOT NULL,
to_date date DEFAULT NULL);

insert into titles_test values (‘1’, ‘10001’, ‘Senior Engineer’, ‘1986-06-26’, ‘9999-01-01’),
(‘2’, ‘10002’, ‘Staff’, ‘1996-08-03’, ‘9999-01-01’),
(‘3’, ‘10003’, ‘Senior Engineer’, ‘1995-12-03’, ‘9999-01-01’),
(‘4’, ‘10004’, ‘Senior Engineer’, ‘1995-12-03’, ‘9999-01-01’),
(‘5’, ‘10001’, ‘Senior Engineer’, ‘1986-06-26’, ‘9999-01-01’),
(‘6’, ‘10002’, ‘Staff’, ‘1996-08-03’, ‘9999-01-01’),
(‘7’, ‘10003’, ‘Senior Engineer’, ‘1995-12-03’, ‘9999-01-01’);
答案：

delete from titles_test where id not in
(select min(id) from titles_test group by emp_no)