1.Create a sample database:
1.cmd
2.mysql -u root -p
3.create database bank;
4.grant all privileges on bank.* to 'vcsql'@'localhost' identified by 'asd321';
#There is a space before 'identified' which is invisiable on the book. It confuses me a lot.
5.quit;#Exit from MySQL toolkit.
6.use bank; #Use the database created just now;
7.mysql -uvcsql -pasd321;#LogIn
2.Some Functions
1.SELECT Now();#Display Time
2.SHOW CHARACTER SET;#Describe the character set
3.
3.Create a table named 'person'
CREATE TABLE person
(person_id SMALLINT UNSIGNED,
fname VARCHAR(20),
lname VARCHAR(20),
gender CHAR(1),
birth_date DATE,
street VARCHAR(30),
city VARCHAR(20),
state VARCHAR(20),
country VARCHAR(20),
postal_code VARCHAR(20),
CONSTRAINT pk_person PRIMARY KEY (person_id)
);
gender CHAR(1) CHECK (gender IN ('M', 'F'));#Add a constraint to gender
gender ENUM('M', 'F')#replace 'gender CHAR(1)' with this
4.Create a table named 'favourite_food'
CREATE TABLE favourite_food
(person_id SMALLINT UNSIGNED,
food VARCHAR(20),
CONSTRAINT pk_favourite_food PRIMARY KEY (person_id, food),#设置两个字段的复合主键
CONSTARINT fk_fav_food_person_id FOREIGN KEY (person_id)#设置索引，将本表（favourite_food)的person_id字段关联到person表的person_id字段，作为外键，此处有大坑，在后面将恶心我好久
REFERENCES person (person_id)
);
5.Check tables
desc person;
desc favourite_food;
6.把主键'person_id'改为自增主键
ALTER TABLE person MODIFY person_id SMALLINT UNSIGNED AUTO_INCREMENT;
#雾草，这里这个大坑，因为之前在favourite表中，已经将本表内复合主键person_id关联到person表的person_id字段作为外键。有点绕。简单地说，就是想要修改的字段，已经被关联作为另一张表的外键了。所以不能随便修改。想要修改，必须先解除外键关系，修改好后，再重新关联外键。
处于数据库安全角度考虑，在解除外键并修改过程中，要将表锁定（LOCK），修改好后，解锁（UNLOCK）。这里坑我好长一段时间，一直报错。
键入ALTER*后
意味着不能这样修改，经搜索后，得到这样一个解决方案：
但是尝试n次后，并不奏效，并一直报错，其中有几次经检查后，是自己代码键入错误，尴尬~！当时检查重拼后，仍有错误，看来是这个解决办法中的代码有问题。
分析代码可知，FOREIGN KEY （person_id)没有指定所在的表。翻阅SQL技术文档，有如下提示：ALTER TABLE Syntax
ALTER TABLE favourite_food
ADD CONSTRAINT fk_fav_food_person_id FOREIGN KEY (person_id)
REFERENCES person (person_id);
UNLOCK TABLES;
让我看一下成果吧：
desc favourite_food;
