对于原文http://www.jianshu.com/p/520d2eda6f3e的sql创建表
存生日而不是年龄的解答：
生日这一个值是不会变动的，而年龄却会随时间增长而变动，很明显，存一个不变动的值在数据库内更有利。
解决方案：
根据用户信息得出出生日期再输入数据库
查询年龄的sql语句：
先创建时间类的表和数据
再进行查询
DATEDIFF函数用于计算时间差（以日记），
再换算为年后取得最小整数即可得到实岁（实际年龄）
select floor(DATEDIFF(CURDATE(),day)/365.25) from test.dates;
cheers！完成！
```
use test;
create table course
(
CNum int(8) primary key,
CName char(20),
CPlace char(30)
);
create table teacher
(
TeaNum int(8) primary key,
TeaName char(20),
TeaBir date
);
create table TeaCla
(
teaNum int(8) ,
CNum int(8),
CTime char(40),
foreign KEY(teaNum) REFERENCES teacher(TeaNum),
foreign KEY(CNum) REFERENCES course(CNum),
primary key(teaNum,CNum)
);
create table student
(
StuNum int(8) primary key,
StuName char(20),
StuBir date,
StuCla int(8),
foreign KEY(StuCla) REFERENCES course(CNum)
);
create table StuC
(
StuNum int(8),
Course int(8),
SelTime datetime,
foreign KEY(Course) REFERENCES course(CNum),
foreign KEY(StuNum) REFERENCES student(StuNum),
primary key(StuNum,Course)
);
create table college
(
collegeID int(8) primary key,
colName char(20),
colPlace int(8)
);
create table institute
(
instituteID int(8) primary key,
insName char(20),
collegeID int(8),
foreign KEY(collegeID) REFERENCES college(collegeID)
);
create table specialty
(
specialtyID int(8) primary key,
speName char(20),
instituteID int(8),
foreign KEY(instituteID) REFERENCES institute(instituteID)
);
create table class
(
ClaNum int(8) primary key,
ClaName char(20),
specialtyID int(8),
foreign KEY(specialtyID) REFERENCES specialty(specialtyID)
);
show tables;
```
接下来进行增删查改
1.增加记录
```
use test;
insert into course values(10000000,'课程','地点');
```
2.查找记录
```
select * from course;
```
3.修改前一条记录
```
update course
set CNum=12000000,CName='数据库',CPlace='二号教学楼101'
where (CNum=10000000);
```
4.删除记录
```
delete from course where (CNum=12000000);
```
至此，数据库基本操作就算基本完成了。