数据库汇总3（SQL with 关系代数）

一. SQL语句

知识点分立：

创建数据库：CREATE DATABASE <database_name>

删除数据库：DROP DATABASE <database_name>

创建表：

CREATE TABLE <table name> (

<attribute name 1> <data type 1>,

...

PRIMARY KEY(,),

FOREIGN KEY(__) REFERENCES <table> (_),

);

删除表：

DROP TABLE <table name>

增加列：

ALTER TABLE <tablename> ADD (column_name datatype);

删除列：

ALTER TABLE table_name DROP column_name;

修改列数据类型：

ALTER TABLE table_name MODIFY column_name datatype;

修改表的约束：

eg1. ALTER TABLE employee ADD CONSTRAINT pk_student PRIMARY KEY (SSN)

eg2. ALTER TABLE student DROP CONSTRAINT pk_student

选择语句整体结构：

SELECT select_list

FROM table_source

[ WHERE search_condition ]

[ GROUP BY group_by_expression ]

[ HAVING search_condition ]

[ ORDER BY order_expression [ ASC | DESC ] ]

注意：聚合函数不能直接跟在where语句后面，所以得在分组后，使用having语句进行进一步限制；一般search_condition可以表示为 aggregate_function(column_name ) operator value。

匹配符：% 匹配多个字符，_匹配单个字符

eg1. SELECT SNAME FROM STUDENT WHERE SNAME LIKE 'a_%c';

eg2. 使用'20#%' ESCAPE '#'来表示20%

聚合函数：count计数，avg数值平均，max最大值，min最小值，sum数值求和

连接：

inner join：（隐式，内连接，自然连接）

eg. SELECT * FROM student, department WHERE student.DeptID = department.DeptID

outer join：（显式，外连接）

**1.**left outer join：（左外连接，保留左边独有的列）

eg. SELECT * FROM student LEFT OUTER JOIN department

ON student.DEPTID = department.DEPTID

2.right outer join：（右外连接，保留右边独有的列）

eg. SELECT * FROM student RIGHT OUTER JOIN department

ON student.DEPTID = department. DEPTID;

3.full outer join：（全外连接，保留左右独有的列）

eg. SELECT * FROM student FULL OUTER JOIN student

ON student.DEPTID = department.DEPTID;

集合操作：

并集union

eg. (select SName from student where ssn=any

(select ssn from sc where cno=21001001) )

union

(select SName from students where ssn=any

(select ssn from sc where cno=22003002) )

交集Intersect

eg. (select SName from student where ssn=any

(select ssn from sc where cno=21001001) )

intersect

(select SName from student where ssn=any

(select ssn from sc where cno=22003002) )

差 except/minus

eg. (select SName from student where ssn=any

(select ssn from sc where cno=21001001) )

minus

(select SName from student where ssn=any

(select ssn from sc where cno=22003002) )

注意：此处的" = any "实际上和" in "是一样的意思，可以相互替换。

子查询（多select语句）：

eg. SELECT e1. SNAME FROM student e1,student e2

WHERE e1. DEPTNO = e2. DEPTNO AND e2. SNAME = 'JONES'

any和all对比：

exists语句和 not exists：
situation1: exists ( ) is true if the set ( ) is not empty.
situation2: exists ( ) is false if the set ( ) is empty.
situation3: not exists ( ) is true if the set ( ) is empty.
situation4: not exists ( ) is false if the set ( ) is not empty.

练笔题目：

1.找出每个学生超过他选课课程平均成绩的课程号.

Select sno,cno From sc x

Where grade>=(select avg(grade) From sc y Where y.sno**=x.sno** )

找出每个学生超过该选课课程平均成绩的课程号.

Select sno,cno From sc x

Where grade>=(select avg(grade) From sc y Where y.cno**=x.cno** )

Find all students who do not take 21003001.

select * from Student s where not exists

(select * from sc

where ssn= s.ssn

and Cno = '21003001.')

This query is equivalent to:

select * from Student where ssn not in

(select ssn from sc

where Cno = '21003001.')

Find the names of all students who study in the same department with a student with sname= 'liucheng' .

SELECT ssn，Sname，DEPTNO

FROM Student S1

WHERE EXISTS

SELECT *

FROM Student S2

WHERE S2.DEPTNO = S1.DEPTNO AND

S2.Sname = ' liucheng '；

Find all the students who take all courses.

select * from Student s where not exists

(select * from Course c where not exists

(select * from sc where ssn= s.ssn and Cno = c.Cno))

注意：此题最重要的是使用双重否定表示肯定，分为全部课程和选的课程相同两部分，对后者进行否定表示选的课程有不同，再从整体进行剔除，表示上了全部课程。

6. Find the names and GPAs of those students who take all courses taken by a student with SID = 2015002001.

select Name, GPA from Student s

where not exists

(select * from Course c where Cno in

(select Cno from sc where ssn= 2015002001)

and not exists

(select * from sc e where ssn=e.ssn and Cno = c.Cno))

exists 、not exists扩展练习

Consider relations:

Employees(Eid, Name, Age, Salary)

Projects(Proj_no, Name, Manager_name)

Works(Eid, Proj_no, Start_Date)
（1）Find the names of those employees who participate all projects.

select name from Employees e where not exists

(select * from Projects p where not exists

(select * from Works w

where e.Eid = w.Eid

and w.Proj_no = p.Proj_no))

（2）Find the names of those projects managed by Smith that are participated by all employees under 40.
selectnamefromProjects pwhere manager_name**=****'Smith'andnot exists**

(select*fromEmployees ewhereAge < 40 and****not exists

(select*fromWorks w****where e.Eid**=w.Eidandw.Proj_no=p.Proj_no))**