Database interim (I)

Student information is stored in the educational administration system , Including student ID 、 full name 、 major ; Course information , Include course number 、 Course name ,

credits , Students' course selection information , Include , Student number 、 Course no. 、 achievement .

1) Write course information 、 Database mode of course selection information ;

Pattern ： The combination of the relationship name and its attribute set is called the pattern of this relationship .

How to express a pattern ： First give a relationship name , Enclose all attributes with parentheses .

Student ( Student number , full name , major ); Course ( Course no. , Course name , credits ); Course selection ( Student number , Course no. , achievement );

2) Write out the command of creating course information ;

Create table Course ( Course no. char(6), Course name varchar(30), credits decimal(2,1), Primary key ( Course no. ) );

3) Write the relational algebraic expression of the following query ：

        a) The result is less than 60 The student number of the student ;

Π Student number (σ achievement <60( Course selection ))

        b) The result is less than 60 The student number of the student 、 full name ;

Π Course selection . Student number , Student . full name (σ Course selection . Student number = Student . Student number ˄ Course selection . achievement <60( Course selection х Student ))

        c) Find out the student ID of students who have never taken elective courses ;

Π Student number ( Student ) - Π Student number ( Course selection )

       d) Find out the student number of the student with the lowest score ;

Π Student number ( Course selection ) - Π Course selection . Student number (σ Course selection . achievement >s1. achievement (( Course selection ) х ρS1( Course selection ))) ——> When a table needs to be used twice , We must rename it to make a new table

       e) At least 2 The number of courses selected by students ;

Π Course selection . Course no. (σ Course selection . Course no. =s1. Course no. ˄ Course selection . Student number <>=s1. Student number (( Course selection ) хρS1( Course selection )))

       f) Find out the student ID of students who have taken all courses

Π Student number ( Course selection ) - Π Student number ( (Π Student number ( Student ) хΠ Course no. ( Course ) - Π Student number , Course no. ( Course selection ))) ——>Π Student number ( Student ) хΠ Course no. ( Course ) - Π Student number , Course no. ( Course selection ): Pair each student with all course numbers （ The practical significance of the generated new relationship is that each student has taken all courses ）, Then subtract the actual course selection of each student recorded in the course selection table , The result is that there are only students who have not completed all courses .