Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

Student Student_ID Student_Name 1022 John Doe 4123 Jill Maher 3456 Matt Samuels

ID: 3598225 • Letter: S

Question

Student

Student_ID

Student_Name

1022

John Doe

4123

Jill Maher

3456

Matt Samuels

3751

Megan Cramer

Advisor_ID

Advisor_Name

Adv_Room

1

Jones

412

2

Smith

216

3

Taylor

322

Course

Course_ID

Course_Name

101-07

Analytics

107-08

Economics

201-01

Statistics

143-01

Biology

211-02

Java

135-03

Sociology

159-02

Physics

214-01

History

147-11

Chemistry

Student_Advisor

Stu_Adv_ID

Student_ID

Advisor_ID

1

1022

1

2

4123

2

3

3456

1

4

3751

3

Student_Course

Stu_Cr_ID

Student_ID

Course_ID

1

1022

101-07

2

1022

143-01

3

1022

159-02

4

4123

201-01

5

4123

211-02

6

4123

214-01

7

3456

101-07

8

3456

143-01

9

3456

159-02

10

3751

107-08

11

3751

135-03

12

3751

147-11

The above tables show the solution for your normalization HW. The data types in the tables are as follows:

Student: (Student_ID Integer PRIMARY KEY, Student_Name Char(50))

Advisor: (Advisor_ID Integer PRIMARY KEY, Advisor_Name Char(50), Adv_Room Integer)

Course: (Course_ID Char(10) PRIMARY KEY, Course_Name Char(50))

Student_Advisor (Stu_Adv_ID Integer PRIMARY KEY, Student_ID Integer, Advisor_ID Integer)

Student_Course (Stu_Cr_ID Integer PRIMARY KEY, Student_ID Integer, Course_ID Char(10))

Write the SQL statements for the following and show the output of the SQL statements where applicable

(No output necessary for Questions 1, 11 and 12)

Create a new student entry for Sam Mills with Student_ID 1077.

Output the names of all students taking Biology.

Output the unique Advisor_Ids from the Student_Advisor table.

Output the names of all students whose name starts with J.

Output all Student_Ids whose Advisor_ID is not 3.

Output the names of all students whose ID is more than 3000.

Show the number of courses in the Course table.

Show the names of all courses in the Course table sorted by the Course name.

Output the names of all students whose ID is in between 3700 and 4000.

Use a sub-query to show the names of all students whose advisor is Jones.

Change the name of “Matt Samuels” to “Kevin Samuels”.

Delete the entry created in Question 1 (Sam Mills) from the Student table.

Student_ID

Student_Name

1022

John Doe

4123

Jill Maher

3456

Matt Samuels

3751

Megan Cramer

Advisor

Advisor_ID

Advisor_Name

Adv_Room

1

Jones

412

2

Smith

216

3

Taylor

322

Explanation / Answer

1-INSERT INTO Student(Student_ID,Student_Name)
Values (1077,"Sam Mills");
2-SELECT Student_Name FROM STUDENT WHERE Student_ID IN (SELECT Student_ID FROM Student_Course WHERE Course_ID=(SELECT Course_ID FROM Course WHERE Course_Name="Biology") );
3-SELECT Distinct Advisor_ID from Student_Advisor;
4-SELECT Student_Name fROM Student WHERE Student_Name like "J%"l
5-SELECT Student_ID from Student_Advisor WHERE Advisor_ID <> 3;
6-SELECT Student_Name FROM Student WHERE Student_ID>3000;
7-SELECT Count(Course_ID) as TotalCourses from Course;
8-SELECT Course_Name From Course Order by Course_Name;
9-SELECT Student_Name FROM Student WHERE Student_ID BETWEEN 3700 AND 4000;
10-SELECT Student_Name FROM Student WHERE ID IN(SELECT Student_ID FROM Student_Advisor WHERE Advisor_ID IN (SELECT Advisor_ID FROM Advisor WHERE Advisor_Name="Jones"));
11-UPDATE Student SET Student_Name="Matt Samuels" WHERE Student_Name="Kevin Samuels";
12-DELETE FROM Student WHERE Student_Name="Sam Mills";