General [IMPORTANT] HW2 SQL

Description

According to HW2 specs, any of the following situations causes a deduction of 0.5 pts.

• Don’t name files properly (naming convention refers to HW2 specs.), However,

• • Error in file extension (e.g. Q1.sql.txt, see Piazza @111) is OK.

• • Because of case insensitiveness, both q1.sql or Q1.SQL should be OK.

• Submit a .zip (required to submit individual files).

The following depend on cases.

• Submit/resubmit after the deadline. (be sure to do it ahead of time, and make sure you did)

Refer to the deadline policy in this course, 10% penalty per day after the deadline.

• A solution is partially wrong/correct or doesn’t explain properly. (See details in upcoming sections).

Give half pts (0.5pts for Q1-4, 1pt for Q5, 0.5pts for each Q5.bonus)

• A solution is totally wrong, doesn’t make any sense (rarely happens). Deduct full pts (1pt for Q1-4, 2pts for Q5).

Other potential cases

• Alternative solutions are VERY possible, better to look into/think of/run their scripts thoroughly.

• A fully correct answer for Q2-5 should contain only A Query statement, not multiple query statements.

• A .sql may not be successfully executable (by “source Q#.sql”), including but not limited to the following:

  • .sql files are not executable due to missing of “;” at the end of any statements.

• • Some students may comment out those DDL/DML in their answers.

Then we can easily fix them. E.g. Kindly copy/paste into the terminal if you would like to run it. If this happens almost everywhere in one’s submissions (making you mad and you don’t know how to fix it), then we propose to deduct 0.5 or 1 pt (depending on how many mistakes and how mad you are) due to the unsuccessful execution of .sql files. Otherwise, be lenient and nice : )

• TBD

Don’t forget to leave proper feedback, which makes regrading easy.

Setup and Cleanup

To run a solution on Mysql Shell, if they don’t Setup or Cleanup (only solution is provided), then we may need to Setup (e.g. create database, use database, etc.) and Cleanup (e.g. drop database, etc.) for grading purposes. Could refer to provided solution files.

Please read the following rubrics along with the solution code and HW2 specs.

DROP DATABASE IF EXISTS testDB;

CREATE DATABASE testDB;

USE testDB;

CREATE TABLE Persons (

PersonID int,

LastName varchar(255),

FirstName varchar(255),

Address varchar(255),

City varchar(255)

);

Insert into Persons

VALUES(23, “Chary”, “Saty”, “USC”, “LA”);

select * from Persons;

DROP DATABASE IF EXISTS testDB;

For Ubuntu 20.04

#Installation

sudo apt-get remove docker docker-engine docker.io containerd runc

sudo apt-get update

sudo mkdir -p /etc/apt/keyrings

curl -fsSL https://download.docker.com/linux/ubuntu/gpg | sudo gpg

–dearmor -o /etc/apt/keyrings/docker.gpg

echo “deb [arch=$(dpkg –print-architecture) signed-by=/etc/apt/keyrings/docker.gpg] https://download.docker.com/linux/ubuntu $(lsb_release -cs) stable” | sudo tee /etc/apt/sources.list.d/docker.list > /dev/null

sudo apt-get update

sudo apt-get install docker-ce docker-ce-cli containerd.io docker-compose-plugin

#Install mysql

sudo docker pull mysql

#Verify

sudo docker images

#Install mysql shell

https://dev.mysql.com/doc/mysql-shell/8.0/en/mysql-shell-install-linux-qui

ck.html

1. Download “Adding the MySQL APT Repository” configuration deb and run

it.

sudo dpkg -i /PATH/version-specific-package-name.deb

2. sudo apt-get update

3. sudo apt-get install mysql-shell

####Installation done####

#Run mysql docker image

sudo docker run -d -e MYSQL_ROOT_PASSWORD=test –name hw2 -p

127.0.0.1:3307:3306 mysql

sudo docker ps

#Make mysql server always restart on boot sudo docker update –restart always hw2

#Connect to mysql

mysqlsh –uri root@127.0.0.1:3307

Q1

• Equivalent or alternative solution, once it makes sense, either SQL statements OR textual explanation is acceptable.

• It’s OK if SQL statements are provided but fail to be successfully executed. However, the explanation must make sense.

For the 1st issue, see if there is a CHECK when creating the table, it could also come with a trigger or a constraint.

startTime < endTime.

For the 2nd issue, see if a trigger (may combine with procedure or function) is created to avoid insertion of the following situation.

(NEW.startTime >= startTime AND NEW.startTime < endTime) OR (NEW.endTime > startTime AND NEW.endTime <= endTime))

Another easy way (without concerning the efficiency) to solve both issues could be Piazza @91, which assumes multiple INSERT statements when booking a period of time. It’s acceptable.

— Setup

DROP DATABASE IF EXISTS HW2Q1DB;

CREATE DATABASE HW2Q1DB;

USE HW2Q1DB;

CREATE TABLE ProjectRoomBookings ( roomNum INTEGER NOT NULL, startTime INTEGER NOT NULL, endTime INTEGER NOT NULL, groupName CHAR(10) NOT NULL, PRIMARY KEY (roomNum, startTime)

);

— Solution

/* To solve the 1st issue */

ALTER TABLE ProjectRoomBookings

ADD CONSTRAINT CHK_BOOKING CHECK (

startTime>=7 AND /* OK if they don’t have */

endTime<=18 AND /* OK if they don’t have */

startTime < endTime /* This check is IMPORTANT */

);

/* To solve the 2nd issue */

DELIMITER $$

CREATE TRIGGER roomAvailability

BEFORE INSERT

ON ProjectRoomBookings FOR EACH ROW

BEGIN

IF ( SELECT COUNT(*) FROM ProjectRoomBookings WHERE roomNum = NEW.roomNum

AND ((NEW.startTime >= startTime AND NEW.startTime < endTime) OR (NEW.endTime > startTime AND NEW.endTime <= endTime)) ) <> 0 /*

This check is IMPORTANT */

THEN

SET @message_text = CONCAT(‘Room ‘, NEW.roomNum, ‘ is occupied during ‘, NEW.startTime, ‘ to ‘, NEW.endTime, ‘.’);

SIGNAL SQLSTATE ‘45000’ SET MESSAGE_TEXT = @message_text; END IF;

END$$

DELIMITER ;

/* Could define a trigger for UPDATE in the same way, but not necessary once an answer show the above

*/

— Test cases (can be ignored) —

/* Test1:

Valid value

*/

INSERT INTO ProjectRoomBookings

VALUES(1, 7, 11, “Team1”);

INSERT INTO ProjectRoomBookings

VALUES(1, 11, 12, “Team2”);

SELECT * FROM ProjectRoomBookings;

DELETE FROM ProjectRoomBookings WHERE groupName = “Team1”; DELETE FROM ProjectRoomBookings WHERE groupName = “Team2”;

/* Test2:

Invalid startTime (or endTime)

*/

INSERT INTO ProjectRoomBookings

VALUES(1, 6, 18, “Team1”);

SELECT * FROM ProjectRoomBookings;

DELETE FROM ProjectRoomBookings WHERE groupName = “Team1”;

/* Test3:

startTime > endTime

*/

INSERT INTO ProjectRoomBookings

VALUES(1, 9, 7, “Team1”);

SELECT * FROM ProjectRoomBookings;

DELETE FROM ProjectRoomBookings WHERE groupName = “Team1”;

/* Test4:

Book unavailable room

*/

INSERT INTO ProjectRoomBookings

VALUES(1, 7, 11, “Team1”);

INSERT INTO ProjectRoomBookings

VALUES(1, 8, 10, “Team2”);

SELECT * FROM ProjectRoomBookings;

DELETE FROM ProjectRoomBookings WHERE groupName = “Team1”; DELETE FROM ProjectRoomBookings WHERE groupName = “Team2”;

— Clean up options

DROP TRIGGER IF EXISTS roomAvailability;

DROP DATABASE IF EXISTS HW2Q1DB;

Q2

• To get full points, an answer must show the correct SQL query statement

• The testing data could be what is shown in HW2.Q2 specs.

• In the output, it’s OK if classNames with the same Total # switch places. (i.e. Javascript and Python, Processing and Java)

— Setup

DROP DATABASE IF EXISTS HW2Q2DB;

CREATE DATABASE HW2Q2DB;

USE HW2Q2DB;

/* A student won’t take a class many times, since it’s within a term. */ CREATE TABLE Enrollment (

SID CHAR(3) NOT NULL,

ClassName CHAR(11) NOT NULL,

Grade CHAR(2),

PRIMARY KEY (SID, ClassName)

);

INSERT INTO Enrollment VALUES (‘123’, ‘Processing’, ‘A’); INSERT INTO Enrollment VALUES (‘123’, ‘Python’, ‘B’); INSERT INTO Enrollment VALUES (‘123’, ‘Scratch’, ‘B’); INSERT INTO Enrollment VALUES (‘662’, ‘Java’, ‘B’); INSERT INTO Enrollment VALUES (‘662’, ‘Python’, ‘A’); INSERT INTO Enrollment VALUES (‘662’, ‘JavaScript’, ‘A’); INSERT INTO Enrollment VALUES (‘662’, ‘Scratch’, ‘B’);

INSERT INTO Enrollment VALUES (‘345’, ‘Scratch’, ‘A’); INSERT INTO Enrollment VALUES (‘345’, ‘JavaScript’, ‘B’); INSERT INTO Enrollment VALUES (‘345’, ‘Python’, ‘A’); INSERT INTO Enrollment VALUES (‘555’, ‘Python’, ‘B’); INSERT INTO Enrollment VALUES (‘555’, ‘JavaScript’, ‘B’); INSERT INTO Enrollment VALUES (‘213’, ‘JavaScript’, ‘A’);

SELECT * FROM Enrollment;

— Solution

SELECT ClassName, COUNT(SID) AS Total

FROM Enrollment

GROUP BY ClassName

ORDER BY Total DESC;

— Clean up option

DROP DATABASE IF EXISTS HW2Q2DB;

Q3

• To get full points, an answer must show the correct SQL query statement.

• NOTE, there may be a project with step 0 only. So in addition to what is shown in HW2.Q3 specs, the testing data should add one more row (refer to the solution code). OR, See if they have an assumption that this kind of project doesn’t exist (i.e. In a project, the # of steps always greater than 0). Otherwise, the query should be able to handle this situation.

— Setup

DROP DATABASE IF EXISTS HW2Q3DB;

CREATE DATABASE HW2Q3DB;

USE HW2Q3DB;

CREATE TABLE ProjectStatus (

PID CHAR(4) NOT NULL,

Step INTEGER NOT NULL,

Status CHAR(1) NOT NULL,

PRIMARY KEY (PID, Step)

);

INSERT INTO ProjectStatus VALUES (‘P100’, 0, ‘C’);

INSERT INTO ProjectStatus VALUES (‘P100’, 1, ‘W’);

INSERT INTO ProjectStatus VALUES (‘P100’, 2, ‘W’);

INSERT INTO ProjectStatus VALUES (‘P201’, 0, ‘C’);

INSERT INTO ProjectStatus VALUES (‘P201’, 1, ‘C’);

INSERT INTO ProjectStatus VALUES (‘P333’, 0, ‘W’);

INSERT INTO ProjectStatus VALUES (‘P333’, 1, ‘W’);

INSERT INTO ProjectStatus VALUES (‘P333’, 2, ‘W’);

INSERT INTO ProjectStatus VALUES (‘P333’, 3, ‘W’);

some intermediate VIEWs have been created as Tables. It’s OK once the design data schema makes sense.

• • Again, a fully correct answer should contain only A Query statement, not multiple query statements.

• Setup

DROP DATABASE IF EXISTS HW2Q4DB;

CREATE DATABASE HW2Q4DB;

USE HW2Q4DB;

— Setup: Using the tables given in this Q2 and Q5

/* Assumptions:

1. A student won’t take a class many times, since it’s within a term.

2. Professors who teach a class always share the same # of students. i.e. no seperated sessions, etc.

3. hourly_rate = 50

*/

CREATE TABLE Enrollment (

SID CHAR(3) NOT NULL,

ClassName CHAR(11) NOT NULL,

Grade CHAR(2),

PRIMARY KEY (SID, ClassName)

);

INSERT INTO Enrollment VALUES (‘123’, ‘Processing’, ‘A’); INSERT INTO Enrollment VALUES (‘123’, ‘Python’, ‘B’); INSERT INTO Enrollment VALUES (‘123’, ‘Scratch’, ‘B’); INSERT INTO Enrollment VALUES (‘662’, ‘Java’, ‘B’); INSERT INTO Enrollment VALUES (‘662’, ‘Python’, ‘A’); INSERT INTO Enrollment VALUES (‘662’, ‘JavaScript’, ‘A’); INSERT INTO Enrollment VALUES (‘662’, ‘Scratch’, ‘B’); INSERT INTO Enrollment VALUES (‘345’, ‘Scratch’, ‘A’); INSERT INTO Enrollment VALUES (‘345’, ‘JavaScript’, ‘B’); INSERT INTO Enrollment VALUES (‘345’, ‘Python’, ‘A’); INSERT INTO Enrollment VALUES (‘555’, ‘Python’, ‘B’); INSERT INTO Enrollment VALUES (‘555’, ‘JavaScript’, ‘B’); INSERT INTO Enrollment VALUES (‘213’, ‘JavaScript’, ‘A’);

CREATE TABLE Instructor ( Instructor CHAR(6) NOT NULL, Subject CHAR(11) NOT NULL, PRIMARY KEY (Instructor, Subject)

);

INSERT INTO Instructor VALUES (‘Aleph’, ‘Scratch’); INSERT INTO Instructor VALUES (‘Aleph’, ‘Java’); INSERT INTO Instructor VALUES (‘Aleph’, ‘Processing’); INSERT INTO Instructor VALUES (‘Bit’, ‘Python’); INSERT INTO Instructor VALUES (‘Bit’, ‘JavaScript’); INSERT INTO Instructor VALUES (‘Bit’, ‘Java’); INSERT INTO Instructor VALUES (‘CRC’, ‘Python’); INSERT INTO Instructor VALUES (‘CRC’, ‘JavaScript’); INSERT INTO Instructor VALUES (‘Dat’, ‘Scratch’); INSERT INTO Instructor VALUES (‘Dat’, ‘Python’); INSERT INTO Instructor VALUES (‘Dat’, ‘JavaScript’); INSERT INTO Instructor VALUES (‘Emscr’, ‘Scratch’); INSERT INTO Instructor VALUES (‘Emscr’, ‘Processing’); INSERT INTO Instructor VALUES (‘Emscr’, ‘JavaScript’); INSERT INTO Instructor VALUES (‘Emscr’, ‘Python’);

— Soulution

/* Step 1: Calculate number of students in each class */ CREATE VIEW ClassCount AS

SELECT ClassName, COUNT(SID) AS Total

FROM Enrollment

GROUP BY ClassName;

SELECT * FROM ClassCount;

/* Step 2: Map professor to # of students in each class */

CREATE VIEW Instructor_ClassCount AS

SELECT I.Instructor AS Instructor,

I.Subject AS Subject,

C.Total AS Total

FROM Instructor I

JOIN ClassCount C

ON I.Subject = C.ClassName;

SELECT * FROM Instructor_ClassCount;

/* Step 3: Calculate bounus for each professor */

CREATE VIEW InstructorBonus AS

SELECT Instructor, SUM(Total) AS sum_of_class_counts FROM Instructor_ClassCount GROUP BY Instructor;

SELECT * FROM InstructorBonus;

/* Step 4: Calculate MAX bounus */

SELECT MAX(50 * sum_of_class_counts * 0.1) AS “Highest Bonus” FROM InstructorBonus;

/* To wrap up everything, this is the final solution */ /* Again, assume hourly_rate = 50 */

SELECT MAX(50 * sum_of_class_counts * 0.1) AS “Highest Bonus” FROM (

SELECT Instructor, SUM(Total) AS sum_of_class_counts FROM (

SELECT I.Instructor AS Instructor,

I.Subject AS Subject,

C.Total AS Total

FROM Instructor I

JOIN (

SELECT ClassName, COUNT(SID) AS Total

FROM Enrollment

GROUP BY ClassName

) C

ON I.Subject = C.ClassName

• AS Instructor_ClassCount GROUP BY Instructor

) AS InstructorBonus;

— Clean up option

DROP DATABASE IF EXISTS HW2Q4DB;

Q5

Grading this one needs more effort than Q1-Q3.

• To get full points, an answer must show the correct SQL query statement and explain what the query does. Otherwise, partially wrong.

• “Can hardcode the subjects just for submission purposes, but your query should work for ANY such table!” If they hardcode the subject names in the query, it’s partially wrong, because if the # of required classes != 3, the query doesn’t work. (See Piazza @80).

• Bonus points.

• • “+1 extra point if an additional solution is in a very different way!” +2 points at most, if they come up with more than 2 different ways. The total score of this

HW2 never exceeds 6 pts.

• • If Q5.sql is partially correct, but Q5_v2.sql is fully correct, one should get 2 (for Q5_v2) + 0.5(for Q5) = 2.5 pts. You know what this means : )

• • • ‘Very different’ means – the approaches do have to be totally distinct, eg. you can’t use NOT to invert an existing solution, or use IN() instead of OR, etc. In other words, it means something semantically different instead of syntactically different only. If a query is not ‘different’, it deserves no points.

  • Again, a fully correct answer should contain only A Query statement, not multiple query statements.

• Setup

DROP DATABASE IF EXISTS HW2Q5DB;

CREATE DATABASE HW2Q5DB;

USE HW2Q5DB;

CREATE TABLE Instructor ( Instructor CHAR(6) NOT NULL, Subject CHAR(11) NOT NULL, PRIMARY KEY (Instructor, Subject)

);

INSERT INTO Instructor VALUES (‘Aleph’, ‘Scratch’); INSERT INTO Instructor VALUES (‘Aleph’, ‘Java’); INSERT INTO Instructor VALUES (‘Aleph’, ‘Processing’); INSERT INTO Instructor VALUES (‘Bit’, ‘Python’); INSERT INTO Instructor VALUES (‘Bit’, ‘JavaScript’); INSERT INTO Instructor VALUES (‘Bit’, ‘Java’); INSERT INTO Instructor VALUES (‘CRC’, ‘Python’); INSERT INTO Instructor VALUES (‘CRC’, ‘JavaScript’); INSERT INTO Instructor VALUES (‘Dat’, ‘Scratch’); INSERT INTO Instructor VALUES (‘Dat’, ‘Python’); INSERT INTO Instructor VALUES (‘Dat’, ‘JavaScript’); INSERT INTO Instructor VALUES (‘Emscr’, ‘Scratch’); INSERT INTO Instructor VALUES (‘Emscr’, ‘Processing’); INSERT INTO Instructor VALUES (‘Emscr’, ‘JavaScript’); INSERT INTO Instructor VALUES (‘Emscr’, ‘Python’);

/* As mentioned in the HW2 specs

You can hardcode the subjects just for submission purposes, but your query should work for ANY such table!

*/

CREATE TABLE RequiredSubject (

Subject CHAR(11) UNIQUE

);

INSERT INTO RequiredSubject VALUES (‘JavaScript’); INSERT INTO RequiredSubject VALUES (‘Scratch’); INSERT INTO RequiredSubject VALUES (‘Python’);

— Solution 1

SELECT Instructor

FROM Instructor I, RequiredSubject R

WHERE I.Subject = R.Subject

/* WHAT THE QUERY DOES:

Find out Instructors who can teach RequiredSubjects Since it’s a many-to-many relation,

in the result up to here, one row shows a Instructor-to-Subject

mapping.

*/

GROUP BY Instructor

HAVING COUNT(*) = (SELECT COUNT(*) FROM RequiredSubject);

/* WHAT THE QUERY DOES:

Aggregate Instructors with restricting condition that

the # of Subject a Instructor can teach should equal to the total #

of RequiredSubjects.

*/

/* Many alternative answers will be simliar to this one.

e.g. to use JOIN is an acceptable different answer.

SELECT Instructor

FROM Instructor I

JOIN RequiredSubject R

ON I.Subject = R.Subject

GROUP BY Instructor

HAVING COUNT(*) = (SELECT COUNT(*) FROM RequiredSubject);

*/

— Solution 2

/* To use a Procedure, then the hard-code RequiredSubject Table is not

necessary */

DELIMITER $$

FROM Instructor

WHERE FIND_IN_SET(Subject, requiredSubjects)

/* WHAT THE QUERY DOES:

if Subject appears in requiredSubjects list, then output who teaches this Subject

*/

GROUP BY Instructor

HAVING COUNT(*) = LENGTH(requiredSubjects)

• LENGTH( REPLACE ( requiredSubjects, “,”, “”) ) + 1;

/* WHAT THE QUERY DOES:

A trick to get the total # of RequiredSubjects =

• • of “,” occurrences + 1

*/

END $$

DELIMITER ;

call searchQualifiedInstructor(‘JavaScript,Scratch,Python’); /* WHAT THE QUERY DOES:

Just call the Procedure to get the result

*/

— Clean up option

DROP DATABASE IF EXISTS HW2Q5DB;