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[Assignment 1 - DB] Cong Son #4

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20 changes: 20 additions & 0 deletions docker-compose.yml
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version: 'latest'
services:
db:
image: mysql:latest
ports:
- 3306:3306
restart: always
volumes:
- db_data:/var/lib/mysql
- ./db/init.sql:/docker-entrypoint-initdb.d/init.sql
environment:
- MYSQL_ROOT_PASSWORD=123456
- MYSQL_DATABASE=BEK05
- MYSQL_USER=congson
- MYSQL_PASSWORD=123456
cap_add:
- SYS_NICE
volumes:
db_data:
driver: local
261 changes: 261 additions & 0 deletions sol-01.sql
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-----------------------------------------------------------------------------
/*
1. Giải bài tập trên leetcode:

https://leetcode.com/problems/capital-gainloss/description/ (gợi ý: sử dụng CASE)
*/
----------------------------------------------------------------------------

/*
My way
*/
SELECT stock_name, SUM(
IF(operation = 'Buy', -price, price)
) as capital_gain_loss
FROM Stocks
GROUP BY stock_name

/*
Submitted way
*/
SELECT stock_name, SUM(
CASE
WHEN operation = 'Buy' THEN -price
ELSE price
END
) AS capital_gain_loss
FROM Stocks
GROUP BY stock_name

-----------------------------------------------------------------------------
/*
https://leetcode.com/problems/count-salary-categories/
*/
-----------------------------------------------------------------------------

/*
My way, but change the structure of the table, if there is no account of a category, it will not be shown in the result table
*/
SELECT (
CASE
WHEN income < 20000 THEN "Low Salary"
WHEN income > 50000 THEN "High Salary"
ELSE "Average Salary"
END
) AS category, COUNT(1) AS accounts_count
FROM Accounts
GROUP BY category

/*
Submitted way
*/
SELECT "Low Salary" AS category, COUNT(1) AS accounts_count
FROM Accounts
WHERE income < 20000
UNION
SELECT "Average Salary" AS category, COUNT(1)
FROM Accounts
WHERE income >= 20000 AND income <= 50000
UNION
SELECT "High Salary" AS category, COUNT(1)
FROM Accounts
WHERE income > 50000

-----------------------------------------------------------------------------
/*
2. Bạn hãy viết một script để tạo các bản cho hệ thống với cấu trúc ở dưới:

- class
- professor: quan hệ one-many với class
- student: quan hệ many-many với class
- course: quan hệ one-many với class
- room: quan hệ one-one với class
*/
----------------------------------------------------------------------------

DROP DATABASE IF EXISTS `UNIVERSITY`;
CREATE DATABASE `UNIVERSITY`;
USE `UNIVERSITY`;
CREATE TABLE IF NOT EXISTS PROFESSOR (
PROF_ID INT PRIMARY KEY AUTO_INCREMENT,
PROF_LNAME VARCHAR(50) NOT NULL,
PROF_FNAME VARCHAR(50) NOT NULL
)

CREATE TABLE IF NOT EXISTS STUDENT (
STUD_ID INT PRIMARY KEY AUTO_INCREMENT,
STUD_LNAME VARCHAR(50) NOT NULL,
STUD_FNAME VARCHAR(50) NOT NULL,
STUD_STREET VARCHAR(255) NOT NULL,
STUD_CITY VARCHAR(50) NOT NULL,
STUD_ZIP VARCHAR(10) NOT NULL
)

CREATE TABLE IF NOT EXISTS COURSE (
COURSE_ID INT PRIMARY KEY AUTO_INCREMENT,
COURSE_NAME VARCHAR(255) NOT NULL
)

CREATE TABLE IF NOT EXISTS CLASS (
CLASS_ID INT PRIMARY KEY AUTO_INCREMENT,
CLASS_NAME VARCHAR(255) NOT NULL,
PROF_ID INT NOT NULL,
COURSE_ID INT NOT NULL,
ROOM_ID INT NOT NULL,
FOREIGN KEY (COURSE_ID) REFERENCES COURSE(COURSE_ID),
FOREIGN KEY (PROF_ID) REFERENCES PROFESSOR(PROF_ID),
FOREIGN KEY (ROOM_ID) REFERENCES ROOM(ROOM_ID)
)

CREATE TABLE IF NOT EXISTS ROOM (
ROOM_ID INT PRIMARY KEY AUTO_INCREMENT,
ROOM_LOC VARCHAR(50) NOT NULL,
ROOM_CAP VARCHAR(50) NOT NULL,
CLASS_ID INT NULL,
FOREIGN KEY (CLASS_ID) REFERENCES CLASS(CLASS_ID)
)

CREATE TABLE IF NOT EXISTS ENROLL (
STUD_ID INT NOT NULL,
CLASS_ID INT NOT NULL,
GRADE VARCHAR(3) NOT NULL,
FOREIGN KEY (STUD_ID) REFERENCES STUDENT(STUD_ID),
FOREIGN KEY (CLASS_ID) REFERENCES CLASS(CLASS_ID),
PRIMARY KEY (STUD_ID, CLASS_ID)
)

-------------------------------------------------------------------------------
/*
3. Hãy viết câu query để tìm:
a. những cặp student-professor có dạy học nhau và số lớp mà họ có liên quan
b. những course (distinct) mà 1 professor cụ thể đang dạy
c. những course (distinct) mà 1 student cụ thể đang học
d. điểm số là A, B, C, D, E, F tương đương với 10, 8, 6, 4, 2, 0
e. điểm số trung bình của 1 học sinh cụ thể (quy ra lại theo chữ cái, và xếp loại học lực (weak nếu avg < 5, average nếu >=5 < 8, good nếu >=8 )
f. điểm số trung bình của các class (quy ra lại theo chữ cái)
g. điểm số trung bình của các course (quy ra lại theo chữ cái)
*/
--------------------------------------------------------------------------------

/* a. những cặp student-professor có dạy học nhau và số lớp mà họ có liên quan
*/
SELECT ENROLL.`STUD_ID`, PROFESSOR.`PROF_ID`, COUNT(1) AS COMMON_CLASS FROM `ENROLL`
INNER JOIN `CLASS` ON CLASS.`CLASS_ID` = ENROLL.`CLASS_ID`
INNER JOIN `PROFESSOR` ON PROFESSOR.`PROF_ID` = CLASS.`PROF_ID`
GROUP BY ENROLL.`STUD_ID`, PROFESSOR.`PROF_ID`;

/* b. những course (distinct) mà 1 professor cụ thể đang dạy
*/
SELECT DISTINCT PROF.`PROF_ID`, COURSE.`COURSE_ID` FROM (
SELECT PROF_ID
FROM `PROFESSOR`
WHERE PROFESSOR.`PROF_ID` = 1
) PROF
INNER JOIN `CLASS` ON CLASS.`PROF_ID` = PROF.`PROF_ID`
INNER JOIN `COURSE` ON COURSE.`COURSE_ID` = CLASS.`COURSE_ID`;

/* c. những course (distinct) mà 1 student cụ thể đang học
*/
SELECT DISTINCT STU.`STUD_ID`, CLASS.`COURSE_ID` FROM (
SELECT STUD_ID
FROM `STUDENT`
WHERE STUDENT.`STUD_ID` = 1
) STU
INNER JOIN `ENROLL` ON ENROLL.`STUD_ID` = STU.`STUD_ID`
INNER JOIN `CLASS` ON CLASS.`CLASS_ID` = ENROLL.`CLASS_ID`;

/* d. điểm số là A, B, C, D, E, F tương đương với 10, 8, 6, 4, 2, 0
*/
SELECT `STUD_ID`, `CLASS_ID`, (CASE
WHEN GRADE = 'F' THEN 0
WHEN GRADE = 'E' THEN 2
WHEN GRADE = 'D' THEN 4
WHEN GRADE = 'C' THEN 6
WHEN GRADE = 'B' THEN 8
ELSE 10
END) AS SCORE
FROM `ENROLL`;

/* e. điểm số trung bình của 1 học sinh cụ thể (quy ra lại theo chữ cái, và xếp loại học lực (weak nếu avg < 5, average nếu >=5 < 8, good nếu >=8 )
*/
SELECT `STUD_ID`, AVG_SCORE, (
CASE
WHEN AVG_SCORE < 5 THEN 'weak'
WHEN AVG_SCORE >= 8 THEN 'good'
ELSE 'average'
END
) AS `TYPE`
FROM (
SELECT `STUD_ID`, AVG(
CASE
WHEN GRADE = 'F' THEN 0
WHEN GRADE = 'E' THEN 2
WHEN GRADE = 'D' THEN 4
WHEN GRADE = 'C' THEN 6
WHEN GRADE = 'B' THEN 8
ELSE 10
END
) AS AVG_SCORE
FROM `ENROLL`
WHERE `STUD_ID` = 1
GROUP BY `STUD_ID`
) STU;

/* f. điểm số trung bình của các class (quy ra lại theo chữ cái)
*/
SELECT `CLASS_ID`, `CLASS_NAME`, AVG_SCORE, (
CASE
WHEN AVG_SCORE < 2 THEN 'F'
WHEN AVG_SCORE >= 2 AND AVG_SCORE < 4 THEN 'E'
WHEN AVG_SCORE >= 4 AND AVG_SCORE < 6 THEN 'D'
WHEN AVG_SCORE >= 6 AND AVG_SCORE < 8 THEN 'C'
WHEN AVG_SCORE >= 8 AND AVG_SCORE < 10 THEN 'B'
ELSE 'A'
END
) AS AVG_GRADE
FROM (
SELECT CLASS.`CLASS_ID`, CLASS.`CLASS_NAME`, AVG(
CASE
WHEN ENROLL.`GRADE` = 'F' THEN 0
WHEN ENROLL.`GRADE` = 'E' THEN 2
WHEN ENROLL.`GRADE` = 'D' THEN 4
WHEN ENROLL.`GRADE` = 'C' THEN 6
WHEN ENROLL.`GRADE` = 'B' THEN 8
ELSE 10
END
) AS AVG_SCORE
FROM `CLASS`
INNER JOIN `ENROLL` ON ENROLL.`CLASS_ID` = CLASS.`CLASS_ID`
GROUP BY CLASS.`CLASS_ID`
) TB;

/* điểm số trung bình của các course (quy ra lại theo chữ cái)
*/
SELECT `COURSE_ID`, `COURSE_NAME`, AVG_SCORE, (
CASE
WHEN AVG_SCORE < 2 THEN 'F'
WHEN AVG_SCORE >= 2 AND AVG_SCORE < 4 THEN 'E'
WHEN AVG_SCORE >= 4 AND AVG_SCORE < 6 THEN 'D'
WHEN AVG_SCORE >= 6 AND AVG_SCORE < 8 THEN 'C'
WHEN AVG_SCORE >= 8 AND AVG_SCORE < 10 THEN 'B'
ELSE 'A'
END
) AS AVG_GRADE
FROM (
SELECT COURSE.`COURSE_ID`, COURSE.`COURSE_NAME`, AVG(
CASE
WHEN ENROLL.`GRADE` = 'F' THEN 0
WHEN ENROLL.`GRADE` = 'E' THEN 2
WHEN ENROLL.`GRADE` = 'D' THEN 4
WHEN ENROLL.`GRADE` = 'C' THEN 6
WHEN ENROLL.`GRADE` = 'B' THEN 8
ELSE 10
END
) AS AVG_SCORE
FROM `COURSE`
INNER JOIN `CLASS` ON CLASS.`COURSE_ID` = COURSE.`COURSE_ID`
INNER JOIN `ENROLL` ON ENROLL.`CLASS_ID` = CLASS.`CLASS_ID`
GROUP BY COURSE.`COURSE_ID`
) TB;