Assignment_1_tuho

capital_gain_loss.sql

@@ -0,0 +1,7 @@
+SELECT 
+    stock_name, 
+    SUM(CASE WHEN operation = 'Sell' THEN price ELSE -price END) AS capital_gain_loss
+FROM 
+    stocks
+GROUP BY 
+    stock_name;

count_salary_categories.sql

@@ -0,0 +1,49 @@
+
+-- cách 1 (cần insert thêm 3 row data, chạy đc ở dev k chạy đc ở leetcode)
+
+insert into Accounts(account_id, income, category) values
+(-1, 0,null),
+(-1, 20001, null),
+(-1, 50001, null);
+
+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 
+CASE 
+    WHEN income < 20000 THEN 'Low Salary'
+    WHEN income > 50000 THEN 'High Salary'
+    ELSE 'Average Salary'
+  END;
+
+-- cách 2(chạy đươc trên leetcode k cần insert )
+
+SELECT
+    c.category
+    , COUNT(a.income) AS accounts_count
+FROM
+    (
+        SELECT 'Low Salary' AS category
+        UNION ALL
+        SELECT 'Average Salary' AS category
+        UNION ALL
+        SELECT 'High Salary' AS category
+    ) c
+LEFT JOIN
+    accounts a ON (
+        (c.category = 'Low Salary' AND a.income < 20000) OR
+        (c.category = 'Average Salary' AND a.income >= 20000 AND a.income <= 50000) OR
+        (c.category = 'High Salary' AND a.income > 50000)
+    )
+GROUP BY
+    c.category
+
+
+
+
+

mission2.sql

@@ -0,0 +1,165 @@
+
+-- create tables  
+CREATE TABLE IF NOT EXISTS professor (
+  id INT PRIMARY KEY,
+  firstName VARCHAR(50) NOT NULL,
+  lastName VARCHAR(50) NOT NULL
+
+);
+
+CREATE TABLE IF NOT EXISTS course (
+  id INT PRIMARY KEY,
+  name VARCHAR(50) NOT NULL
+);
+
+
+CREATE TABLE IF NOT EXISTS student (
+  id INT PRIMARY KEY,
+  firstName VARCHAR(50) NOT NULL,
+  lastName VARCHAR(50) NOT NULL,
+  street varchar(255),
+  city varchar(50),
+  zip varchar(10)
+);
+
+CREATE TABLE IF NOT EXISTS class (
+  id INT PRIMARY KEY,
+  name VARCHAR(50) NOT NULL,
+  professor_id INT NOT NULL,
+  course_id INT NOT NULL,
+  room_id INT NOT NULL,
+  FOREIGN KEY (professor_id) REFERENCES professor(id),
+  FOREIGN KEY (course_id) REFERENCES course(id)
+  -- FOREIGN KEY (room_id) REFERENCES room(id)  
+);
+
+
+CREATE TABLE IF NOT EXISTS room (
+  id INT PRIMARY KEY,
+  name VARCHAR(50) NOT NULL,
+  loc varchar(50),
+  cap varchar(50),
+  class_id INT UNIQUE NOT NULL,
+  FOREIGN KEY (class_id) REFERENCES class(id)
+);
+
+-- vì đây là quan hệ n-n nên không khai báo fk trong cả 2 đc vì chạy k đc, nên bỏ 1 trong 2 tạo bảng rồi mới add fk  
+alter table class 
+add constraint class_room
+foreign key(room_id) references room(id);
+
+
+CREATE TABLE IF NOT EXISTS class_student (
+  class_id INT NOT NULL,
+  student_id INT NOT NULL,
+
+  grade varchar(3),
+  PRIMARY KEY (class_id, student_id),
+  FOREIGN KEY (class_id) REFERENCES class(id),
+  FOREIGN KEY (student_id) REFERENCES student(id)
+);
+
+-- những cặp student-professor có dạy học nhau và số lớp mà họ có liên quan 
+-- tạo thêm table class_professor để tối ưu query 
+
+CREATE TABLE IF NOT EXISTS class_professor (
+  class_id INT NOT NULL,
+  professor_id INT NOT NULL,
+  PRIMARY KEY (class_id, professor_id),
+  FOREIGN KEY (class_id) REFERENCES class(id),
+  FOREIGN KEY (professor_id) REFERENCES professor(id)
+);
+
+
+select t.id as student_id, t.lastName as student_name,
+	   p.id as professor_id, p.lastName as professor_name,
+       count(distinct(class_id)) as numClass
+from student as t
+join class_student as c_s
+on t.id= c_s.student_id
+join class_professor as c_f
+on c_s.class_id= c_f.class_id
+join professor as p 
+on c_f.professor_id= p.id
+group by student_id,professor_id;
+
+
+-- những course (distinct) mà 1 professor cụ thể đang dạy
+
+select p.id, c.course_id
+
+from professor as p
+join class as c on p.id= c.professor_id
+group by p.id, c.course_id;
+
+
+-- những course (distinct) mà 1 student cụ thể đang học 
+SELECT s.id AS student_id, c.name AS course_name
+FROM class c
+JOIN class_student cs ON c.id = cs.class_id
+JOIN student s ON cs.student_id = s.id
+GROUP BY student_id, course_name;
+
+
+-- điểm số là A, B, C, D, E, F tương đương với 10, 8, 6, 4, 2, 0 
+-- 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 cs.student_id,
+	   AVG(CASE
+             WHEN cs.grade =10 THEN 'A'
+             WHEN cs.grade >= 8 THEN 'B'
+             WHEN cs.grade >= 6 THEN 'C'
+             WHEN cs.grade >= 4 THEN 'D'
+             WHEN cs.grade >= 2 THEN 'E'
+             ELSE 'F'
+           END) AS letter_grade,
+       CASE
+         WHEN AVG(cs.grade) < 5 THEN 'weak'
+         WHEN AVG(cs.grade) < 8 THEN 'average'
+         ELSE 'good'
+       END AS performance_level
+FROM class_student cs
+group by cs.student_id;
+
+--  điểm số trung bình của các class (quy ra lại theo chữ cái) 
+
+SELECT c.id AS class_id, c.name AS class_name,
+       AVG(CASE
+			 WHEN cs.grade =10 THEN 'A'
+             WHEN cs.grade >= 8 THEN 'B'
+             WHEN cs.grade >= 6 THEN 'C'
+             WHEN cs.grade >= 4 THEN 'D'
+             WHEN cs.grade >= 2 THEN 'E'
+             ELSE 'F'
+           END) AS letter_grade
+FROM class c
+JOIN class_student cs ON c.id = cs.class_id
+GROUP BY class_id, class_name;
+
+-- điểm số trung bình của các course (quy ra lại theo chữ cái)
+
+SELECT c.course_id AS course_id,
+       AVG(CASE
+			 WHEN cs.grade =10 THEN 'A'
+             WHEN cs.grade >= 8 THEN 'B'
+             WHEN cs.grade >= 6 THEN 'C'
+             WHEN cs.grade >= 4 THEN 'D'
+             WHEN cs.grade >= 2 THEN 'E'
+             ELSE 'F'
+           END) AS letter_grade
+FROM class c
+JOIN class_student cs ON c.id = cs.class_id
+GROUP BY course_id
+
+
+
+
+
+
+
+
+
+
+
+
+