[SQL-156] - Storing Data in MySQL as JSON

mysql-queries/mysql-store-json/creating-table-example.sql

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+CREATE TABLE Departments ( 
+    id INT PRIMARY KEY NOT NULL, 
+    name VARCHAR(50) NOT NULL, 
+    info JSON 
+);

mysql-queries/mysql-store-json/inserting-json-data.sql

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+INSERT INTO Departments (id, name, info) VALUES 
+(1, 'Computer Science', '{"head": "Dr. Smith", "location": "Building 1", "courses": ["Algorithms", "Data Structures"]}');
+
+INSERT INTO Departments (id, name, info) VALUES 
+(2, 'Mathematics', JSON_OBJECT('head', 'Dr. Johnson', 'location', 'Building 2', 'courses', JSON_ARRAY('Calculus', 'Linear Algebra')));
+
+INSERT INTO Departments (id, name, info) VALUES 
+(3, 'Physics', JSON_OBJECT('head', 'Dr. Clark', 'location', 'Building 3', 'courses', JSON_ARRAY('Quantum Mechanics', 'Thermodynamics')));
+
+SET @head = 'Dr. Miller'; 
+SET @location = 'Building 4'; 
+SET @courses = JSON_ARRAY('Classical Mechanics', 'Electromagnetism'); 
+INSERT INTO Departments (id, name, info) VALUES 
+(4, 'Engineering', JSON_OBJECT('head', @head, 'location', @location, 'courses', @courses));

mysql-queries/mysql-store-json/retrieving-json-data.sql

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+SELECT id, name, info FROM Departments;
+
+SELECT id, name, JSON_EXTRACT(info, '$.head') AS head FROM Departments;
+
+SELECT id, name, info->'$.head' AS head FROM Departments;

mysql-queries/mysql-store-json/updating-json-data.sql

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+UPDATE Departments 
+SET info = JSON_SET(info, '$.head', 'Dr. Parker') 
+WHERE id = 1;
+SELECT id, name, info FROM Departments WHERE id = 1;
+
+UPDATE Departments 
+SET info = JSON_REPLACE(info, '$.location', 'Building 2A')
+WHERE id = 2;
+SELECT id, name, info FROM Departments WHERE id = 2;
+
+UPDATE Departments 
+SET info = JSON_REPLACE(info, '$.dean', 'Dr. Brown') 
+WHERE id = 2;
+SELECT id, name, info FROM Departments WHERE id = 2;
+
+UPDATE Departments 
+SET info = JSON_SET(info, '$.dean', 'Dr. Brown') 
+WHERE id = 2;
+SELECT id, name, info FROM Departments WHERE id = 2;
+
+UPDATE Departments 
+SET info = JSON_REMOVE(info, '$.courses')
+WHERE id = 3;
+SELECT id, name, info FROM Departments WHERE id = 3;
+
+UPDATE Departments
+SET info = JSON_ARRAY_APPEND(info, '$.courses', 'Artificial Intelligence') 
+WHERE id = 4;
+SELECT id, name, info FROM Departments WHERE id = 4;