Practical/Practical Exam/SQL/S5 - SQL Queries.md

@@ -21,8 +21,8 @@ employees who work for 'TechM' and earn more than $10,000.
 
 ## Creating the database
 ```sql
-CREATE DATABASE Companies1;
-USE Companies1;
+CREATE DATABASE Companies;
+USE Companies;
 
 ```
 

Practical/Practical Exam/SQL/S6 - SQL Queries.md

@@ -1,149 +0,0 @@
-# S6 - SQL Queries (in MySQL)
-
-**Problem Statement:**
-Consider following Relation
-Employee(emp_id,employee_name,street,city)
-Works(employee_name,company_name,salary)
-Company(company_name,city)
-Manages(employee_name,manager_name)
-Create above tables with appropriate constraints like primary key,
-foreign key, not null etc.
-1. Change the city of employee working with InfoSys to ‘Bangalore’
-2. Find the names of all employees who earn more than the average
-salary of all employees of their company. Assume that all people
-work for at most one company.
-3. Find the names, street address, and cities of residence for all
-employees who work for 'TechM' and earn more than $10,000.
-4. Change name of table Manages to Management.
-5. Create Simple and Unique index on employee table.
-6. Display index Information
-
----
-
-## Creating the database
-```sql
-CREATE DATABASE Companies2;
-USE Companies2;
-
-```
-
-## Creating tables:
-
-```sql
-CREATE TABLE Employee (
-  emp_id INT UNIQUE NOT NULL, -- can be set to auto increment using AUTO_INCREMENT
-  employee_name VARCHAR(255),
-  street VARCHAR(255),
-  city VARCHAR(255),
-  PRIMARY KEY (employee_name)
-);
-
-CREATE TABLE Works (
-  employee_name VARCHAR(255),
-  company_name VARCHAR(255),
-  salary INT -- use FLOAT if you are feeling fancy and pay your employees in Paise
-);
-
-CREATE TABLE Company (
-  company_name VARCHAR(255),
-  city VARCHAR(255),
-  PRIMARY KEY (company_name)
-);
-
-CREATE TABLE Manages (
-  employee_name VARCHAR(255),
-  manager_name VARCHAR(255)
-);
-
-```
-
-## Declaring foreign keys
-
-```sql
-ALTER TABLE Works ADD FOREIGN KEY (employee_name) REFERENCES Employee (employee_name);
-ALTER TABLE Works ADD FOREIGN KEY (company_name) REFERENCES Company (company_name);
-ALTER TABLE Manages ADD FOREIGN KEY (employee_name) REFERENCES Employee (employee_name);
-
-
-```
-
-## Inserting data
-
-```sql
-INSERT INTO Employee VALUES
-(1, 'Mehul', 'Street 42', 'Pune'),
-(2, 'Himanshu', 'Street 74', 'Mumbai'),
-(3, 'Gundeti', 'Street 14', 'Pune'),
-(4, 'Salvi', 'Street 38', 'Pune'),
-(5, 'Afan', 'Steet 98', 'Pune'),
-(6, 'Jambo', 'Street 23', 'Mumbai');
-
-INSERT INTO Company VALUES
-('TCS', 'Pune'),
-('Infosys', 'Mumbai'),
-('TechM', 'Pune'),
-('MEPA', 'Pune');
-
-INSERT INTO Works VALUES
-('Mehul', 'MEPA', 15000),
-('Himanshu', 'TCS', 25000),
-('Gundeti', 'TCS', 9000),
-('Salvi', 'TechM', 8000),
-('Afan', 'Infosys', 13000),
-('Jambo', 'MEPA', 28000);
-
-
-INSERT INTO Manages VALUES
-('Mehul', 'Kalas'),
-('Himanshu', 'Kshitij'),
-('Gundeti', 'Macho'),
-('Salvi', 'Kshitij'),
-('Afan', 'Kalas'),
-('Jambo', 'Macho');
-
-```
-
-## Queries
-
-1. Change the city of employee working with InfoSys to ‘Bangalore’
-```sql
-UPDATE Company SET city = "Bangalore" WHERE company_name = "Infosys";
-
-```
-
-2. Find the names of all employees who earn more than the average salary of all employees of their company. Assume that all people work for at most one company.
-```sql
-SELECT employee_name, salary, company_name FROM Works as W WHERE salary > (SELECT AVG(salary) FROM Works WHERE company_name = W.company_name);
-
-```
-
-3. Find the names, street address, and cities of residence for all employees who work for 'TechM' and earn more than $10,000.
-```sql
-SELECT Employee.employee_name, street, city FROM Employee INNER JOIN Works ON Employee.employee_name = Works.employee_name WHERE salary > 10000;
-
-```
-
-4. Change name of table Manages to Management.
-```sql
-ALTER TABLE Manages RENAME TO Management;
-SHOW TABLES;
-
-```
-
-5. Create Simple and Unique index on employee table.
-```sql
--- Simple Index
-CREATE INDEX emp_index ON Employee(employee_name);
-
--- Unique Index
-CREATE UNIQUE INDEX emp_uniqueIndex ON Employee(emp_id);
-
-```
-
-6. Display index Information
-```sql
-SHOW INDEX FROM Employee;
-
-```
-
----