138 lines
3.4 KiB
Markdown
138 lines
3.4 KiB
Markdown
# S5 - SQL Queries (in MySQL)
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**Problem Statement:**
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Consider following Relation
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Employee(emp_id,employee_name,street,city)
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Works(employee_name,company_name,salary)
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Company(company_name,city)
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Manages(employee_name,manager_name)
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Create above tables with appropriate constraints like primary key,
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foreign key, not null etc.
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1. Find the names of all employees who work for ‘TCS’.
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2. Find the names and company names of all employees sorted in
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ascending order of company name and descending order of employee
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names of that company.
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3. Change the city of employee working with InfoSys to ‘Bangalore’
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4. Find the names, street address, and cities of residence for all
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employees who work for 'TechM' and earn more than $10,000.
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5. Add Column Asset to Company table.
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---
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## Creating the database
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```sql
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CREATE DATABASE Companies1;
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USE Companies1;
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```
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## Creating tables:
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```sql
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CREATE TABLE Employee (
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emp_id INT UNIQUE NOT NULL, -- can be set to auto increment using AUTO_INCREMENT
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employee_name VARCHAR(255),
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street VARCHAR(255),
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city VARCHAR(255),
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PRIMARY KEY (employee_name)
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);
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CREATE TABLE Works (
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employee_name VARCHAR(255),
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company_name VARCHAR(255),
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salary INT -- use FLOAT if you are feeling fancy and pay your employees in Paise
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);
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CREATE TABLE Company (
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company_name VARCHAR(255),
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city VARCHAR(255),
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PRIMARY KEY (company_name)
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);
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CREATE TABLE Manages (
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employee_name VARCHAR(255),
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manager_name VARCHAR(255)
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);
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```
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## Declaring foreign keys
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```sql
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ALTER TABLE Works ADD FOREIGN KEY (employee_name) REFERENCES Employee (employee_name);
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ALTER TABLE Works ADD FOREIGN KEY (company_name) REFERENCES Company (company_name);
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ALTER TABLE Manages ADD FOREIGN KEY (employee_name) REFERENCES Employee (employee_name);
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```
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## Inserting data
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```sql
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INSERT INTO Employee VALUES
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(1, 'Mehul', 'Street 42', 'Pune'),
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(2, 'Himanshu', 'Street 74', 'Mumbai'),
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(3, 'Gundeti', 'Street 14', 'Pune'),
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(4, 'Salvi', 'Street 38', 'Pune'),
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(5, 'Afan', 'Steet 98', 'Pune');
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INSERT INTO Company VALUES
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('TCS', 'Pune'),
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('Infosys', 'Mumbai'),
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('TechM', 'Pune'),
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('MEPA', 'Pune');
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INSERT INTO Works VALUES
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('Mehul', 'MEPA', 15000),
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('Himanshu', 'TCS', 25000),
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('Gundeti', 'TCS', 21500),
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('Salvi', 'TechM', 11000),
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('Afan', 'Infosys', 13000);
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INSERT INTO Manages VALUES
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('Mehul', 'Kalas'),
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('Himanshu', 'Kshitij'),
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('Gundeti', 'Macho'),
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('Salvi', 'Kshitij'),
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('Afan', 'Kalas');
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```
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## Queries
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1. Find the names of all employees who work for ‘TCS’.
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```sql
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SELECT employee_name FROM Works WHERE company_name = "TCS";
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```
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2. Find the names and company names of all employees sorted in ascending order of company name and descending order of employee names of that company.
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```sql
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SELECT company_name, employee_name FROM Works ORDER BY company_name ASC, employee_name DESC;
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```
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3. Change the city of employee working with InfoSys to ‘Bangalore’
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```sql
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update Employee set city = "Banglore" where employee_name in (select employee_name from Works where company_name = "Infosys");
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select * from Employee;
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```
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4. Find the names, street address, and cities of residence for all employees who work for 'TechM' and earn more than $10,000.
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```sql
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SELECT Employee.employee_name, Employee.street, Employee.city FROM Employee INNER JOIN Works ON Employee.employee_name = Works.employee_name WHERE Works.company_name = "TechM" AND Works.salary > 10000;
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```
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5. Add Column Asset to Company table.
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```sql
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ALTER TABLE Company ADD assets INT;
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DESCRIBE Company;
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```
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---
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