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SQL Exercises/Employee management

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Relational Schema

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Exercises

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1. Select the last name of all employees.

Click to see solution
SELECT LastName FROM Employees;


2. Select the last name of all employees, without duplicates.

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SELECT DISTINCT LastName FROM Employees;


3. Select all the data of employees whose last name is "Smith".

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SELECT * FROM Employees WHERE LastName = 'Smith';


4. Select all the data of employees whose last name is "Smith" or "Doe".

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/* With OR */
SELECT * FROM Employees
  WHERE LastName = 'Smith' OR LastName = 'Doe';

/* With IN */
SELECT * FROM Employees
  WHERE LastName IN ('Smith' , 'Doe');


5. Select all the data of employees that work in department 14.

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SELECT * FROM Employees WHERE Department = 14;


6. Select all the data of employees that work in department 37 or department 77.

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/* With OR */
SELECT * FROM Employees
  WHERE Department = 37 OR Department = 77;

/* With IN */
SELECT * FROM Employees
  WHERE Department IN (37,77);


7. Select all the data of employees whose last name begins with an "S".

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SELECT * FROM Employees
  WHERE LastName LIKE 'S%';


8. Select the sum of all the departments' budgets.

Click to see solution
SELECT SUM(Budget) FROM Departments;


9. Select the number of employees in each department (you only need to show the department code and the number of employees).

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SELECT Department, COUNT(*)
  FROM Employees
  GROUP BY Department;


10. Select all the data of employees, including each employee's department's data.

Click to see solution
SELECT *
 FROM Employees E INNER JOIN Departments D
 ON E.Department = D.Code;


11. Select the name and last name of each employee, along with the name and budget of the employee's department.

Click to see solution
/* Without labels */
SELECT Employees.Name, LastName, Departments.Name AS DepartmentsName, Budget
  FROM Employees INNER JOIN Departments
  ON Employees.Department = Departments.Code;

/* With labels */
SELECT E.Name, LastName, D.Name AS DepartmentsName, Budget
  FROM Employees E INNER JOIN Departments D
  ON E.Department = D.Code;


12. Select the name and last name of employees working for departments with a budget greater than $60,000.

Click to see solution
/* Without subquery */
SELECT Employees.Name, LastName
  FROM Employees INNER JOIN Departments
  ON Employees.Department = Departments.Code
    AND Departments.Budget > 60000;

/* With subquery */
SELECT Name, LastName FROM Employees
  WHERE Department IN
  (SELECT Code FROM Departments WHERE Budget > 60000);


13. Select the departments with a budget larger than the average budget of all the departments.

Click to see solution
SELECT *
  FROM Departments
  WHERE Budget >
  (
    SELECT AVG(Budget)
    FROM Departments
  );


14. Select the names of departments with more than two employees.

Click to see solution
/*With subquery*/
SELECT D.Name FROM Departments D
  WHERE 2 < 
  (
   SELECT COUNT(*) 
     FROM Employees
     WHERE Department = D.Code
  );

/* With IN and subquery */
SELECT Name FROM Departments
  WHERE Code IN
  (
    SELECT Department
      FROM Employees
      GROUP BY Department
      HAVING COUNT(*) > 2
  );

/* With UNION. This assumes that no two departments have
   the same name */
SELECT Departments.Name
  FROM Employees INNER JOIN Departments
  ON Department = Code
  GROUP BY Departments.Name
  HAVING COUNT(*) > 2;


15. Select the name and last name of employees working for departments with second lowest budget.

Click to see solution
/* With subquery */
SELECT e.Name, e.LastName
FROM Employees e 
WHERE e.Department = (
       SELECT sub.Code 
       FROM (SELECT * FROM Departments d ORDER BY d.budget LIMIT 2) sub 
       ORDER BY budget DESC LIMIT 1);

/* With subquery */
SELECT Name, LastName 
FROM Employees 
WHERE Department IN (
  SELECT Code 
  FROM Departments 
  WHERE Budget = (
    SELECT TOP 1 Budget 
    FROM Departments 
    WHERE Budget IN (
      SELECT DISTINCT TOP 2 Budget 
      FROM Departments 
     ORDER BY Budget ASC
    ) 
    ORDER BY Budget DESC
  )
);


16. Add a new department called "Quality Assurance", with a budget of $40,000 and departmental code 11. Add an employee called "Mary Moore" in that department, with SSN 847-21-9811.

Click to see solution
INSERT INTO Departments
  VALUES ( 11 , 'Quality Assurance' , 40000);

INSERT INTO Employees
  VALUES ( '847219811' , 'Mary' , 'Moore' , 11);


/*Note: Quoting numbers in SQL works but is bad practice. SSN should not be quoted it is an integer.*/


17. Reduce the budget of all departments by 10%.

Click to see solution
UPDATE Departments SET Budget = Budget * 0.9;


18. Reassign all employees from the Research department (code 77) to the IT department (code 14).

Click to see solution
UPDATE Employees SET Department = 14 WHERE Department = 77;


19. Delete from the table all employees in the IT department (code 14).

Click to see solution
DELETE FROM Employees
  WHERE Department = 14;


20. Delete from the table all employees who work in departments with a budget greater than or equal to $60,000.

Click to see solution
DELETE FROM Employees
  WHERE Department IN
  (
    SELECT Code FROM Departments
      WHERE Budget >= 60000
  );


21. Delete from the table all employees.

Click to see solution
DELETE FROM Employees;

Table creation code

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 CREATE TABLE Departments (
   Code INTEGER PRIMARY KEY NOT NULL,
   Name VARCHAR NOT NULL ,
   Budget REAL NOT NULL 
 );
 
 CREATE TABLE Employees (
   SSN INTEGER PRIMARY KEY NOT NULL,
   Name TEXT NOT NULL ,
   LastName VARCHAR NOT NULL ,--since question 2 asks about removing duplicate - text must be converted if the answer is using distinct
   Department INTEGER NOT NULL , 
   CONSTRAINT fk_Departments_Code FOREIGN KEY(Department) 
   REFERENCES Departments(Code)
 );


Click to see MySQL syntax.
CREATE TABLE Departments (
  Code INTEGER PRIMARY KEY,
  Name varchar(255) NOT NULL ,
  Budget decimal NOT NULL 
);

CREATE TABLE Employees (
  SSN INTEGER PRIMARY KEY,
  Name varchar(255) NOT NULL ,
  LastName varchar(255) NOT NULL ,
  Department INTEGER NOT NULL , 
  foreign key (department) references Departments(Code) 
) ENGINE=INNODB;
Click to see Oracle syntax.
CREATE TABLE Departments (
  Code INT PRIMARY KEY NOT NULL,
  Name VARCHAR(100) NOT NULL ,
  Budget NUMBER NOT NULL 
);

CREATE TABLE Employees (
  SSN INT PRIMARY KEY NOT NULL,
  Name VARCHAR(30) NOT NULL ,
  LastName VARCHAR(30) NOT NULL ,
  Department INT NOT NULL , 
  CONSTRAINT fk_Departments_Code FOREIGN KEY(Department) REFERENCES Departments(Code)
);

Sample dataset

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INSERT INTO Departments(Code,Name,Budget) VALUES(14,'IT',65000);
INSERT INTO Departments(Code,Name,Budget) VALUES(37,'Accounting',15000);
INSERT INTO Departments(Code,Name,Budget) VALUES(59,'Human Resources',240000);
INSERT INTO Departments(Code,Name,Budget) VALUES(77,'Research',55000);

INSERT INTO Employees(SSN,Name,LastName,Department) VALUES('123234877','Michael','Rogers',14);
INSERT INTO Employees(SSN,Name,LastName,Department) VALUES('152934485','Anand','Manikutty',14);
INSERT INTO Employees(SSN,Name,LastName,Department) VALUES('222364883','Carol','Smith',37);
INSERT INTO Employees(SSN,Name,LastName,Department) VALUES('326587417','Joe','Stevens',37);
INSERT INTO Employees(SSN,Name,LastName,Department) VALUES('332154719','Mary-Anne','Foster',14);
INSERT INTO Employees(SSN,Name,LastName,Department) VALUES('332569843','George','O''Donnell',77);
INSERT INTO Employees(SSN,Name,LastName,Department) VALUES('546523478','John','Doe',59);
INSERT INTO Employees(SSN,Name,LastName,Department) VALUES('631231482','David','Smith',77);
INSERT INTO Employees(SSN,Name,LastName,Department) VALUES('654873219','Zacary','Efron',59);
INSERT INTO Employees(SSN,Name,LastName,Department) VALUES('745685214','Eric','Goldsmith',59);
INSERT INTO Employees(SSN,Name,LastName,Department) VALUES('845657245','Elizabeth','Doe',14);
INSERT INTO Employees(SSN,Name,LastName,Department) VALUES('845657246','Kumar','Swamy',14);
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