Relational database concepts in MySQL - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to select all columns from the table named 'employees'.

MySQL

SELECT [1] FROM employees;

Drag options to blanks, or click blank then click option'

Acolumns

BALL

C*

Deverything

Attempts:

3 left

💡 Hint

Common Mistakes

Writing 'ALL' instead of '*'

Using words like 'columns' or 'everything' which are not valid SQL keywords

Using * selects all columns from the table.

2fill in blank

medium

Complete the code to filter employees whose age is greater than 30.

MySQL

SELECT * FROM employees WHERE age [1] 30;

Drag options to blanks, or click blank then click option'

A>

B<

C=

D<=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<' which means less than

Using '=' which means equal to

The '>' operator filters rows where age is greater than 30.

3fill in blank

hard

Fix the error in the code to count the number of employees.

MySQL

SELECT COUNT([1]) FROM employees;

Drag options to blanks, or click blank then click option'

Aage

B*

Cname

Dsalary

Attempts:

3 left

💡 Hint

Common Mistakes

Counting a specific column which may ignore NULL values

Using a column name instead of '*'

Using COUNT(*) counts all rows regardless of column values.

4fill in blank

hard

Fill both blanks to select distinct departments and order them alphabetically.

MySQL

SELECT [1] department FROM employees ORDER BY department [2];

Drag options to blanks, or click blank then click option'

ADISTINCT

BDESC

CASC

DALL

Attempts:

3 left

💡 Hint

Common Mistakes

Using ALL instead of DISTINCT

Ordering with DESC which sorts in reverse alphabetical order

DISTINCT selects unique departments, and ASC orders them from A to Z.

5fill in blank

hard

Fill all three blanks to create a new table named 'departments' with columns 'id' as integer primary key and 'name' as text.

MySQL

CREATE TABLE [1] ([2] INT PRIMARY KEY, [3] TEXT);

Drag options to blanks, or click blank then click option'

Adepartments

Bid

Cname

Demployees

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong table name like 'employees'

Swapping column names or missing PRIMARY KEY

This creates a table called 'departments' with 'id' as an integer primary key and 'name' as a text column.

Practice

1. What is the main purpose of a primary key in a relational database table?

easy

A. To speed up database backups

B. To store large text data

C. To link two tables together

D. To uniquely identify each record in the table

Solution

1. Step 1: Understand the role of primary keys

   A primary key is a column or set of columns that uniquely identifies each row in a table.

2. Step 2: Compare options with the definition

   Only To uniquely identify each record in the table describes this unique identification role correctly.

3. Final Answer:

   To uniquely identify each record in the table -> Option D

4. Quick Check:

   Primary key = unique record ID [OK]

Hint: Primary key means unique ID for each row [OK]

Common Mistakes:

• Confusing primary key with foreign key
• Thinking primary key stores large data
• Assuming primary key speeds backups

2. Which of the following is the correct syntax to declare a primary key on the column id in a MySQL table?

easy

A. PRIMARY KEY (id);

B. KEY PRIMARY id;

C. PRIMARY KEY id;

D. PRIMARY id KEY;

Solution

1. Step 1: Recall MySQL primary key syntax

   In MySQL, the primary key is declared using PRIMARY KEY (column_name).

2. Step 2: Match syntax with options

   PRIMARY KEY (id); matches the correct syntax with parentheses around the column name.

3. Final Answer:

   PRIMARY KEY (id); -> Option A

4. Quick Check:

   Primary key syntax uses parentheses [OK]

Hint: Primary key syntax uses parentheses around column [OK]

Common Mistakes:

• Omitting parentheses around column name
• Swapping order of keywords
• Using incorrect keywords like KEY PRIMARY

3. Given two tables: Customers(id, name) and Orders(order_id, customer_id), which SQL query correctly retrieves all orders with the customer names?

medium

A. SELECT Orders.order_id, Customers.name FROM Orders JOIN Customers ON Orders.id = Customers.customer_id;

B. SELECT Orders.order_id, Customers.name FROM Orders JOIN Customers ON Orders.customer_id = Customers.id;

C. SELECT order_id, name FROM Orders, Customers WHERE Orders.id = Customers.id;

D. SELECT order_id, name FROM Orders LEFT JOIN Customers ON Orders.customer_id = Customers.order_id;

Solution

1. Step 1: Identify the correct join condition

   The foreign key customer_id in Orders links to id in Customers.

2. Step 2: Check each option's join condition

   SELECT Orders.order_id, Customers.name FROM Orders JOIN Customers ON Orders.customer_id = Customers.id; correctly joins on Orders.customer_id = Customers.id. Others use wrong columns or join types.

3. Final Answer:

   SELECT Orders.order_id, Customers.name FROM Orders JOIN Customers ON Orders.customer_id = Customers.id; -> Option B

4. Quick Check:

   Join on foreign key = primary key [OK]

Hint: Join tables on foreign key = primary key [OK]

Common Mistakes:

• Swapping join columns incorrectly
• Using WHERE instead of JOIN condition properly
• Mixing up LEFT JOIN with INNER JOIN

4. You wrote this SQL to create a foreign key: FOREIGN KEY (customer_id) REFERENCES Customers(customer_id). But it gives an error. What is the likely cause?

medium

A. You must declare foreign keys before primary keys

B. Foreign keys cannot reference columns in the same table

C. The referenced column customer_id does not exist or is not a primary key in Customers

D. Foreign keys must reference columns named 'id' only

Solution

1. Step 1: Understand foreign key requirements

   A foreign key must reference a column that exists and is a primary or unique key in the referenced table.

2. Step 2: Analyze the error cause

   If Customers has no customer_id column or it is not a key, the foreign key declaration fails.

3. Final Answer:

   The referenced column customer_id does not exist or is not a primary key in Customers -> Option C

4. Quick Check:

   Foreign key must reference existing primary key [OK]

Hint: Foreign key must reference existing primary key column [OK]

Common Mistakes:

• Referencing non-existent columns
• Assuming foreign keys must reference 'id' column
• Incorrect order of key declarations

5. You have two tables: Employees(emp_id, name) and Departments(dept_id, dept_name, manager_id). The manager_id in Departments references emp_id in Employees. Which of these statements is TRUE about this relationship?

hard

A. An employee can manage multiple departments, but a department can have only one manager

B. Each department must have exactly one manager who is an employee

C. A department can have multiple managers if multiple employees share the same emp_id

D. The manager_id column in Departments can be NULL only if emp_id is NULL in Employees

Solution

1. Step 1: Understand foreign key relationship

   The manager_id in Departments points to emp_id in Employees, linking each department to one employee as manager.

2. Step 2: Analyze relationship cardinality

   One department has one manager, but an employee can manage many departments (one-to-many).

3. Step 3: Evaluate options

   An employee can manage multiple departments, but a department can have only one manager correctly states this one-to-many relationship. Each department must have exactly one manager who is an employee is wrong because departments can have one manager, but not necessarily exactly one if NULL allowed. A department can have multiple managers if multiple employees share the same emp_id is wrong because emp_id is unique. The manager_id column in Departments can be NULL only if emp_id is NULL in Employees is incorrect because NULL in manager_id is independent of emp_id NULL.

4. Final Answer:

   An employee can manage multiple departments, but a department can have only one manager -> Option A

5. Quick Check:

   One-to-many foreign key = An employee can manage multiple departments, but a department can have only one manager [OK]

Hint: Foreign key one-to-many means one employee, many departments [OK]

Common Mistakes:

• Assuming multiple managers per department
• Confusing NULL constraints
• Thinking emp_id can be duplicated