Subqueries vs JOINs comparison in MySQL - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to select all employees who work in the 'Sales' department using a subquery.

MySQL

SELECT name FROM employees WHERE department_id = (SELECT [1] FROM departments WHERE name = 'Sales');

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

Aemployee_id

Bname

Cdepartment_id

Did

Attempts:

3 left

💡 Hint

Common Mistakes

Using the department name instead of its id in the subquery.

Selecting a column that does not exist in the departments table.

The subquery must return the department's id to match with employees' department_id.

2fill in blank

medium

Complete the code to join employees with departments to get employee names and their department names.

MySQL

SELECT employees.name, departments.[1] FROM employees JOIN departments ON employees.department_id = departments.id;

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

Aname

Bdepartment_id

Cid

Demployee_id

Attempts:

3 left

💡 Hint

Common Mistakes

Selecting the department id instead of the name.

Using a column that belongs to employees instead of departments.

We want to select the name of the department to show alongside employee names.

3fill in blank

hard

Fix the error in the subquery to find employees with salary greater than the average salary.

MySQL

SELECT name FROM employees WHERE salary > (SELECT [1](salary) FROM employees);

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

ACOUNT

BSUM

CAVG

DMAX

Attempts:

3 left

💡 Hint

Common Mistakes

Using SUM instead of AVG.

Using COUNT which counts rows, not salary values.

The query needs the average salary, so the function AVG is correct.

4fill in blank

hard

Fill both blanks to write a JOIN that selects employee names and their manager names.

MySQL

SELECT e.name AS employee, m.[1] AS manager FROM employees e JOIN employees m ON e.[2] = m.id;

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

Aname

Bmanager_id

Cid

Ddepartment_id

Attempts:

3 left

💡 Hint

Common Mistakes

Selecting the wrong column for manager's name.

Using department_id instead of manager_id for the join.

The manager's name is selected, and employees' manager_id links to managers' id.

5fill in blank

hard

Fill all three blanks to write a subquery that finds departments with more than 5 employees.

MySQL

SELECT name FROM departments WHERE id IN (SELECT [1] FROM employees GROUP BY [2] HAVING COUNT([3]) > 5);

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

Adepartment_id

Bemployee_id

C*

Did

Attempts:

3 left

💡 Hint

Common Mistakes

Using employee_id instead of department_id in GROUP BY.

Counting department_id instead of all rows with *.

The subquery selects department_id from employees, groups by it, and counts all rows with * to find departments with more than 5 employees.

Practice

1. Which of the following best describes the main difference between a JOIN and a subquery in MySQL?

easy

A. JOIN runs inside another query; subqueries combine tables side-by-side.

B. JOIN is used only for filtering; subqueries are used only for combining tables.

C. JOIN combines tables side-by-side; subqueries run inside another query to filter or find values.

D. JOIN and subqueries are exactly the same in functionality and performance.

Solution

1. Step 1: Understand the role of JOIN

   A JOIN combines rows from two or more tables side-by-side based on a related column between them.

2. Step 2: Understand the role of subqueries

   A subquery is a query nested inside another query, often used to filter or find specific values before the main query runs.

3. Final Answer:

   JOIN combines tables side-by-side; subqueries run inside another query to filter or find values. -> Option C

4. Quick Check:

   JOIN = combine tables, Subquery = nested query [OK]

Hint: JOINs combine tables; subqueries nest queries inside [OK]

Common Mistakes:

• Confusing JOINs as nested queries
• Thinking subqueries combine tables side-by-side
• Believing JOINs and subqueries have identical performance always

2. Which of the following is the correct syntax to write a JOIN between tables employees and departments on the column department_id?

easy

A. SELECT * FROM employees JOIN departments USING employees.department_id;

B. SELECT * FROM employees JOIN departments WHERE employees.department_id = departments.department_id;

C. SELECT * FROM employees, departments ON employees.department_id = departments.department_id;

D. SELECT * FROM employees JOIN departments ON employees.department_id = departments.department_id;

Solution

1. Step 1: Identify correct JOIN syntax

   The correct JOIN syntax uses the ON keyword to specify the matching columns between tables.

2. Step 2: Check each option

   SELECT * FROM employees JOIN departments ON employees.department_id = departments.department_id; uses ON correctly. SELECT * FROM employees JOIN departments WHERE employees.department_id = departments.department_id; incorrectly uses WHERE instead of ON. SELECT * FROM employees, departments ON employees.department_id = departments.department_id; uses comma join but incorrectly adds ON. SELECT * FROM employees JOIN departments USING employees.department_id; misuses USING with full column name.

3. Final Answer:

   SELECT * FROM employees JOIN departments ON employees.department_id = departments.department_id; -> Option D

4. Quick Check:

   JOIN requires ON for matching columns [OK]

Hint: Use ON to specify join condition, not WHERE [OK]

Common Mistakes:

• Using WHERE instead of ON for JOIN condition
• Using comma join with ON clause
• Incorrect USING syntax with full column names

3. Given the tables orders(order_id, customer_id, amount) and customers(customer_id, name), what will the following query return?

SELECT name FROM customers WHERE customer_id IN (SELECT customer_id FROM orders WHERE amount > 100);

medium

A. Names of customers who have orders with amount greater than 100.

B. All customer names regardless of orders.

C. Names of customers who have no orders.

D. Syntax error due to subquery.

Solution

1. Step 1: Analyze the subquery

   The subquery selects customer_id from orders where amount is greater than 100.

2. Step 2: Analyze the main query

   The main query selects names from customers where customer_id is in the list returned by the subquery.

3. Final Answer:

   Names of customers who have orders with amount greater than 100. -> Option A

4. Quick Check:

   Subquery filters customers by order amount [OK]

Hint: IN with subquery filters matching rows [OK]

Common Mistakes:

• Assuming all customers are returned
• Thinking it returns customers without orders
• Believing subquery causes syntax error

4. Identify the error in the following query that tries to list all employees with their department names:

SELECT e.name, d.department_name FROM employees e JOIN departments d WHERE e.department_id = d.department_id;

medium

A. Using WHERE instead of HAVING for join condition.

B. Missing ON keyword before the join condition.

C. Incorrect table aliases used.

D. No error; query is correct.

Solution

1. Step 1: Check JOIN syntax

   JOIN requires the ON keyword to specify the join condition between tables.

2. Step 2: Identify the mistake

   The query uses WHERE instead of ON to specify the join condition, which is incorrect syntax.

3. Final Answer:

   Missing ON keyword before the join condition. -> Option B

4. Quick Check:

   JOIN needs ON, not WHERE, for conditions [OK]

Hint: JOIN conditions must use ON, not WHERE [OK]

Common Mistakes:

• Using WHERE instead of ON for JOIN
• Confusing HAVING with JOIN condition
• Assuming aliases cause error

5. You want to find all customers who have placed orders but only want to use a JOIN instead of a subquery. Which query correctly replaces this subquery?

SELECT name FROM customers WHERE customer_id IN (SELECT customer_id FROM orders);

hard

A. SELECT DISTINCT c.name FROM customers c JOIN orders o ON c.customer_id = o.customer_id;

B. SELECT c.name FROM customers c JOIN orders o WHERE c.customer_id = o.customer_id;

C. SELECT c.name FROM customers c, orders o WHERE c.customer_id = o.customer_id;

D. SELECT name FROM customers LEFT JOIN orders ON customer_id = customer_id;

Solution

1. Step 1: Understand the original subquery

   The subquery selects customer_ids from orders, filtering customers who placed orders.

2. Step 2: Translate to JOIN

   Using JOIN on customer_id between customers and orders returns matching customers. DISTINCT is needed to avoid duplicates.

3. Step 3: Evaluate options

   SELECT DISTINCT c.name FROM customers c JOIN orders o ON c.customer_id = o.customer_id; uses JOIN with ON and DISTINCT correctly. SELECT c.name FROM customers c JOIN orders o WHERE c.customer_id = o.customer_id; misses ON keyword. SELECT c.name FROM customers c, orders o WHERE c.customer_id = o.customer_id; uses old-style join but is valid but may return duplicates. SELECT name FROM customers LEFT JOIN orders ON customer_id = customer_id; uses LEFT JOIN but ON condition is ambiguous.

4. Final Answer:

   SELECT DISTINCT c.name FROM customers c JOIN orders o ON c.customer_id = o.customer_id; -> Option A

5. Quick Check:

   JOIN with ON and DISTINCT matches subquery result [OK]

Hint: Use JOIN with ON and DISTINCT to replace IN subquery [OK]

Common Mistakes:

• Omitting ON keyword in JOIN
• Not using DISTINCT causing duplicates
• Using ambiguous ON conditions