Why subqueries nest queries in MySQL - Test Your Understanding

Practice - 5 Tasks

1fill in blank

easy

Complete the code to select all employees who work in the department with ID 5.

MySQL

SELECT * FROM employees WHERE department_id = [1];

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

Aemployee_id

B5

C10

Ddepartment_id

Attempts:

3 left

💡 Hint

Common Mistakes

Using the wrong column name in the WHERE clause.

Using a value other than 5.

The query filters employees whose department_id equals 5.

2fill in blank

medium

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

MySQL

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

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

Aid

Bname

Cdepartment_id

Demployee_id

Attempts:

3 left

💡 Hint

Common Mistakes

Selecting the department name instead of its ID.

Selecting employee columns in the subquery.

The subquery must return the id of the department named 'Sales' to match department_id in employees.

3fill in blank

hard

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

MySQL

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

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

Asalary

BMAX(salary)

CAVG(salary)

DMIN(salary)

Attempts:

3 left

💡 Hint

Common Mistakes

Selecting salary directly without aggregation.

Using MAX or MIN instead of AVG.

The subquery must calculate the average salary using AVG(salary) to compare each employee's salary.

4fill in blank

hard

Fill both blanks to select employees who work in departments with more than 10 employees.

MySQL

SELECT * FROM employees WHERE department_id IN (SELECT [1] FROM departments WHERE [2] > 10);

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

Aid

Bemployee_count

Cnum_employees

Ddepartment_id

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong column names for department ID or employee count.

Using department_id in the WHERE clause instead of employee count.

The subquery selects department id where the number of employees (num_employees) is greater than 10.

5fill in blank

hard

Fill all three blanks to select employees whose salary is above the average salary in their department.

MySQL

SELECT * FROM employees e WHERE salary > (SELECT AVG([1]) FROM employees WHERE [2] = e.[3]);

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

Asalary

Bdepartment_id

Demployee_id

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong columns for average or matching condition.

Confusing employee_id with department_id.

The subquery calculates the average salary for employees in the same department_id as the outer query employee.

Practice

1. Why do we use subqueries in MySQL?
SELECT * FROM employees WHERE department_id IN (SELECT id FROM departments WHERE location = 'NY');

easy

A. To nest one query inside another and use its results

B. To speed up the database server hardware

C. To create new tables automatically

D. To change the database schema

Solution

1. Step 1: Understand subquery usage

   The inner query SELECT id FROM departments WHERE location = 'NY' finds department IDs in New York.

2. Step 2: Use subquery results in outer query

   The outer query selects employees whose department_id matches those IDs from the subquery.

3. Final Answer:

   To nest one query inside another and use its results -> Option A

4. Quick Check:

   Subqueries = nested queries using results [OK]

Hint: Subqueries let you use one query's results inside another [OK]

Common Mistakes:

• Thinking subqueries change database structure
• Confusing subqueries with hardware upgrades
• Believing subqueries create tables automatically

2. Which of the following is the correct syntax for a subquery in MySQL?

easy

A. SELECT * FROM table1 WHERE id = SELECT id FROM table2 WHERE condition;

B. SELECT * FROM table1 WHERE id == (SELECT id FROM table2 WHERE condition);

C. SELECT * FROM table1 WHERE id IN SELECT id FROM table2 WHERE condition;

D. SELECT * FROM table1 WHERE id = (SELECT id FROM table2 WHERE condition);

Solution

1. Step 1: Check subquery syntax

   Subqueries must be enclosed in parentheses and used with operators like = or IN.

2. Step 2: Identify correct syntax

   SELECT * FROM table1 WHERE id = (SELECT id FROM table2 WHERE condition); uses parentheses correctly and the = operator, which is valid syntax.

3. Final Answer:

   SELECT * FROM table1 WHERE id = (SELECT id FROM table2 WHERE condition); -> Option D

4. Quick Check:

   Subquery syntax requires parentheses [OK]

Hint: Always put subqueries inside parentheses [OK]

Common Mistakes:

• Using double equals (==) instead of single equals (=)
• Omitting parentheses around subquery
• Using IN without parentheses

3. What will be the output of this query?

SELECT name FROM employees WHERE department_id IN (SELECT id FROM departments WHERE location = 'LA');

Given:
- departments table has ids 1,2,3 with locations NY, LA, LA respectively
- employees table has:
(name: Alice, department_id: 1),
(name: Bob, department_id: 2),
(name: Carol, department_id: 3),
(name: Dave, department_id: 4)

medium

A. Bob, Carol

B. Alice, Bob, Carol

C. Alice, Dave

D. Dave only

Solution

1. Step 1: Identify department IDs in LA

   From departments, IDs with location 'LA' are 2 and 3.

2. Step 2: Find employees in those departments

   Employees with department_id 2 and 3 are Bob and Carol.

3. Final Answer:

   Bob, Carol -> Option A

4. Quick Check:

   Subquery filters departments = Bob, Carol [OK]

Hint: Match outer query filter with subquery results [OK]

Common Mistakes:

• Including employees from other departments
• Ignoring subquery filtering
• Assuming all employees are returned

4. Identify the error in this query:

SELECT name FROM employees WHERE department_id = SELECT id FROM departments WHERE location = 'NY';

medium

A. Missing WHERE clause in outer query

B. Using = instead of IN for multiple results

C. Missing parentheses around the subquery

D. Wrong table name 'employees'

Solution

1. Step 1: Check subquery syntax

   The subquery must be enclosed in parentheses to be valid.

2. Step 2: Identify missing parentheses

   The query lacks parentheses around SELECT id FROM departments WHERE location = 'NY'.

3. Final Answer:

   Missing parentheses around the subquery -> Option C

4. Quick Check:

   Subqueries need parentheses [OK]

Hint: Always wrap subqueries in parentheses [OK]

Common Mistakes:

• Forgetting parentheses causes syntax error
• Confusing = and IN without checking subquery results
• Assuming table names are incorrect without checking

5. You want to find all employees who earn more than the average salary in their department. Which query correctly uses a subquery to achieve this?

hard

A. SELECT name, salary FROM employees WHERE salary > AVG(salary);

B. SELECT name, salary FROM employees e WHERE salary > (SELECT AVG(salary) FROM employees WHERE department_id = e.department_id);

C. SELECT name, salary FROM employees WHERE salary > (SELECT salary FROM employees);

D. SELECT name, salary FROM employees WHERE salary > ALL (SELECT AVG(salary) FROM employees GROUP BY department_id);

Solution

1. Step 1: Understand the problem

   We want employees earning more than their department's average salary.

2. Step 2: Analyze subquery usage

   SELECT name, salary FROM employees e WHERE salary > (SELECT AVG(salary) FROM employees WHERE department_id = e.department_id); uses a correlated subquery that calculates average salary per employee's department.

3. Step 3: Check other options

   SELECT name, salary FROM employees WHERE salary > AVG(salary); lacks grouping, C compares salary to all salaries, D compares to all averages without correlation.

4. Final Answer:

   SELECT name, salary FROM employees e WHERE salary > (SELECT AVG(salary) FROM employees WHERE department_id = e.department_id); -> Option B

5. Quick Check:

   Correlated subquery per department = SELECT name, salary FROM employees e WHERE salary > (SELECT AVG(salary) FROM employees WHERE department_id = e.department_id); [OK]

Hint: Use correlated subqueries to compare per group [OK]

Common Mistakes:

• Using aggregate functions without grouping
• Not correlating subquery with outer query
• Comparing salary to unrelated values