Correlated subqueries in MySQL - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is a correlated subquery in SQL?

A correlated subquery is a subquery that uses values from the outer query. It runs once for each row processed by the outer query.

beginner

How does a correlated subquery differ from a regular subquery?

A regular subquery runs once and returns a result set. A correlated subquery runs repeatedly, once for each row of the outer query, because it depends on outer query values.

intermediate

Example: What does this query do?

SELECT e1.name FROM employees e1 WHERE salary > (SELECT AVG(salary) FROM employees e2 WHERE e2.department = e1.department);

This query finds employees whose salary is greater than the average salary of their own department. The subquery uses e1.department from the outer query, so it is correlated.

intermediate

Why can correlated subqueries be slower than regular subqueries?

Because the correlated subquery runs once for each row of the outer query, it can be slower especially on large tables. It repeats work multiple times.

intermediate

Can a correlated subquery appear in the SELECT clause?

Yes, a correlated subquery can appear in the SELECT clause to calculate a value for each row, like a computed column based on related data.

What does a correlated subquery depend on?

A Only constants

B Values from the outer query

C Only the inner query's tables

D No dependencies

A correlated subquery depends on values from the outer query to run for each row.

Which of these is true about correlated subqueries?

A They run once per row of the outer query

B They always return multiple rows

C They cannot use outer query columns

D They run once for the entire query

Correlated subqueries run once for each row of the outer query because they depend on outer values.

In MySQL, where can a correlated subquery appear?

A Only in JOIN conditions

B Only in FROM clause

C Only in WHERE clause

D In WHERE, SELECT, or HAVING clauses

Correlated subqueries can appear in WHERE, SELECT, or HAVING clauses to filter or compute values.

What is a common use case for correlated subqueries?

A Creating indexes

B Joining two unrelated tables

C Comparing each row to an aggregate of related rows

D Updating multiple tables at once

Correlated subqueries often compare each row to aggregates like averages or counts from related rows.

Why might you avoid correlated subqueries on large tables?

A They can be slow due to repeated execution

B They use too much disk space

C They do not work with indexes

D They cause syntax errors

Correlated subqueries can be slow because they run once per outer row, repeating work many times.

Explain what a correlated subquery is and how it works in a SQL query.

Think about how the inner query uses data from the outer query.

Describe a scenario where using a correlated subquery is helpful and what to watch out for.

Consider real-life examples like comparing a student's score to class average.

Practice

1. What is a correlated subquery in MySQL?

easy

A. A subquery that uses values from the outer query for each row.

B. A subquery that runs only once before the outer query.

C. A query that joins two tables without any subqueries.

D. A subquery that returns a fixed constant value.

Solution

1. Step 1: Understand subquery behavior

   A correlated subquery depends on the outer query's current row values to execute.

2. Step 2: Compare options

   The option describing a subquery that uses values from the outer query for each row correctly defines a correlated subquery.

3. Final Answer:

   A subquery that uses values from the outer query for each row. -> Option A

4. Quick Check:

   Correlated subquery = uses outer row values [OK]

Hint: Correlated subqueries depend on outer query row values [OK]

Common Mistakes:

• Thinking subqueries always run once
• Confusing joins with subqueries
• Assuming subqueries return constants only

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

easy

A. SELECT * FROM employees WHERE salary > (SELECT AVG(salary) FROM employees e2 WHERE e2.department_id = employees.department_id);

B. SELECT * FROM employees WHERE salary IN (SELECT salary FROM employees);

C. SELECT * FROM employees WHERE salary > (SELECT AVG(salary) FROM employees);

D. SELECT * FROM employees WHERE salary = 50000;

Solution

1. Step 1: Identify correlated subquery syntax

   A correlated subquery references a column from the outer query inside the subquery, like e2.department_id = employees.department_id.

2. Step 2: Check each option

   SELECT * FROM employees WHERE salary > (SELECT AVG(salary) FROM employees e2 WHERE e2.department_id = employees.department_id); uses the outer table's column inside the subquery condition, making it a correlated subquery.

3. Final Answer:

   SELECT * FROM employees WHERE salary > (SELECT AVG(salary) FROM employees e2 WHERE e2.department_id = employees.department_id); -> Option A

4. Quick Check:

   Correlated subquery syntax = SELECT * FROM employees WHERE salary > (SELECT AVG(salary) FROM employees e2 WHERE e2.department_id = employees.department_id); [OK]

Hint: Look for subquery referencing outer query table alias [OK]

Common Mistakes:

• Missing outer table reference in subquery
• Using subquery without WHERE clause for correlation
• Confusing correlated with simple subqueries

3. Given the tables:
employees(id, name, department_id, salary)
and
departments(id, name)
What will this query return?

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

medium

A. Employees whose salary is above the average salary of all departments combined.

B. Employees whose salary is above the average salary of all employees.

C. Employees whose salary is above the average salary of their department.

D. Syntax error due to subquery referencing outer query.

Solution

1. Step 1: Understand the subquery condition

   The subquery calculates the average salary for the department of the current employee (using e.department_id).

2. Step 2: Compare employee salary with department average

   The outer query selects employees whose salary is greater than their department's average salary.

3. Final Answer:

   Employees whose salary is above the average salary of their department. -> Option C

4. Quick Check:

   Correlated subquery filters by department average salary [OK]

Hint: Check if subquery filters by outer row's department_id [OK]

Common Mistakes:

• Thinking subquery calculates overall average salary
• Assuming syntax error due to correlation
• Confusing department and employee averages

4. Identify the error in this correlated subquery:

SELECT e1.name FROM employees e1 WHERE salary > (SELECT AVG(salary) FROM employees e2 WHERE e2.department = e1.department_id);

medium

A. Subquery cannot reference outer query columns.

B. Alias 'e1' is not allowed in subquery.

C. Missing GROUP BY clause in subquery.

D. Column 'department' does not exist in employees table.

Solution

1. Step 1: Check column names in subquery condition

   The subquery compares e2.department to e1.department_id, but 'department' column does not exist in employees table.

2. Step 2: Confirm correct column usage

   The correct column should be e2.department_id to match e1.department_id for correlation.

3. Final Answer:

   Column 'department' does not exist in employees table. -> Option D

4. Quick Check:

   Invalid column name causes error [OK]

Hint: Verify column names match exactly in both queries [OK]

Common Mistakes:

• Using wrong column names for correlation
• Assuming alias usage causes error
• Thinking GROUP BY is required in subquery

5. You want to find all departments where every employee earns more than the average salary of all employees in the company. Which query correctly uses a correlated subquery to achieve this?

hard

A. SELECT d.name FROM departments d WHERE ALL (SELECT salary FROM employees WHERE department_id = d.id) > (SELECT AVG(salary) FROM employees);

B. SELECT d.name FROM departments d WHERE NOT EXISTS (SELECT 1 FROM employees e WHERE e.department_id = d.id AND e.salary <= (SELECT AVG(salary) FROM employees));

C. SELECT d.name FROM departments d WHERE EXISTS (SELECT 1 FROM employees e WHERE e.department_id = d.id AND e.salary > (SELECT AVG(salary) FROM employees));

D. SELECT d.name FROM departments d WHERE EXISTS (SELECT 1 FROM employees e WHERE e.department_id = d.id AND e.salary > (SELECT AVG(salary) FROM employees WHERE department_id = d.id));

Solution

1. Step 1: Understand the requirement

   We want departments where every employee earns more than the company-wide average salary.

2. Step 2: Analyze the NOT EXISTS query

   SELECT d.name FROM departments d WHERE NOT EXISTS (SELECT 1 FROM employees e WHERE e.department_id = d.id AND e.salary <= (SELECT AVG(salary) FROM employees)); uses NOT EXISTS to ensure no employee in the department earns less or equal to the company average, meaning all earn more.

3. Step 3: Check other options

   The options using EXISTS check for the existence of at least one qualifying employee, not all. The option attempting to use ALL with a subquery on the left side uses invalid syntax in MySQL.

4. Final Answer:

   SELECT d.name FROM departments d WHERE NOT EXISTS (SELECT 1 FROM employees e WHERE e.department_id = d.id AND e.salary <= (SELECT AVG(salary) FROM employees)); -> Option B

5. Quick Check:

   NOT EXISTS ensures all employees earn above average [OK]

Hint: Use NOT EXISTS to check all employees meet condition [OK]

Common Mistakes:

• Using EXISTS instead of NOT EXISTS for 'all' condition
• Misusing ALL keyword in MySQL
• Comparing with department average instead of company average