Correlated subqueries in MySQL

Introduction

Correlated subqueries let you compare each row in one table with rows in another table. They help find related data by checking conditions row by row.

When you want to find rows in one table that depend on values in another table for each row.

When you need to compare each employee's salary to the average salary in their department.

When you want to find customers who placed orders larger than their average order amount.

When you want to filter products that cost more than the average price in their category.

Syntax

MySQL

SELECT column1, column2
FROM table1 t1
WHERE column3 > (
  SELECT AVG(column3)
  FROM table2 t2
  WHERE t2.related_column = t1.related_column
);

The subquery uses a column from the outer query to compare values row by row.

The subquery runs once for each row in the outer query, so it can be slower on large data.

Examples

Find employees who earn more than the average salary in their own department.

MySQL

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

Find customers who have at least one order with total greater than 100.

MySQL

SELECT c.customer_id, c.name
FROM customers c
WHERE EXISTS (
  SELECT 1
  FROM orders o
  WHERE o.customer_id = c.customer_id
  AND o.total > 100
);

Sample Program

This query finds employees who earn more than the average salary in their department.

MySQL

CREATE TABLE departments (
  department_id INT PRIMARY KEY,
  department_name VARCHAR(50)
);

CREATE TABLE employees (
  employee_id INT PRIMARY KEY,
  name VARCHAR(50),
  salary DECIMAL(10,2),
  department_id INT,
  FOREIGN KEY (department_id) REFERENCES departments(department_id)
);

INSERT INTO departments VALUES
(1, 'Sales'),
(2, 'Engineering');

INSERT INTO employees VALUES
(101, 'Alice', 70000, 1),
(102, 'Bob', 60000, 1),
(103, 'Charlie', 80000, 2),
(104, 'Diana', 90000, 2);

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

Important Notes

Correlated subqueries can be slower than joins because the subquery runs for each row in the outer query.

Use correlated subqueries when you need row-by-row comparison that joins cannot easily do.

Make sure the subquery references the outer query table with an alias to avoid confusion.

Summary

Correlated subqueries compare each row in one table with related rows in another.

They run the subquery once per outer row, so they can be slower on big data.

Useful for finding rows based on related data conditions that depend on each row.

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