Why subqueries are needed in PostgreSQL - 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.

PostgreSQL

SELECT * FROM employees WHERE department_id = [1];

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

A5

Bemployee_id

Cdepartment_id

D10

Attempts:

3 left

💡 Hint

Common Mistakes

Using the wrong column name instead of the department ID.

Using a wrong number for the department.

The department ID 5 is used to filter employees working in that department.

2fill in blank

medium

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

PostgreSQL

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

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

Aemployee_id

Bdepartment_id

Cname

Did

Attempts:

3 left

💡 Hint

Common Mistakes

Selecting the department name instead of its ID in the subquery.

Using employee columns inside the subquery.

The subquery selects the department's ID where the name is 'Sales'. This ID is used to find employees in that department.

3fill in blank

hard

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

PostgreSQL

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

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

Asalary

BAVG(salary)

CMAX(salary)

DMIN(salary)

Attempts:

3 left

💡 Hint

Common Mistakes

Using salary column 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 against it.

4fill in blank

hard

Fill both blanks to select employees who have the highest salary in their department.

PostgreSQL

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

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

AMAX(salary)

BMIN(salary)

Cdepartment_id

Demployee_id

Attempts:

3 left

💡 Hint

Common Mistakes

Using MIN instead of MAX.

Comparing with employee_id instead of department_id.

The subquery finds the maximum salary in the employee's department. The outer query compares each employee's salary to that maximum.

5fill in blank

hard

Fill all three blanks to select departments with more than 10 employees.

PostgreSQL

SELECT name FROM departments WHERE id IN (SELECT [1] FROM employees GROUP BY [2] HAVING COUNT([3]) > 10);

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

Adepartment_id

C*

Demployee_id

Attempts:

3 left

💡 Hint

Common Mistakes

Using employee_id instead of department_id for grouping.

Counting employee_id instead of all rows with COUNT(*).

The subquery groups employees by department_id and counts all employees (*) in each group. Departments with more than 10 employees are selected.

Practice

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

easy

A. To change data types of columns

B. To speed up the database server automatically

C. To ask a question inside another query and filter results based on it

D. To create new tables from existing ones

Solution

1. Step 1: Understand the role of the subquery

   The subquery (SELECT id FROM departments WHERE location = 'NY') finds department IDs located in NY.

2. Step 2: See how the main query uses the subquery

   The main query selects employees whose department_id matches any ID from the subquery result.

3. Final Answer:

   To ask a question inside another query and filter results based on it -> Option C

4. Quick Check:

   Subqueries filter data using inner query results [OK]

Hint: Subqueries answer inner questions to filter outer query [OK]

Common Mistakes:

• Thinking subqueries speed up queries automatically
• Confusing subqueries with table creation
• Using subqueries to change data types

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

easy

A. SELECT * FROM table WHERE column == (SELECT MAX(column) FROM table);

B. SELECT * FROM table WHERE column = (SELECT MAX(column) FROM table);

C. SELECT * FROM table WHERE column = SELECT MAX(column) FROM table;

D. SELECT * FROM table WHERE column IN SELECT MAX(column) FROM table;

Solution

1. Step 1: Identify correct subquery syntax

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

2. Step 2: Check each option

   SELECT * FROM table WHERE column = (SELECT MAX(column) FROM table); correctly uses parentheses and = with a subquery. The other options have syntax errors like using == instead of =, omitting parentheses entirely, or missing them after IN.

3. Final Answer:

   SELECT * FROM table WHERE column = (SELECT MAX(column) FROM table); -> Option B

4. Quick Check:

   Subqueries need parentheses and proper operators [OK]

Hint: Subqueries always go inside parentheses [OK]

Common Mistakes:

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

3. What will be the output of this query?

SELECT name FROM employees WHERE salary > (SELECT AVG(salary) FROM employees);

medium

A. Syntax error due to subquery

B. Names of employees earning less than the average salary

C. All employee names

D. Names of employees earning more than the average salary

Solution

1. Step 1: Understand the subquery

   The subquery (SELECT AVG(salary) FROM employees) calculates the average salary of all employees.

2. Step 2: Apply the condition in the main query

   The main query selects employee names where their salary is greater than this average.

3. Final Answer:

   Names of employees earning more than the average salary -> Option D

4. Quick Check:

   Subquery returns average; main query filters salaries above it [OK]

Hint: Subqueries can return single values to compare in main query [OK]

Common Mistakes:

• Thinking it returns employees below average
• Assuming syntax error due to subquery
• Confusing AVG with SUM

4. Identify the error in this query:

SELECT name FROM employees WHERE department_id = SELECT id FROM departments WHERE name = 'Sales';

medium

A. Missing parentheses around the subquery

B. Wrong comparison operator used

C. Subquery returns multiple rows causing error

D. Incorrect table name used

Solution

1. Step 1: Check subquery syntax

   The subquery must be enclosed in parentheses to be valid in the WHERE clause.

2. Step 2: Identify the missing parentheses

   The query misses parentheses around SELECT id FROM departments WHERE name = 'Sales', causing syntax error.

3. Final Answer:

   Missing parentheses around the subquery -> Option A

4. Quick Check:

   Subqueries need parentheses to be valid [OK]

Hint: Always put subqueries inside parentheses [OK]

Common Mistakes:

• Forgetting parentheses around subqueries
• Assuming wrong operator causes error
• Thinking table name is incorrect

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

hard

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

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

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

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

Solution

1. Step 1: Understand the requirement

   We need to compare each employee's salary to the average salary of their own department.

2. Step 2: Analyze each option

   SELECT name, salary, department_id FROM employees WHERE salary > (SELECT AVG(salary) FROM employees WHERE department_id = employees.department_id); uses a correlated subquery that calculates average salary per employee's department. SELECT name, salary FROM employees WHERE salary > (SELECT AVG(salary) FROM employees); compares to overall average, ignoring departments. SELECT name FROM employees WHERE salary > ALL (SELECT salary FROM employees); compares salary to all salaries, not averages. SELECT name FROM employees WHERE salary > ANY (SELECT AVG(salary) FROM employees GROUP BY department_id); compares to a list of averages but uses ANY incorrectly.

3. Final Answer:

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

4. Quick Check:

   Correlated subquery compares salary to department average [OK]

Hint: Use correlated subqueries to compare within groups [OK]

Common Mistakes:

• Comparing to overall average instead of department average
• Using ALL or ANY incorrectly with averages
• Not correlating subquery with outer query