Recursive CTEs in MySQL - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to start a recursive CTE with the base query.

MySQL

WITH RECURSIVE cte AS (SELECT id, parent_id FROM employees WHERE parent_id IS [1])

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

AFALSE

BNOT NULL

CTRUE

DNULL

Attempts:

3 left

💡 Hint

Common Mistakes

Using NOT NULL instead of NULL causes no rows to be selected.

Using TRUE or FALSE is not valid syntax for this condition.

The base query selects rows where parent_id is NULL, which usually means top-level employees without a manager.

2fill in blank

medium

Complete the recursive part to join the CTE with the employees table.

MySQL

SELECT id, parent_id FROM employees UNION ALL SELECT e.id, e.parent_id FROM employees e JOIN cte ON e.parent_id = cte.[1]

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

Aid

Bparent_id

Cname

Demployee_id

Attempts:

3 left

💡 Hint

Common Mistakes

Joining on parent_id = parent_id causes a self-join with no hierarchy.

Using name or employee_id which are not part of the CTE columns.

The recursive part joins employees where their parent_id matches the id from the CTE, to find children.

3fill in blank

hard

Fix the error in the recursive CTE by completing the anchor member.

MySQL

WITH RECURSIVE cte AS (SELECT id, parent_id FROM employees WHERE [1]) SELECT * FROM cte;

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

Aparent_id IS NULL

Bid = 0

Cparent_id = id

Did IS NULL

Attempts:

3 left

💡 Hint

Common Mistakes

Using id = 0 may not match any rows.

Using parent_id = id causes a wrong condition.

Using id IS NULL is usually false for primary keys.

The anchor member must select the root rows where parent_id is NULL to start recursion properly.

4fill in blank

hard

Fill both blanks to complete the recursive CTE that finds all descendants of employee 5.

MySQL

WITH RECURSIVE descendants AS (SELECT id, parent_id FROM employees WHERE id = [1] UNION ALL SELECT e.id, e.parent_id FROM employees e JOIN descendants d ON e.[2] = d.id) SELECT * FROM descendants;

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

A5

Bparent_id

Cid

Demployee_id

Attempts:

3 left

💡 Hint

Common Mistakes

Using e.id = d.id causes no recursion.

Using wrong employee id in anchor member.

Joining on e.employee_id which is not a valid column.

The anchor selects employee with id = 5. The recursive part joins on e.parent_id = d.id to find children.

5fill in blank

hard

Fill all three blanks to create a recursive CTE that calculates the depth level of each employee in the hierarchy.

MySQL

WITH RECURSIVE hierarchy AS (SELECT id, parent_id, 1 AS level FROM employees WHERE parent_id IS [1] UNION ALL SELECT e.id, e.parent_id, h.level [2] 1 FROM employees e JOIN hierarchy h ON e.parent_id = h.[3]) SELECT * FROM hierarchy;

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

ANULL

B+

Cid

D-

Attempts:

3 left

💡 Hint

Common Mistakes

Using parent_id IS NOT NULL in anchor selects wrong rows.

Using - instead of + decreases level incorrectly.

Joining on wrong column like parent_id instead of id.

The anchor selects root employees with parent_id IS NULL. The recursive part increments the level by 1 and joins on h.id.

Practice

1. What is the main purpose of a recursive CTE in MySQL?

easy

A. To create temporary tables for storing large datasets

B. To repeatedly execute a query to explore hierarchical or connected data

C. To speed up simple SELECT queries without joins

D. To permanently store data in the database

Solution

1. Step 1: Understand recursive CTE concept

   A recursive CTE runs a base query and then repeatedly runs a recursive part referring back to itself to explore connected data.

2. Step 2: Identify the purpose

   This process is mainly used to handle hierarchical or tree-like data, such as organizational charts or folder structures.

3. Final Answer:

   To repeatedly execute a query to explore hierarchical or connected data -> Option B

4. Quick Check:

   Recursive CTE = hierarchical data exploration [OK]

Hint: Recursive CTEs repeat queries to explore connected data [OK]

Common Mistakes:

• Confusing recursive CTEs with temporary tables
• Thinking recursive CTEs speed up all queries
• Assuming recursive CTEs permanently store data

2. Which of the following is the correct basic syntax to start a recursive CTE in MySQL?

easy

A. WITH cte_name RECURSIVE (anchor_query UNION recursive_query) SELECT * FROM cte_name;

B. CREATE RECURSIVE TABLE cte_name AS (anchor_query UNION recursive_query);

C. WITH RECURSIVE cte_name AS (anchor_query UNION ALL recursive_query) SELECT * FROM cte_name;

D. SELECT RECURSIVE * FROM cte_name WHERE condition;

Solution

1. Step 1: Recall MySQL recursive CTE syntax

   MySQL uses the keyword WITH RECURSIVE followed by the CTE name and a query combining an anchor and recursive part with UNION ALL.

2. Step 2: Match syntax options

   WITH RECURSIVE cte_name AS (anchor_query UNION ALL recursive_query) SELECT * FROM cte_name; correctly shows this syntax, including UNION ALL and the final SELECT from the CTE.

3. Final Answer:

   WITH RECURSIVE cte_name AS (anchor_query UNION ALL recursive_query) SELECT * FROM cte_name; -> Option C

4. Quick Check:

   WITH RECURSIVE + UNION ALL = correct syntax [OK]

Hint: Look for WITH RECURSIVE and UNION ALL keywords [OK]

Common Mistakes:

• Using UNION instead of UNION ALL in recursive part
• Trying to CREATE RECURSIVE TABLE (not valid)
• Misplacing RECURSIVE keyword after CTE name

3. Given the table employees(id, manager_id, name), what will this query return?

WITH RECURSIVE subordinates AS (
  SELECT id, name FROM employees WHERE id = 2
  UNION ALL
  SELECT e.id, e.name FROM employees e
  JOIN subordinates s ON e.manager_id = s.id
)
SELECT * FROM subordinates;

medium

A. All employees who report directly or indirectly to employee with id 2

B. Only the employee with id 2

C. All employees who manage employee with id 2

D. Syntax error due to missing anchor query

Solution

1. Step 1: Analyze anchor query

   The anchor selects the employee with id = 2 as the starting point.

2. Step 2: Analyze recursive part

   The recursive part joins employees whose manager_id matches the current subordinates' id, thus finding direct reports, then their reports, and so on.

3. Step 3: Understand final output

   The query returns employee 2 and all employees reporting to them directly or indirectly.

4. Final Answer:

   All employees who report directly or indirectly to employee with id 2 -> Option A

5. Quick Check:

   Recursive join on manager_id finds all subordinates [OK]

Hint: Recursive join on manager_id finds all subordinates [OK]

Common Mistakes:

• Confusing manager and subordinate roles
• Thinking it returns only direct reports
• Assuming syntax error due to anchor query presence

4. Identify the error in this recursive CTE:

WITH RECURSIVE cte AS (
  SELECT id FROM items WHERE parent_id IS NULL
  UNION
  SELECT i.id FROM items i JOIN cte ON i.parent_id = cte.id
)
SELECT * FROM cte;

medium

A. Anchor query should use WHERE parent_id = 0 instead of IS NULL

B. Recursive part should use LEFT JOIN instead of JOIN

C. CTE name cannot be 'cte', must be more descriptive

D. Missing UNION ALL causes infinite recursion or wrong results

Solution

1. Step 1: Check recursive CTE syntax

   Recursive CTEs require UNION ALL to combine anchor and recursive parts to avoid duplicate elimination.

2. Step 2: Understand effect of UNION vs UNION ALL

   Using UNION removes duplicates each iteration, which can cause infinite recursion or incorrect results in recursive queries.

3. Final Answer:

   Missing UNION ALL causes infinite recursion or wrong results -> Option D

4. Quick Check:

   Recursive CTEs must use UNION ALL [OK]

Hint: Always use UNION ALL in recursive CTEs to avoid infinite loops [OK]

Common Mistakes:

• Using UNION instead of UNION ALL in recursive CTE
• Assuming NULL vs 0 matters for recursion logic
• Thinking JOIN type affects recursion correctness

5. You want to list all ancestors of an employee in a company hierarchy stored in employees(id, manager_id, name). Which recursive CTE query correctly finds all managers up the chain for employee with id 5?

hard

A. WITH RECURSIVE ancestors AS (
SELECT id, manager_id, name FROM employees WHERE id = 5
UNION ALL
SELECT e.id, e.manager_id, e.name FROM employees e
JOIN ancestors a ON e.id = a.manager_id
)
SELECT * FROM ancestors WHERE id != 5;

B. WITH RECURSIVE ancestors AS (
SELECT id, manager_id, name FROM employees WHERE id = 5
UNION ALL
SELECT e.id, e.manager_id, e.name FROM employees e
JOIN ancestors a ON e.manager_id = a.id
)
SELECT * FROM ancestors WHERE id != 5;

C. WITH RECURSIVE ancestors AS (
SELECT id, manager_id, name FROM employees WHERE manager_id = 5
UNION ALL
SELECT e.id, e.manager_id, e.name FROM employees e
JOIN ancestors a ON e.manager_id = a.id
)
SELECT * FROM ancestors;

D. WITH RECURSIVE ancestors AS (
SELECT id, manager_id, name FROM employees WHERE id = 5
UNION
SELECT e.id, e.manager_id, e.name FROM employees e
JOIN ancestors a ON e.id = a.manager_id
)
SELECT * FROM ancestors WHERE id != 5;

Solution

1. Step 1: Understand the goal

   We want to find all managers (ancestors) of employee 5, moving up the chain via manager_id.

2. Step 2: Analyze join condition

   To move up, join employees where employee id matches the manager_id of current ancestor (e.id = a.manager_id).

3. Step 3: Check anchor and recursive parts

   Anchor selects employee 5; recursive part finds their manager, then that manager's manager, etc.

4. Final Answer:

   WITH RECURSIVE ancestors AS (SELECT id, manager_id, name FROM employees WHERE id = 5 UNION ALL SELECT e.id, e.manager_id, e.name FROM employees e JOIN ancestors a ON e.id = a.manager_id) SELECT * FROM ancestors WHERE id != 5; -> Option A

5. Quick Check:

   Join on e.id = a.manager_id moves up hierarchy [OK]

Hint: Join on e.id = a.manager_id to move up the hierarchy [OK]

Common Mistakes:

• Reversing join condition to move down instead of up
• Using UNION instead of UNION ALL causing errors
• Starting anchor with manager_id = 5 instead of id = 5