Self join patterns in PostgreSQL - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to select employee names and their managers' names using a self join.

PostgreSQL

SELECT e.name AS employee, m.name AS manager FROM employees e JOIN employees m ON e.manager_id = [1];

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

Am.id

Be.id

Cm.manager_id

De.manager_id

Attempts:

3 left

💡 Hint

Common Mistakes

Using employee's own id instead of manager's id.

Mixing up the join condition columns.

The self join matches each employee's manager_id with the manager's id. So, e.manager_id = m.id links employees to their managers.

2fill in blank

medium

Complete the code to find pairs of employees who share the same manager.

PostgreSQL

SELECT e1.name AS employee1, e2.name AS employee2 FROM employees e1 JOIN employees e2 ON e1.[1] = e2.[1] WHERE e1.id <> e2.id;

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

Adepartment_id

Bmanager_id

Cname

Did

Attempts:

3 left

💡 Hint

Common Mistakes

Joining on employee ids instead of manager ids.

Not excluding pairs where employee1 equals employee2.

Employees who share the same manager have the same manager_id. The join condition matches e1.manager_id = e2.manager_id.

3fill in blank

hard

Fix the error in the self join to correctly list employees and their managers.

PostgreSQL

SELECT e.name, m.name FROM employees e JOIN employees m ON e.[1] = m.[2];

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

Aid

Bmanager_id

Cname

Ddepartment_id

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping the columns in the join condition.

Joining on non-key columns like name or department_id.

The join should match the employee's manager_id to the manager's id. So, e.manager_id = m.id is correct.

4fill in blank

hard

Fill both blanks to select employees and their managers' departments using a self join.

PostgreSQL

SELECT e.name, m.name, m.[1] FROM employees e JOIN employees m ON e.[2] = m.id;

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

Adepartment

Bmanager_id

Cid

Dname

Attempts:

3 left

💡 Hint

Common Mistakes

Using employee's id in place of manager_id in join.

Selecting wrong column for department.

The query selects the manager's department by joining on e.manager_id = m.id.

5fill in blank

hard

Fill all three blanks to find employees who manage others and count how many they manage.

PostgreSQL

SELECT m.name AS manager, COUNT(e.[1]) AS team_size FROM employees e JOIN employees m ON e.[2] = m.[3] GROUP BY m.name HAVING COUNT(e.id) > 0;

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

Aid

Bmanager_id

Dname

Attempts:

3 left

💡 Hint

Common Mistakes

Counting manager ids instead of employee ids.

Mixing join columns.

The count is on employee id. The join matches e.manager_id = m.id to find employees managed by each manager.

Practice

1. What is the main purpose of using a self join in PostgreSQL?

easy

A. To join a table to itself to compare rows within the same table

B. To join two different tables based on a common column

C. To create a backup of a table

D. To delete duplicate rows from a table

Solution

1. Step 1: Understand the concept of self join

   A self join is used when you want to compare rows within the same table by joining the table to itself.

2. Step 2: Differentiate from other joins

   Other joins combine different tables, but self join works on one table with aliases to distinguish instances.

3. Final Answer:

   To join a table to itself to compare rows within the same table -> Option A

4. Quick Check:

   Self join = join table to itself [OK]

Hint: Self join means joining table to itself using aliases [OK]

Common Mistakes:

• Confusing self join with joining two different tables
• Thinking self join creates a new table
• Assuming self join deletes duplicates

2. Which of the following is the correct syntax to perform a self join on a table named employees with aliases e1 and e2?

easy

A. SELECT * FROM employees e1 LEFT JOIN employees e2 ON e1.id = e2.manager_id;

B. SELECT * FROM employees JOIN employees ON employees.manager_id = employees.id;

C. SELECT * FROM employees AS e1, employees AS e2 WHERE e1.id = e2.id;

D. SELECT * FROM employees e1 JOIN employees e2 ON e1.manager_id = e2.id;

Solution

1. Step 1: Identify correct alias usage

   Using aliases e1 and e2 for the same table employees is necessary to distinguish instances.

2. Step 2: Check join condition

   The join condition e1.manager_id = e2.id correctly links employees to their managers.

3. Final Answer:

   SELECT * FROM employees e1 JOIN employees e2 ON e1.manager_id = e2.id; -> Option D

4. Quick Check:

   Use aliases and proper ON condition [OK]

Hint: Use aliases and ON clause for self join syntax [OK]

Common Mistakes:

• Not using aliases causing ambiguity
• Wrong join condition columns
• Using comma join without ON clause

3. Given the table employees(id, name, manager_id) with data:

id | name  | manager_id
---+-------+-----------
1  | Alice | NULL
2  | Bob   | 1
3  | Carol | 1
4  | Dave  | 2

What will be the output of this query?

SELECT e1.name AS employee, e2.name AS manager
FROM employees e1
LEFT JOIN employees e2 ON e1.manager_id = e2.id
ORDER BY e1.id;

medium

A. [{"employee": "Alice", "manager": null}, {"employee": "Bob", "manager": null}, {"employee": "Carol", "manager": null}, {"employee": "Dave", "manager": null}]

B. [{"employee": "Alice", "manager": null}, {"employee": "Bob", "manager": "Alice"}, {"employee": "Carol", "manager": "Alice"}, {"employee": "Dave", "manager": "Bob"}]

C. [{"employee": "Alice", "manager": "Alice"}, {"employee": "Bob", "manager": "Bob"}, {"employee": "Carol", "manager": "Carol"}, {"employee": "Dave", "manager": "Dave"}]

D. [{"employee": "Alice", "manager": "Bob"}, {"employee": "Bob", "manager": "Carol"}, {"employee": "Carol", "manager": "Dave"}, {"employee": "Dave", "manager": null}]

Solution

1. Step 1: Understand the join logic

   The LEFT JOIN matches each employee's manager_id to the manager's id, returning manager's name or null if none.

2. Step 2: Map each employee to their manager

   Alice has no manager (null), Bob and Carol have Alice as manager, Dave has Bob as manager.

3. Final Answer:

   [{"employee": "Alice", "manager": null}, {"employee": "Bob", "manager": "Alice"}, {"employee": "Carol", "manager": "Alice"}, {"employee": "Dave", "manager": "Bob"}] -> Option B

4. Quick Check:

   LEFT JOIN matches manager_id to id [OK]

Hint: Match manager_id to id with LEFT JOIN for manager names [OK]

Common Mistakes:

• Confusing employee and manager columns
• Using INNER JOIN losing employees without managers
• Assuming manager_id is employee id

4. Consider this query intended to find pairs of employees who share the same manager:

SELECT e1.name, e2.name
FROM employees e1
JOIN employees e2 ON e1.manager_id = e2.manager_id
WHERE e1.id != e2.id;

What is the main issue with this query?

medium

A. It will return duplicate pairs in both orders (e.g., (Alice, Bob) and (Bob, Alice))

B. It will not return any rows because of the WHERE clause

C. It causes a syntax error due to missing aliases

D. It only returns employees with no manager

Solution

1. Step 1: Analyze join condition

   The join matches employees with the same manager_id, including reversed pairs.

2. Step 2: Check WHERE clause effect

   WHERE e1.id != e2.id avoids self-pairing but does not prevent duplicate pairs in reverse order.

3. Final Answer:

   It will return duplicate pairs in both orders (e.g., (Alice, Bob) and (Bob, Alice)) -> Option A

4. Quick Check:

   Self join can produce reversed duplicate pairs [OK]

Hint: Self join pairs may appear twice reversed; use filtering [OK]

Common Mistakes:

• Assuming WHERE clause removes duplicates
• Thinking query causes syntax error
• Believing it returns only employees without managers

5. You want to find all employees who are managers themselves, i.e., employees who have at least one other employee reporting to them. Which self join query correctly achieves this?

hard

A. SELECT e1.name FROM employees e1 JOIN employees e2 ON e1.manager_id = e2.id;

B. SELECT e1.name FROM employees e1 LEFT JOIN employees e2 ON e1.id = e2.manager_id WHERE e2.id IS NULL;

C. SELECT DISTINCT e1.name FROM employees e1 JOIN employees e2 ON e1.id = e2.manager_id;

D. SELECT e1.name FROM employees e1 WHERE e1.manager_id IS NOT NULL;

Solution

1. Step 1: Understand the requirement

   Managers are employees whose id appears as manager_id for others.

2. Step 2: Analyze each option

   SELECT DISTINCT e1.name FROM employees e1 JOIN employees e2 ON e1.id = e2.manager_id; joins employees where e1.id = e2.manager_id, selecting distinct managers correctly.

3. Final Answer:

   SELECT DISTINCT e1.name FROM employees e1 JOIN employees e2 ON e1.id = e2.manager_id; -> Option C

4. Quick Check:

   Join on id = manager_id finds managers [OK]

Hint: Join employees where id = manager_id to find managers [OK]

Common Mistakes:

• Using LEFT JOIN with IS NULL returns non-managers
• Joining on wrong columns (manager_id = id reversed)
• Filtering on manager_id IS NOT NULL returns employees with managers