INNER JOIN execution in PostgreSQL - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to select all columns from two tables joined on the "id" column.

PostgreSQL

SELECT * FROM employees [1] departments ON employees.id = departments.id;

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

AINNER JOIN

BRIGHT JOIN

CLEFT JOIN

DFULL JOIN

Attempts:

3 left

💡 Hint

Common Mistakes

Using LEFT JOIN instead of INNER JOIN returns unmatched rows from the left table.

Using FULL JOIN returns all rows from both tables, not just matches.

The INNER JOIN returns rows when there is a match in both tables based on the join condition.

2fill in blank

medium

Complete the code to join "orders" and "customers" tables on the "customer_id" column.

PostgreSQL

SELECT orders.order_id, customers.name FROM orders [1] customers ON orders.customer_id = customers.customer_id;

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

ALEFT JOIN

BCROSS JOIN

CRIGHT JOIN

DINNER JOIN

Attempts:

3 left

💡 Hint

Common Mistakes

Using CROSS JOIN returns all combinations, not just matches.

LEFT JOIN includes unmatched rows from the left table.

Using INNER JOIN ensures only orders with matching customers are selected.

3fill in blank

hard

Fix the error in the join condition to correctly join "products" and "categories" on "category_id".

PostgreSQL

SELECT * FROM products INNER JOIN categories ON products.[1] = categories.category_id;

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

Acategory

Bcategory_id

Cid

Dproduct_id

Attempts:

3 left

💡 Hint

Common Mistakes

Using a wrong column like 'category' or 'product_id' causes join errors or wrong results.

Using 'id' instead of 'category_id' mismatches the join keys.

The join condition must use the correct column category_id from the products table to match categories.

4fill in blank

hard

Fill both blanks to create a join that selects employee names and their department names.

PostgreSQL

SELECT employees.name, departments.[1] FROM employees [2] departments ON employees.department_id = departments.id;

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

Aname

BLEFT JOIN

CINNER JOIN

Dsalary

Attempts:

3 left

💡 Hint

Common Mistakes

Using LEFT JOIN returns employees without departments.

Selecting 'salary' instead of 'name' shows wrong data.

The query selects the department name and uses an INNER JOIN to get matching employees and departments.

5fill in blank

hard

Fill all three blanks to join "sales" and "stores" tables and select store name and total sales.

PostgreSQL

SELECT stores.[1], SUM(sales.amount) AS total_sales FROM sales [2] stores ON sales.[3] = stores.id GROUP BY stores.name;

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

Aname

BINNER JOIN

Cstore_id

Ddate

Attempts:

3 left

💡 Hint

Common Mistakes

Using LEFT JOIN includes sales without matching stores.

Joining on 'date' instead of 'store_id' causes wrong matches.

The query selects the store name, joins with INNER JOIN on store_id to match sales to stores, and sums sales amounts.

Practice

1. What does an INNER JOIN do in PostgreSQL?

easy

A. Combines rows from two tables where matching column values exist in both tables.

B. Returns all rows from the first table, and matching rows from the second table.

C. Returns all rows from both tables, matching or not.

D. Deletes rows from both tables where values match.

Solution

1. Step 1: Understand INNER JOIN purpose

   INNER JOIN returns rows only when there is a match in both tables based on the join condition.

2. Step 2: Compare options with INNER JOIN behavior

   Combines rows from two tables where matching column values exist in both tables. correctly describes this behavior. Options B and C describe LEFT JOIN and FULL OUTER JOIN respectively, and D is unrelated.

3. Final Answer:

   Combines rows from two tables where matching column values exist in both tables. -> Option A

4. Quick Check:

   INNER JOIN = matched rows only [OK]

Hint: INNER JOIN returns only rows with matches in both tables [OK]

Common Mistakes:

• Confusing INNER JOIN with LEFT JOIN or FULL JOIN
• Thinking INNER JOIN returns unmatched rows
• Assuming INNER JOIN deletes data

2. Which of the following is the correct syntax for an INNER JOIN in PostgreSQL?

easy

A. SELECT * FROM table1 INNER JOIN table2 ON table1.id = table2.id;

B. SELECT * FROM table1 INNER JOIN table2 WHERE table1.id = table2.id;

C. SELECT * FROM table1 INNER JOIN table2 USING (id);

D. SELECT * FROM table1 JOIN table2 ON table1.id == table2.id;

Solution

1. Step 1: Review INNER JOIN syntax

   The correct syntax uses INNER JOIN with ON clause specifying the join condition using a single equals sign (=).

2. Step 2: Check each option

   SELECT * FROM table1 INNER JOIN table2 ON table1.id = table2.id; uses correct INNER JOIN syntax with ON and =. SELECT * FROM table1 INNER JOIN table2 WHERE table1.id = table2.id; missing ON or USING clause, syntax error. SELECT * FROM table1 INNER JOIN table2 USING (id); uses correct USING syntax with parentheses. SELECT * FROM table1 JOIN table2 ON table1.id == table2.id; uses double equals (==) which is invalid in SQL.

3. Final Answer:

   SELECT * FROM table1 INNER JOIN table2 ON table1.id = table2.id; -> Option A

4. Quick Check:

   INNER JOIN uses ON with = [OK]

Hint: Use ON with single = for INNER JOIN conditions [OK]

Common Mistakes:

• Using WHERE instead of ON for join condition
• Using double equals (==) instead of single =
• Confusing USING clause applicability

3. Given the tables:
employees(id, name, dept_id)
departments(id, dept_name)
What will this query return?

SELECT e.name, d.dept_name FROM employees e INNER JOIN departments d ON e.dept_id = d.id;

medium

A. All employees, even if they have no matching department.

B. An error because of alias usage.

C. All departments with employee names, even if no employees exist.

D. All employees with their department names, only if dept_id matches a department id.

Solution

1. Step 1: Understand INNER JOIN effect on employees and departments

   The INNER JOIN returns rows where employees.dept_id matches departments.id, so only employees with a valid department appear.

2. Step 2: Analyze query output

   The query selects employee names and their department names for matching rows only. Employees without a matching department are excluded.

3. Final Answer:

   All employees with their department names, only if dept_id matches a department id. -> Option D

4. Quick Check:

   INNER JOIN filters unmatched rows [OK]

Hint: INNER JOIN shows only matching rows from both tables [OK]

Common Mistakes:

• Assuming unmatched employees appear
• Confusing aliases causing errors
• Thinking departments without employees appear

4. Identify the error in this INNER JOIN query:

SELECT * FROM orders INNER JOIN customers ON orders.customer_id = customers.id WHERE orders.date > '2023-01-01'

medium

A. Missing table alias causing ambiguity.

B. WHERE clause cannot be used with INNER JOIN.

C. No error; query is valid and will run correctly.

D. Incorrect join condition syntax; should use USING instead of ON.

Solution

1. Step 1: Check JOIN syntax and WHERE clause

   The INNER JOIN uses ON with a valid condition. The WHERE clause filters rows after the join, which is allowed.

2. Step 2: Verify if any syntax or semantic errors exist

   There is no ambiguity or syntax error. Aliases are optional. USING is an alternative but not required.

3. Final Answer:

   No error; query is valid and will run correctly. -> Option C

4. Quick Check:

   INNER JOIN with ON and WHERE is valid [OK]

Hint: WHERE filters after JOIN; no syntax error here [OK]

Common Mistakes:

• Thinking WHERE cannot be used with JOIN
• Confusing USING and ON as mandatory
• Assuming aliases are always required

5. You have two tables:
products(id, name)
sales(product_id, quantity)
Write a query to find product names with total quantity sold, but only for products that have sales.
Which query correctly uses INNER JOIN to achieve this?

hard

A. SELECT p.name, SUM(s.quantity) FROM sales s LEFT JOIN products p ON s.product_id = p.id GROUP BY p.name;

B. SELECT p.name, SUM(s.quantity) FROM products p INNER JOIN sales s ON p.id = s.product_id GROUP BY p.name;

C. SELECT p.name, SUM(s.quantity) FROM products p LEFT JOIN sales s ON p.id = s.product_id GROUP BY p.name;

D. SELECT p.name, SUM(s.quantity) FROM products p INNER JOIN sales s ON p.id = s.product_id WHERE p.name > 0 GROUP BY p.name;

Solution

1. Step 1: Understand requirement for products with sales only

   INNER JOIN returns only products that have matching sales records, excluding products without sales.

2. Step 2: Analyze each query

   SELECT p.name, SUM(s.quantity) FROM sales s LEFT JOIN products p ON s.product_id = p.id GROUP BY p.name; uses LEFT JOIN, includes unmatched sales (NULL p.name). SELECT p.name, SUM(s.quantity) FROM products p LEFT JOIN sales s ON p.id = s.product_id GROUP BY p.name; uses LEFT JOIN, includes products without sales. SELECT p.name, SUM(s.quantity) FROM products p INNER JOIN sales s ON p.id = s.product_id GROUP BY p.name; uses INNER JOIN correctly. SELECT p.name, SUM(s.quantity) FROM products p INNER JOIN sales s ON p.id = s.product_id WHERE p.name > 0 GROUP BY p.name; WHERE p.name > 0 causes type error.

3. Step 3: Choose best answer matching INNER JOIN usage

   SELECT p.name, SUM(s.quantity) FROM products p INNER JOIN sales s ON p.id = s.product_id GROUP BY p.name; is correct.

4. Final Answer:

   SELECT p.name, SUM(s.quantity) FROM products p INNER JOIN sales s ON p.id = s.product_id GROUP BY p.name; -> Option B

5. Quick Check:

   INNER JOIN + GROUP BY sums sales per product [OK]

Hint: Use INNER JOIN and GROUP BY to sum only matching sales [OK]

Common Mistakes:

• Using LEFT JOIN includes products without sales
• Forgetting GROUP BY causes errors
• Confusing table order in JOIN