INNER JOIN in MySQL - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to select all columns from two tables using INNER JOIN.

MySQL

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

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

AFULL JOIN

BLEFT JOIN

CRIGHT JOIN

DINNER JOIN

Attempts:

3 left

💡 Hint

Common Mistakes

Using LEFT JOIN or RIGHT JOIN instead of INNER JOIN changes the result set.

Forgetting the ON clause after the join keyword.

The INNER JOIN keyword selects records that have matching values in both tables.

2fill in blank

medium

Complete the code to join tables employees and salaries on employee_id.

MySQL

SELECT employees.name, salaries.amount FROM employees [1] salaries ON employees.id = salaries.employee_id;

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

AINNER JOIN

BLEFT JOIN

CRIGHT JOIN

DFULL JOIN

Attempts:

3 left

💡 Hint

Common Mistakes

Using the wrong join type.

Joining on wrong columns.

Use INNER JOIN to get matching employee names and their salaries by joining on employee ID.

3fill in blank

hard

Fix the error in the INNER JOIN syntax to correctly join customers and orders on customer_id.

MySQL

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

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

AINNER JOIN ON

BINNER JOIN

CJOIN ON

DINNER JOIN WHERE

Attempts:

3 left

💡 Hint

Common Mistakes

Omitting ON keyword.

Using WHERE instead of ON for join condition.

The correct syntax requires INNER JOIN followed by the table and ON for the join condition.

4fill in blank

hard

Fill both blanks to join products and categories on category_id and select product name and category name.

MySQL

SELECT products.name, categories.[1] FROM products [2] categories ON products.category_id = categories.id;

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

Aname

BINNER JOIN

Ccategory_name

DLEFT JOIN

Attempts:

3 left

💡 Hint

Common Mistakes

Selecting wrong column from categories.

Using LEFT JOIN instead of INNER JOIN.

We select category_name from categories and use INNER JOIN to join on category_id.

5fill in blank

hard

Fill all three blanks to join orders, customers, and payments to select order id, customer name, and payment amount.

MySQL

SELECT orders.id, customers.[1], payments.[2] FROM orders [3] customers ON orders.customer_id = customers.id INNER JOIN payments ON orders.id = payments.order_id;

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

Aname

Bamount

CINNER JOIN

DLEFT JOIN

Attempts:

3 left

💡 Hint

Common Mistakes

Using LEFT JOIN instead of INNER JOIN.

Selecting wrong column names.

Select customer name and payment amount. Use INNER JOIN to join orders and customers.

Practice

1. What does an INNER JOIN do in MySQL?

easy

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

B. Returns all rows from the first table regardless of matches.

C. Returns all rows from the second table regardless of matches.

D. Combines rows even if there is no matching value in either table.

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 definition

   Combines rows from two tables where matching values exist in both tables. correctly describes this behavior. Options A, C, and D describe other join types or incorrect behavior.

3. Final Answer:

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

4. Quick Check:

   INNER JOIN = matching rows only [OK]

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

Common Mistakes:

• Confusing INNER JOIN with LEFT or RIGHT JOIN
• Thinking INNER JOIN returns unmatched rows
• Assuming INNER JOIN returns all rows from one table

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

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 table1.id;

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

Solution

1. Step 1: Review INNER JOIN syntax

   The correct syntax uses INNER JOIN with ON and a single equals sign (=) for comparison.

2. Step 2: Analyze each option

   SELECT * FROM table1 INNER JOIN table2 ON table1.id = table2.id; uses correct syntax. SELECT * FROM table1 INNER JOIN table2 WHERE table1.id = table2.id; uses WHERE instead of ON, which is incorrect for JOIN conditions. SELECT * FROM table1 INNER JOIN table2 USING table1.id; uses USING with table prefix which is invalid; USING requires unqualified column names. SELECT * FROM table1 INNER 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 syntax 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 equals (=)
• Confusing USING syntax with ON syntax

3. Given two tables:
Employees(id, name, dept_id)
Departments(dept_id, dept_name)
What will this query return?

SELECT Employees.name, Departments.dept_name FROM Employees INNER JOIN Departments ON Employees.dept_id = Departments.dept_id;

medium

A. All departments with employees, even if no employees exist.

B. All employees, even if they have no department.

C. An error because dept_id is ambiguous.

D. All employees with their department names, only if department exists.

Solution

1. Step 1: Understand INNER JOIN effect on these tables

   The INNER JOIN returns rows where Employees.dept_id matches Departments.dept_id, so only employees with existing departments appear.

2. Step 2: Analyze query output

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

3. Final Answer:

   All employees with their department names, only if department exists. -> Option D

4. Quick Check:

   INNER JOIN filters to matching dept_id rows [OK]

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

Common Mistakes:

• Thinking all employees appear even without departments
• Confusing INNER JOIN with LEFT JOIN
• Assuming ambiguous column error without aliasing

4. Identify the error in this query:

SELECT e.name, d.dept_name FROM Employees e INNER JOIN Departments d ON e.dept_id = d.id;

medium

A. Missing WHERE clause for join condition.

B. Column d.id does not exist; should be d.dept_id.

C. Alias 'e' is not allowed in INNER JOIN.

D. Using INNER JOIN without SELECT * causes error.

Solution

1. Step 1: Check join condition columns

   The join condition uses d.id, but Departments table has dept_id column, not id.

2. Step 2: Identify correct column name

   It should be d.dept_id to match Employees.dept_id. Other options are incorrect because aliasing is allowed, WHERE is not needed for JOIN, and SELECT * is not mandatory.

3. Final Answer:

   Column d.id does not exist; should be d.dept_id. -> Option B

4. Quick Check:

   Join columns must exist and match [OK]

Hint: Check column names in JOIN condition carefully [OK]

Common Mistakes:

• Using wrong column names in ON clause
• Thinking WHERE is required for JOIN
• Believing aliases are not allowed

5. You have two tables:
Orders(order_id, customer_id, amount)
Customers(customer_id, name)
You want to find customers who placed orders over 100 dollars. Which query correctly uses INNER JOIN to get customer names and order amounts over 100?

hard

A. SELECT Customers.name, Orders.amount FROM Orders INNER JOIN Customers ON Orders.customer_id = Customers.customer_id WHERE Orders.amount < 100;

B. SELECT Customers.name, Orders.amount FROM Customers LEFT JOIN Orders ON Customers.customer_id = Orders.customer_id WHERE Orders.amount > 100;

C. SELECT Customers.name, Orders.amount FROM Customers INNER JOIN Orders ON Customers.customer_id = Orders.customer_id WHERE Orders.amount > 100;

D. SELECT Customers.name, Orders.amount FROM Orders INNER JOIN Customers ON Orders.customer_id = Customers.customer_id WHERE Orders.amount >= 100;

Solution

1. Step 1: Understand the requirement

   We want customers with orders over 100 dollars, so filter Orders.amount > 100.

2. Step 2: Check INNER JOIN usage and filter

   SELECT Customers.name, Orders.amount FROM Customers INNER JOIN Orders ON Customers.customer_id = Orders.customer_id WHERE Orders.amount > 100; correctly joins Customers and Orders on customer_id and filters orders with amount > 100. SELECT Customers.name, Orders.amount FROM Customers LEFT JOIN Orders ON Customers.customer_id = Orders.customer_id WHERE Orders.amount > 100; uses LEFT JOIN instead of INNER JOIN. SELECT Customers.name, Orders.amount FROM Orders INNER JOIN Customers ON Orders.customer_id = Customers.customer_id WHERE Orders.amount >= 100; uses >= 100 which includes 100, not strictly over 100. SELECT Customers.name, Orders.amount FROM Orders INNER JOIN Customers ON Orders.customer_id = Customers.customer_id WHERE Orders.amount < 100; filters for amount < 100, opposite of requirement.

3. Final Answer:

   SELECT Customers.name, Orders.amount FROM Customers INNER JOIN Orders ON Customers.customer_id = Orders.customer_id WHERE Orders.amount > 100; -> Option C

4. Quick Check:

   INNER JOIN + filter amount > 100 = SELECT Customers.name, Orders.amount FROM Customers INNER JOIN Orders ON Customers.customer_id = Orders.customer_id WHERE Orders.amount > 100; [OK]

Hint: Use INNER JOIN with WHERE to filter matching rows [OK]

Common Mistakes:

• Using LEFT JOIN instead of INNER JOIN
• Using wrong comparison operator in WHERE
• Filtering on wrong table or column