INNER JOIN with ON condition in SQL - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to join two tables on the customer_id column.

SQL

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

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

Acustomer_id

Border_id

Cname

Ddate

Attempts:

3 left

💡 Hint

Common Mistakes

Using a column that does not exist in both tables.

Joining on order_id instead of customer_id.

The INNER JOIN uses the customer_id column to match rows from both tables.

2fill in blank

medium

Complete the code to select product names and order quantities by joining orders and products tables.

SQL

SELECT products.product_name, orders.quantity FROM orders INNER JOIN products ON orders.[1] = products.product_id;

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

Aquantity

Bproduct_id

Corder_id

Dcustomer_id

Attempts:

3 left

💡 Hint

Common Mistakes

Joining on order_id instead of product_id.

Using quantity as join condition.

The join matches orders.product_id with products.product_id to get product details for each order.

3fill in blank

hard

Fix the error in the join condition to correctly join employees and departments tables.

SQL

SELECT employees.name, departments.department_name FROM employees INNER JOIN departments ON employees.[1] = departments.dept_id;

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

Adepartment

Bdept_id

Cdepartment_id

Demployee_id

Attempts:

3 left

💡 Hint

Common Mistakes

Using employee_id instead of department_id.

Using department instead of department_id.

The correct join column in employees is department_id, which matches departments.dept_id.

4fill in blank

hard

Fill both blanks to join sales and customers tables and select sale_id and customer name.

SQL

SELECT sales.sale_id, customers.[1] FROM sales INNER JOIN customers ON sales.[2] = customers.customer_id;

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

Aname

Bsale_id

Ccustomer_id

Ddate

Attempts:

3 left

💡 Hint

Common Mistakes

Using sale_id instead of name for the first blank.

Joining on sale_id instead of customer_id.

The query selects the customer's name and joins on sales.customer_id matching customers.customer_id.

5fill in blank

hard

Fill all three blanks to join orders, customers, and products tables and select order_id, customer name, and product name.

SQL

SELECT orders.[1], customers.[2], products.[3] FROM orders INNER JOIN customers ON orders.customer_id = customers.customer_id INNER JOIN products ON orders.product_id = products.product_id;

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

Aorder_id

Bname

Cproduct_name

Dquantity

Attempts:

3 left

💡 Hint

Common Mistakes

Selecting quantity instead of product_name.

Using wrong column names for customer or product.

The query selects order_id from orders, name from customers, and product_name from products.

Practice

1. What does an INNER JOIN do in SQL when used with an ON condition?

easy

A. It returns all rows from the first table regardless of matches.

B. It returns all rows from both tables, matching or not.

C. It returns all rows from the second table regardless of matches.

D. It returns only rows where the join condition matches in both tables.

Solution

1. Step 1: Understand INNER JOIN behavior

   INNER JOIN returns rows only when the join condition matches rows in both tables.

2. Step 2: Compare with other join types

   Unlike LEFT or RIGHT JOIN, INNER JOIN excludes rows without matches.

3. Final Answer:

   It returns only rows where the join condition matches in both tables. -> Option D

4. Quick Check:

   INNER JOIN = matching rows only [OK]

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

Common Mistakes:

• Thinking INNER JOIN returns unmatched rows
• Confusing INNER JOIN with LEFT JOIN
• Ignoring the ON condition effect

2. Which of the following is the correct syntax for an INNER JOIN with an ON condition between tables Employees and Departments on DepartmentID?

easy

A. SELECT * FROM Employees INNER JOIN Departments ON Employees.DepartmentID = Departments.DepartmentID;

B. SELECT * FROM Employees INNER JOIN Departments ON Employees.ID = Departments.ID;

C. SELECT * FROM Employees JOIN Departments USING DepartmentID;

D. SELECT * FROM Employees INNER JOIN Departments WHERE Employees.DepartmentID = Departments.DepartmentID;

Solution

1. Step 1: Identify correct INNER JOIN syntax

   The INNER JOIN requires the ON keyword followed by the join condition.

2. Step 2: Check the join condition correctness

   The join should be on Employees.DepartmentID = Departments.DepartmentID, not just ID or WHERE clause.

3. Final Answer:

   SELECT * FROM Employees INNER JOIN Departments ON Employees.DepartmentID = Departments.DepartmentID; -> Option A

4. Quick Check:

   INNER JOIN uses ON with condition [OK]

Hint: Use ON keyword with join condition for INNER JOIN [OK]

Common Mistakes:

• Using WHERE instead of ON for join condition
• Joining on wrong columns
• Omitting ON keyword

3. Given these tables:

Employees
ID | Name | DeptID
1 | Alice | 10
2 | Bob | 20
3 | Carol | 30

Departments
DeptID | DeptName
10 | Sales
20 | HR
40 | IT

What is the result of this query?

SELECT Employees.Name, Departments.DeptName FROM Employees INNER JOIN Departments ON Employees.DeptID = Departments.DeptID;

medium

A. [('Alice', 'Sales'), ('Bob', 'HR')]

B. [('Alice', 'Sales'), ('Bob', 'HR'), ('Carol', 'IT')]

C. [('Alice', 'Sales'), ('Carol', 'IT')]

D. [('Bob', 'HR'), ('Carol', 'IT')]

Solution

1. Step 1: Match Employees.DeptID with Departments.DeptID

   Employees have DeptIDs 10, 20, 30; Departments have 10, 20, 40.

2. Step 2: Find matching DeptIDs

   Matches are 10 and 20 only; 30 (Carol) has no matching department.

3. Final Answer:

   [('Alice', 'Sales'), ('Bob', 'HR')] -> Option A

4. Quick Check:

   INNER JOIN returns only matching DeptIDs [OK]

Hint: Only rows with matching DeptID appear in INNER JOIN result [OK]

Common Mistakes:

• Including unmatched rows like Carol's
• Confusing DeptID with ID
• Assuming all departments join

4. Consider this query:

SELECT e.Name, d.DeptName FROM Employees e INNER JOIN Departments d ON e.DeptID = d.ID;

Given Departments has column DeptID but no ID column, what is the issue?

medium

A. The query will join on wrong columns but still return rows.

B. The query will run but return no rows.

C. The query will cause a syntax error due to wrong column name.

D. The query will join correctly because aliases fix column names.

Solution

1. Step 1: Check column names in ON condition

   The query uses d.ID but Departments table has DeptID, not ID.

2. Step 2: Understand SQL error on invalid column

   Referencing a non-existent column causes a syntax or runtime error.

3. Final Answer:

   The query will cause a syntax error due to wrong column name. -> Option C

4. Quick Check:

   Wrong column in ON causes error [OK]

Hint: Verify column names in ON condition to avoid errors [OK]

Common Mistakes:

• Assuming aliases rename columns automatically
• Using wrong column names in ON clause
• Expecting empty result instead of error

5. You have two tables:

Orders
OrderID | CustomerID | Amount
1 | 101 | 50
2 | 102 | 75
3 | 103 | 100
4 | 101 | 25

Customers
CustomerID | Name
101 | John
102 | Jane
104 | Mike

Write an INNER JOIN query to find total order amount per customer name, including only customers with orders. Which query is correct?

hard

A. SELECT c.Name, SUM(o.Amount) FROM Customers c INNER JOIN Orders o ON c.CustomerID = o.CustomerID GROUP BY c.CustomerID;

B. SELECT c.Name, SUM(o.Amount) FROM Orders o INNER JOIN Customers c ON o.CustomerID = c.CustomerID GROUP BY c.Name;

C. SELECT c.Name, SUM(o.Amount) FROM Customers c LEFT JOIN Orders o ON c.CustomerID = o.CustomerID GROUP BY c.Name;

D. SELECT c.Name, SUM(o.Amount) FROM Orders o INNER JOIN Customers c ON o.CustomerID = c.CustomerID GROUP BY o.CustomerID;

Solution

1. Step 1: Understand requirement - total per customer with orders only

   INNER JOIN keeps only customers with matching orders; GROUP BY customer name to sum amounts.

2. Step 2: Check query correctness

   SELECT c.Name, SUM(o.Amount) FROM Orders o INNER JOIN Customers c ON o.CustomerID = c.CustomerID GROUP BY c.Name; joins Orders to Customers on CustomerID and groups by c.Name, matching requirement exactly.

3. Step 3: Compare other options

   SELECT c.Name, SUM(o.Amount) FROM Customers c INNER JOIN Orders o ON c.CustomerID = o.CustomerID GROUP BY c.CustomerID; groups by CustomerID but c.Name not grouped (SQL error). SELECT c.Name, SUM(o.Amount) FROM Customers c LEFT JOIN Orders o ON c.CustomerID = o.CustomerID GROUP BY c.Name; uses LEFT JOIN (includes customers without orders). SELECT c.Name, SUM(o.Amount) FROM Orders o INNER JOIN Customers c ON o.CustomerID = c.CustomerID GROUP BY o.CustomerID; groups by o.CustomerID (not customer name).

4. Final Answer:

   SELECT c.Name, SUM(o.Amount) FROM Orders o INNER JOIN Customers c ON o.CustomerID = c.CustomerID GROUP BY c.Name; -> Option B

5. Quick Check:

   INNER JOIN with GROUP BY customer name sums orders [OK]

Hint: Join Orders to Customers, group by customer name for totals [OK]

Common Mistakes:

• Using LEFT JOIN includes customers without orders
• Grouping by wrong column
• Joining tables in wrong order causing confusion