INNER JOIN syntax in SQL - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

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

SQL

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

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

AINNER JOIN

BOUTER JOIN

CLEFT JOIN

DJOIN

Attempts:

3 left

💡 Hint

Common Mistakes

Using LEFT JOIN instead of INNER JOIN returns unmatched rows too.

Using JOIN alone may work but INNER JOIN is clearer.

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

2fill in blank

medium

Complete the code to join orders with customers on customer_id.

SQL

SELECT orders.id, customers.name FROM orders [1] customers ON orders.customer_id = customers.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 LEFT JOIN returns all orders even without customers.

Using FULL JOIN returns unmatched rows from both tables.

INNER JOIN returns only orders that have a matching customer.

3fill in blank

hard

Fix the error in the INNER JOIN syntax to correctly join products and categories.

SQL

SELECT products.name, categories.name FROM products [1] categories ON products.category_id = categories.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 causes syntax error.

Using WHERE instead of ON for join condition is incorrect.

The INNER JOIN must be followed by ON to specify the join condition.

4fill in blank

hard

Fill both blanks to join sales and stores on store_id and select sale amount and store name.

SQL

SELECT sales.amount, stores.[1] FROM sales [2] stores ON sales.store_id = stores.id;

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

Aname

BINNER JOIN

CLEFT JOIN

Dlocation

Attempts:

3 left

💡 Hint

Common Mistakes

Selecting a wrong column from stores.

Using LEFT JOIN instead of INNER JOIN changes the result.

We select the name column from stores and use INNER JOIN to combine matching rows.

5fill in blank

hard

Fill all three blanks to join employees and salaries on employee_id and select employee name, salary, and department.

SQL

SELECT employees.[1], salaries.[2], employees.[3] FROM employees INNER JOIN salaries ON employees.id = salaries.employee_id;

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

AINNER JOIN

Bsalary

Cname

Ddepartment

Attempts:

3 left

💡 Hint

Common Mistakes

Mixing up column names or tables.

Using wrong join type or missing ON keyword.

We select name and department from employees, salary from salaries, and use INNER JOIN to combine matching rows.

Practice

1. What does an INNER JOIN do in SQL?

easy

A. It returns all rows from the first table only.

B. It returns rows that have matching values in both tables.

C. It returns all rows from the second table only.

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

Solution

1. Step 1: Understand the purpose of INNER JOIN

   INNER JOIN combines rows from two tables where the join condition matches in both tables.

2. Step 2: Compare options with INNER JOIN behavior

   Only the description "It returns rows that have matching values in both tables." correctly states that it returns rows with matching values in both tables.

3. Final Answer:

   It returns rows that have matching values in both tables. -> Option B

4. Quick Check:

   INNER JOIN = matching rows only [OK]

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

Common Mistakes:

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

2. Which of the following is the correct syntax for an INNER JOIN between tables Employees and Departments on the column DeptID using the ON clause?

easy

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

B. SELECT * FROM Employees JOIN Departments WHERE Employees.DeptID = Departments.DeptID;

C. SELECT * FROM Employees INNER JOIN Departments USING DeptID;

D. SELECT * FROM Employees INNER JOIN Departments ON Employees.DeptID = Departments.DeptID;

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: Check each option

   SELECT * FROM Employees INNER JOIN Departments ON Employees.DeptID = Departments.DeptID; uses correct INNER JOIN syntax with ON and =. SELECT * FROM Employees JOIN Departments WHERE Employees.DeptID = Departments.DeptID; uses WHERE instead of ON. SELECT * FROM Employees INNER JOIN Departments USING DeptID; uses USING instead of ON. SELECT * FROM Employees INNER JOIN Departments ON Employees.DeptID == Departments.DeptID; uses double equals (==), which is not valid in SQL.

3. Final Answer:

   SELECT * FROM Employees INNER JOIN Departments ON Employees.DeptID = Departments.DeptID; -> Option D

4. Quick Check:

   INNER JOIN syntax = ON with single = [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 (=)
• Using USING instead of ON

3. Given the tables:
Employees(id, name, dept_id)
Departments(dept_id, dept_name)
What is the result of this query?

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

Assuming:
Employees: (1, 'Alice', 10), (2, 'Bob', 20), (3, 'Carol', 30)
Departments: (10, 'HR'), (20, 'Sales')

medium

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

B. [('Alice', 'HR'), ('Bob', 'Sales'), ('Carol', '30')]

C. [('Alice', 'HR'), ('Bob', 'Sales'), ('Carol', NULL)]

D. [('Alice', 'HR'), ('Bob', 'Sales'), ('Carol', 'Finance')]

Solution

1. Step 1: Identify matching rows by dept_id

   Employees with dept_id 10 and 20 match Departments with same dept_id. Carol's dept_id 30 has no match.

2. Step 2: Understand INNER JOIN output

   INNER JOIN returns only rows with matching dept_id in both tables, so Carol is excluded.

3. Final Answer:

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

4. Quick Check:

   INNER JOIN excludes unmatched rows [OK]

Hint: INNER JOIN excludes rows without matching keys [OK]

Common Mistakes:

• Including unmatched rows in result
• Assuming NULL values appear for unmatched rows
• Confusing INNER JOIN with LEFT JOIN behavior

4. Identify the error in this SQL query:

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

medium

A. The join condition uses '==' instead of '='.

B. Using alias names for tables is not allowed.

C. Missing WHERE clause for filtering.

D. INNER JOIN requires USING instead of ON.

Solution

1. Step 1: Check join condition syntax

   SQL uses a single equals sign (=) for comparison, not double equals (==).

2. Step 2: Verify other parts of the query

   Aliases e and d are valid. WHERE clause is optional. INNER JOIN can use ON.

3. Final Answer:

   The join condition uses '==' instead of '='. -> Option A

4. Quick Check:

   Use = for join conditions, not == [OK]

Hint: Use single = in ON clause, not double == [OK]

Common Mistakes:

• Using == instead of = in join condition
• Thinking aliases are disallowed
• Confusing ON with WHERE clause necessity

5. You have two tables:
Orders(order_id, customer_id, amount)
Customers(customer_id, customer_name)
You want to find all customers who have placed orders and the total amount they spent.
Which query correctly uses INNER JOIN and aggregation to get this result?

hard

A. SELECT c.customer_name, SUM(o.amount) AS total_spent FROM Customers c LEFT JOIN Orders o ON c.customer_id = o.customer_id GROUP BY c.customer_name;

B. SELECT c.customer_name, o.amount FROM Customers c INNER JOIN Orders o ON c.customer_id = o.customer_id;

C. SELECT c.customer_name, SUM(o.amount) AS total_spent FROM Customers c INNER JOIN Orders o ON c.customer_id = o.customer_id GROUP BY c.customer_name;

D. SELECT customer_name, amount FROM Customers INNER JOIN Orders ON customer_id = customer_id;

Solution

1. Step 1: Understand the requirement

   We want customers who placed orders and the total amount spent, so INNER JOIN with aggregation is needed.

2. Step 2: Analyze each option

   SELECT c.customer_name, SUM(o.amount) AS total_spent FROM Customers c INNER JOIN Orders o ON c.customer_id = o.customer_id GROUP BY c.customer_name; correctly uses INNER JOIN on customer_id and sums amounts grouped by customer_name. SELECT c.customer_name, SUM(o.amount) AS total_spent FROM Customers c LEFT JOIN Orders o ON c.customer_id = o.customer_id GROUP BY c.customer_name; uses LEFT JOIN, which includes customers without orders. SELECT c.customer_name, o.amount FROM Customers c INNER JOIN Orders o ON c.customer_id = o.customer_id; does not aggregate amounts. SELECT customer_name, amount FROM Customers INNER JOIN Orders ON customer_id = customer_id; has ambiguous join condition and no aggregation.

3. Final Answer:

   SELECT c.customer_name, SUM(o.amount) AS total_spent FROM Customers c INNER JOIN Orders o ON c.customer_id = o.customer_id GROUP BY c.customer_name; -> Option C

4. Quick Check:

   INNER JOIN + GROUP BY + SUM = total spent per customer [OK]

Hint: Use INNER JOIN with GROUP BY and SUM for totals [OK]

Common Mistakes:

• Using LEFT JOIN instead of INNER JOIN when only matching rows needed
• Forgetting GROUP BY with aggregation
• Incorrect join condition syntax