LEFT JOIN in MySQL - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to select all customers and their orders using a LEFT JOIN.

MySQL

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

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

ALEFT JOIN

BINNER JOIN

CRIGHT JOIN

DFULL JOIN

Attempts:

3 left

💡 Hint

Common Mistakes

Using INNER JOIN will exclude customers without orders.

RIGHT JOIN returns all orders even if no customer matches.

The LEFT JOIN returns all customers even if they have no orders.

2fill in blank

medium

Complete the code to get all employees and their department names, including employees without a department.

MySQL

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

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

ARIGHT JOIN

BINNER JOIN

CCROSS JOIN

DLEFT JOIN

Attempts:

3 left

💡 Hint

Common Mistakes

Using INNER JOIN excludes employees without departments.

RIGHT JOIN would include all departments, not all employees.

LEFT JOIN ensures all employees appear, even those without a department.

3fill in blank

hard

Fix the error in the LEFT JOIN query to include all products and their categories.

MySQL

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'

ARIGHT JOIN

BINNER JOIN

CLEFT JOIN

DFULL JOIN

Attempts:

3 left

💡 Hint

Common Mistakes

Using INNER JOIN excludes products without categories.

FULL JOIN is not supported in MySQL.

LEFT JOIN returns all products, even those without a category.

4fill in blank

hard

Fill both blanks to select all students and their grades, including students without grades.

MySQL

SELECT students.name, grades.score FROM students [1] grades ON students.id [2] grades.student_id;

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

ALEFT JOIN

B=

C!=

DRIGHT JOIN

Attempts:

3 left

💡 Hint

Common Mistakes

Using '!=' in ON condition breaks the join logic.

RIGHT JOIN would keep all grades, not all students.

LEFT JOIN keeps all students, and the ON condition uses '=' to match student IDs.

5fill in blank

hard

Fill all three blanks to select all orders, their customers, and order dates, including orders without customers.

MySQL

SELECT orders.id, [1], orders.order_date FROM orders [2] customers ON orders.customer_id [3] customers.id;

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

Acustomers.name

BLEFT JOIN

C=

DINNER JOIN

Attempts:

3 left

💡 Hint

Common Mistakes

Using INNER JOIN excludes orders without customers.

Using '!=' in ON condition breaks the join.

LEFT JOIN keeps all orders, matching customers by customer_id with '='.

Practice

1. What does a LEFT JOIN do in SQL?

easy

A. Deletes rows from the left table that do not have matches in the right table.

B. Returns only rows that have matching values in both tables.

C. Returns all rows from the right table and matching rows from the left table.

D. Returns all rows from the left table and matching rows from the right table, NULL if no match.

Solution

1. Step 1: Understand LEFT JOIN behavior

   A LEFT JOIN keeps all rows from the left table regardless of matches in the right table.

2. Step 2: Check what happens with unmatched rows

   If no matching row exists in the right table, the result shows NULL for right table columns.

3. Final Answer:

   Returns all rows from the left table and matching rows from the right table, NULL if no match. -> Option D

4. Quick Check:

   LEFT JOIN = all left rows + matched right rows [OK]

Hint: LEFT JOIN keeps all left rows, fills right with NULL if no match [OK]

Common Mistakes:

• Confusing LEFT JOIN with INNER JOIN
• Thinking it returns all right table rows
• Assuming unmatched rows are dropped

2. Which of the following is the correct syntax for a LEFT JOIN in MySQL?

easy

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

B. SELECT * FROM table1 JOIN LEFT table2 ON table1.id = table2.id;

C. SELECT * FROM table1 LEFT OUTER JOIN table2 WHERE table1.id = table2.id;

D. SELECT * FROM table1 LEFT JOIN table2 WHERE table1.id = table2.id;

Solution

1. Step 1: Review standard LEFT JOIN syntax

   The correct syntax uses LEFT JOIN followed by ON condition to specify matching columns.

2. Step 2: Check each option

   SELECT * FROM table1 LEFT JOIN table2 ON table1.id = table2.id; uses correct syntax: LEFT JOIN with ON clause. SELECT * FROM table1 JOIN LEFT table2 ON table1.id = table2.id; has JOIN LEFT which is invalid. SELECT * FROM table1 LEFT OUTER JOIN table2 WHERE table1.id = table2.id; uses WHERE instead of ON, which is incorrect for JOIN condition. SELECT * FROM table1 LEFT JOIN table2 WHERE table1.id = table2.id; uses WHERE instead of ON, invalid syntax.

3. Final Answer:

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

4. Quick Check:

   LEFT JOIN syntax = LEFT JOIN ... ON ... [OK]

Hint: Use LEFT JOIN ... ON ... for correct syntax [OK]

Common Mistakes:

• Using JOIN LEFT instead of LEFT JOIN
• Using WHERE instead of ON for join condition
• Putting join condition directly in WHERE after table

3. Given these tables:

Employees(id, name)
Departments(id, dept_name, emp_id)

What is the output of this query?

SELECT e.name, d.dept_name FROM Employees e LEFT JOIN Departments d ON e.id = d.emp_id ORDER BY e.id;

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

medium

A. [('Alice', NULL), ('Bob', NULL), ('Carol', NULL)]

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

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

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

Solution

1. Step 1: Match Employees with Departments using LEFT JOIN

   LEFT JOIN keeps all Employees. For each employee, it tries to find a matching department where e.id = d.emp_id.

2. Step 2: Check matches and NULLs

   'Alice' (id=1) matches department 'HR' (emp_id=1). 'Bob' (id=2) has no matching department, so dept_name is NULL. 'Carol' (id=3) matches 'IT' (emp_id=3).

3. Final Answer:

   [('Alice', 'HR'), ('Bob', NULL), ('Carol', 'IT')] -> Option C

4. Quick Check:

   LEFT JOIN keeps all left rows, unmatched right columns NULL [OK]

Hint: LEFT JOIN shows all left rows; unmatched right columns are NULL [OK]

Common Mistakes:

• Assuming unmatched rows are dropped
• Mixing up emp_id and id columns
• Ignoring ORDER BY effect

4. Identify the error in this LEFT JOIN query:

SELECT a.id, b.value FROM tableA a LEFT JOIN tableB b ON a.id = b.id WHERE b.value > 10;

medium

A. The WHERE clause filters out rows where b.value is NULL, defeating LEFT JOIN purpose.

B. The ON clause is missing a join condition.

C. LEFT JOIN should be replaced with INNER JOIN for this query.

D. The SELECT statement is missing table aliases.

Solution

1. Step 1: Understand effect of WHERE on LEFT JOIN

   LEFT JOIN keeps all rows from tableA, but WHERE b.value > 10 excludes rows where b.value is NULL (no match), effectively turning it into INNER JOIN.

2. Step 2: Identify the error

   The error is using WHERE to filter on right table columns after LEFT JOIN, which removes unmatched rows.

3. Final Answer:

   The WHERE clause filters out rows where b.value is NULL, defeating LEFT JOIN purpose. -> Option A

4. Quick Check:

   WHERE on right table column after LEFT JOIN removes unmatched rows [OK]

Hint: Use ON for filtering right table, not WHERE, to keep unmatched rows [OK]

Common Mistakes:

• Using WHERE instead of ON for right table filters
• Confusing LEFT JOIN with INNER JOIN
• Ignoring NULLs in join columns

5. You have two tables:

Products(product_id, name)
Sales(sale_id, product_id, quantity)

You want to list all products with total quantity sold, including products with zero sales.
Which query correctly uses LEFT JOIN to achieve this?

hard

A. SELECT p.name, SUM(s.quantity) FROM Products p INNER JOIN Sales s ON p.product_id = s.product_id GROUP BY p.product_id;

B. SELECT p.name, SUM(s.quantity) FROM Products p LEFT JOIN Sales s ON p.product_id = s.product_id GROUP BY p.product_id;

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

D. SELECT p.name, COUNT(s.sale_id) FROM Products p LEFT JOIN Sales s ON p.product_id = s.product_id WHERE s.quantity > 0 GROUP BY p.product_id;

Solution

1. Step 1: Use LEFT JOIN from Products to Sales

   To include all products even with no sales, LEFT JOIN Sales to Products on product_id.

2. Step 2: Aggregate sales quantity and group by product

   Use SUM(s.quantity) to total sales per product. Group by p.product_id to get one row per product.

3. Final Answer:

   SELECT p.name, SUM(s.quantity) FROM Products p LEFT JOIN Sales s ON p.product_id = s.product_id GROUP BY p.product_id; -> Option B

4. Quick Check:

   LEFT JOIN + SUM + GROUP BY lists all products with total sales [OK]

Hint: LEFT JOIN products to sales, SUM quantity, GROUP BY product [OK]

Common Mistakes:

• Using INNER JOIN excludes products with zero sales
• Joining Sales to Products reverses LEFT JOIN logic
• Filtering in WHERE removes unmatched rows