UPDATE with JOIN in MySQL - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ query_result

intermediate

What is the output after this UPDATE with JOIN?

Given two tables, employees and departments, the query updates employee salaries by adding 1000 for those in the 'Sales' department. What will be the salary of employee with id=3 after running this?

MySQL

UPDATE employees e
JOIN departments d ON e.department_id = d.id
SET e.salary = e.salary + 1000
WHERE d.name = 'Sales';

-- employees table before update:
-- id | name    | department_id | salary
-- 1  | Alice   | 1             | 5000
-- 2  | Bob     | 2             | 6000
-- 3  | Charlie | 1             | 5500

-- departments table:
-- id | name
-- 1  | Sales
-- 2  | HR

A 5500

B 6500

C 6000

D SyntaxError

💡 Hint

Check which employees belong to the 'Sales' department and how their salary changes.

The employee with id=3 is in department 1, which is 'Sales'. The query adds 1000 to their salary of 5500, resulting in 6500.

📝 Syntax

intermediate

Which UPDATE with JOIN syntax is correct in MySQL?

Choose the correct syntax to update a table orders by joining with customers to set orders.status to 'shipped' where customers.region is 'West'.

A UPDATE orders SET status = 'shipped' WHERE customer_id IN (SELECT id FROM customers WHERE region = 'West');

B UPDATE orders SET status = 'shipped' FROM orders o JOIN customers c ON o.customer_id = c.id WHERE c.region = 'West';

C UPDATE orders o, customers c SET o.status = 'shipped' WHERE o.customer_id = c.id AND c.region = 'West';

D UPDATE orders o JOIN customers c ON o.customer_id = c.id SET status = 'shipped' WHERE c.region = 'West';

💡 Hint

MySQL uses JOIN syntax directly in UPDATE, not FROM.

Option D uses the correct MySQL syntax with JOIN in UPDATE. Option D uses invalid FROM clause. Option D uses comma syntax instead of JOIN. Option D uses subquery, not JOIN.

❓ optimization

advanced

How to optimize this UPDATE with JOIN for large tables?

You have a large products table and categories table. You want to update products.discount to 10 where categories.name is 'Clearance'. The current query is:

UPDATE products p
JOIN categories c ON p.category_id = c.id
SET p.discount = 10
WHERE c.name = 'Clearance';

Which optimization is best to improve performance?

A Add an index on categories.name and products.category_id columns.

B Rewrite the query using a subquery instead of JOIN.

C Remove the WHERE clause to update all products.

D Use a LEFT JOIN instead of JOIN.

💡 Hint

Indexes help speed up JOIN and WHERE conditions.

Adding indexes on the join and filter columns helps MySQL quickly find matching rows, improving update speed. Other options either reduce precision or do not improve performance.

🔧 Debug

advanced

Why does this UPDATE with JOIN fail with 'Unknown column' error?

Consider this query:

UPDATE orders o
JOIN customers c ON o.customer_id = c.id
SET status = 'delivered'
WHERE region = 'East';

It raises an error: Unknown column 'region' in 'where clause'. Why?

A Because 'region' column does not exist in orders table and is not prefixed with 'c.'.

B Because 'region' is ambiguous and needs table alias prefix.

C Because JOIN syntax is invalid in UPDATE.

D Because SET clause must come after WHERE clause.

💡 Hint

Check which table 'region' belongs to and how it is referenced.

The WHERE clause references 'region' without specifying the table alias. Since 'region' is in customers table, it must be referenced as 'c.region'. Without prefix, MySQL looks for 'region' in orders table and fails.

🧠 Conceptual

expert

What is the effect of this UPDATE with JOIN query?

Given tables students and scores, what does this query do?

UPDATE students s
JOIN scores sc ON s.id = sc.student_id
SET s.status = 'passed'
WHERE sc.score >= 60;

A Raises an error because UPDATE with JOIN cannot use WHERE on joined table.

B Marks all students as 'passed' regardless of scores.

C Marks students as 'passed' if they have at least one score 60 or above.

D Marks students as 'passed' only if all their scores are 60 or above.

💡 Hint

Think about how JOIN and WHERE filter rows for update.

The JOIN matches students with their scores. The WHERE filters scores >= 60. So students with at least one such score get updated. It does not check all scores.

Practice

1. What is the main purpose of using UPDATE with JOIN in MySQL?

easy

A. To select data from multiple tables without updating

B. To delete rows from two tables at once

C. To update rows in one table based on matching rows in another table

D. To create a new table by joining two tables

Solution

1. Step 1: Understand the purpose of UPDATE with JOIN

   UPDATE with JOIN is used to change data in one table using related data from another table by matching rows.

2. Step 2: Compare options

   Only To update rows in one table based on matching rows in another table describes updating one table based on another, which matches the purpose.

3. Final Answer:

   To update rows in one table based on matching rows in another table -> Option C

4. Quick Check:

   UPDATE with JOIN = update one table using another [OK]

Hint: UPDATE with JOIN updates one table using another table's data [OK]

Common Mistakes:

• Confusing UPDATE with JOIN as a SELECT operation
• Thinking UPDATE with JOIN deletes data
• Believing it creates new tables

2. Which of the following is the correct syntax to update table1 using table2 with a JOIN on id in MySQL?

easy

A. UPDATE table1 JOIN table2 ON table1.id = table2.id SET table1.value = table2.value;

B. UPDATE table1 SET table1.value = table2.value JOIN table2 ON table1.id = table2.id;

C. UPDATE table1 SET value = table2.value WHERE table1.id = table2.id JOIN table2;

D. UPDATE table1, table2 SET table1.value = table2.value WHERE table1.id = table2.id;

Solution

1. Step 1: Recall correct UPDATE with JOIN syntax

   The correct syntax places JOIN after UPDATE, then ON condition, then SET clause.

2. Step 2: Check each option

   UPDATE table1 JOIN table2 ON table1.id = table2.id SET table1.value = table2.value; matches the correct syntax: UPDATE table1 JOIN table2 ON ... SET ...; others misuse JOIN placement or syntax.

3. Final Answer:

   UPDATE table1 JOIN table2 ON table1.id = table2.id SET table1.value = table2.value; -> Option A

4. Quick Check:

   UPDATE + JOIN + ON + SET order = B [OK]

Hint: JOIN comes right after UPDATE, then ON, then SET [OK]

Common Mistakes:

• Placing JOIN after SET clause
• Using WHERE instead of ON for JOIN condition
• Separating tables with commas instead of JOIN

3. Given these tables:

users(id, name, city_id)
cities(id, city_name)

What will this query do?

UPDATE users u JOIN cities c ON u.city_id = c.id SET u.name = CONCAT(u.name, ' - ', c.city_name) WHERE c.city_name = 'Paris';

medium

A. Delete users who live in Paris

B. Replace all user names with 'Paris'

C. Add a new user with name 'Paris'

D. Append ' - Paris' to the name of users living in Paris

Solution

1. Step 1: Understand the JOIN and SET clauses

   The query joins users with cities on city_id, then updates user names by appending city name for users in Paris.

2. Step 2: Analyze WHERE condition

   Only users with city_name 'Paris' are updated, so their names get ' - Paris' appended.

3. Final Answer:

   Append ' - Paris' to the name of users living in Paris -> Option D

4. Quick Check:

   UPDATE with JOIN + WHERE filters = append city name [OK]

Hint: JOIN matches rows; SET changes only filtered rows [OK]

Common Mistakes:

• Thinking it replaces all names with 'Paris'
• Confusing UPDATE with INSERT
• Assuming it deletes rows

4. Identify the error in this query:

UPDATE orders o JOIN customers c SET o.status = 'shipped' ON o.customer_id = c.id WHERE c.region = 'West';

medium

A. The ON clause is after SET instead of after JOIN

B. Missing WHERE clause

C. Using JOIN instead of INNER JOIN

D. No error, query is correct

Solution

1. Step 1: Check JOIN syntax order

   ON clause must come immediately after JOIN, before SET clause.

2. Step 2: Locate ON clause in query

   ON is incorrectly placed after SET, causing syntax error.

3. Final Answer:

   The ON clause is after SET instead of after JOIN -> Option A

4. Quick Check:

   ON must follow JOIN, not SET [OK]

Hint: ON clause always follows JOIN, never SET [OK]

Common Mistakes:

• Placing ON after SET clause
• Confusing WHERE and ON clauses
• Assuming JOIN needs INNER keyword

5. You have two tables:

products(id, price, category_id)
categories(id, discount_percent)

You want to reduce the price of all products by their category discount. Which query correctly updates products.price applying the discount using UPDATE with JOIN?

hard

A. UPDATE products p, categories c SET p.price = p.price - (p.price * c.discount_percent / 100) WHERE p.category_id = c.id;

B. UPDATE products p JOIN categories c ON p.category_id = c.id SET p.price = p.price - (p.price * c.discount_percent / 100);

C. UPDATE products p SET p.price = p.price - (p.price * c.discount_percent / 100) JOIN categories c ON p.category_id = c.id;

D. UPDATE products p INNER JOIN categories c WHERE p.category_id = c.id SET p.price = p.price * c.discount_percent;

Solution

1. Step 1: Write correct UPDATE with JOIN syntax

   JOIN must come after UPDATE, then ON condition, then SET clause with calculation.

2. Step 2: Verify discount calculation

   Price is reduced by price * discount_percent / 100, which is correct formula.

3. Step 3: Check other options

   Other options misuse JOIN placement, use comma syntax instead of explicit JOIN, or have incorrect syntax/order and discount formulas.

4. Final Answer:

   UPDATE products p JOIN categories c ON p.category_id = c.id SET p.price = p.price - (p.price * c.discount_percent / 100); -> Option B

5. Quick Check:

   UPDATE + JOIN + ON + SET with correct formula = C [OK]

Hint: JOIN after UPDATE, ON before SET, formula subtracts discount [OK]

Common Mistakes:

• Placing JOIN after SET clause
• Using WHERE instead of ON for JOIN
• Incorrect discount formula