Bird
Raised Fist0
MySQLquery~5 mins

UPDATE with JOIN in MySQL

Choose your learning style10 modes available
Introduction
You use UPDATE with JOIN to change data in one table based on matching data in another table. It helps update related information easily.
When you want to update a customer's address using data from a separate address table.
When you need to fix product prices based on new rates stored in another table.
When you want to update employee department names using a department info table.
When you want to correct order statuses by joining with a status reference table.
Syntax
MySQL
UPDATE table1
JOIN table2 ON table1.common_field = table2.common_field
SET table1.field_to_update = new_value
WHERE some_condition;
The JOIN clause connects the tables using a common field.
The WHERE clause limits which rows get updated.
Examples
Increase salary by 10% for employees in the Sales department.
MySQL
UPDATE employees
JOIN departments ON employees.dept_id = departments.id
SET employees.salary = employees.salary * 1.1
WHERE departments.name = 'Sales';
Mark orders as shipped for customers in the West region.
MySQL
UPDATE orders
JOIN customers ON orders.customer_id = customers.id
SET orders.status = 'Shipped'
WHERE customers.region = 'West';
Sample Program
This example creates two tables, inserts data, then increases salary by 10% for employees in Sales department. Finally, it shows updated salaries.
MySQL
CREATE TABLE employees (id INT, name VARCHAR(20), dept_id INT, salary DECIMAL(10,2));
CREATE TABLE departments (id INT, name VARCHAR(20));

INSERT INTO employees VALUES (1, 'Alice', 10, 50000), (2, 'Bob', 20, 60000), (3, 'Carol', 10, 55000);
INSERT INTO departments VALUES (10, 'Sales'), (20, 'HR');

UPDATE employees
JOIN departments ON employees.dept_id = departments.id
SET employees.salary = employees.salary * 1.1
WHERE departments.name = 'Sales';

SELECT id, name, salary FROM employees ORDER BY id;
OutputSuccess
Important Notes
Always back up your data before running UPDATE queries with JOIN to avoid accidental changes.
Make sure the JOIN condition correctly matches rows to update the right records.
You can join multiple tables if needed by adding more JOIN clauses.
Summary
UPDATE with JOIN lets you update one table using data from another table.
Use JOIN to connect tables on common fields and WHERE to filter rows.
It is useful for updating related data efficiently in one query.

Practice

(1/5)
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
Recommended

Ready to prove it?

Pass the test and this pattern moves unknown strong while earning +1 Grip toward your certificate.

10 questions~5 mineasy hard