UPDATE with JOIN in MySQL - Cheat Sheet & Quick Revision

Recall & Review

beginner

What does the UPDATE with JOIN statement do in SQL?

It updates rows in one table based on matching rows in another table using a join condition.

beginner

Write the basic syntax for UPDATE with JOIN in MySQL.

UPDATE table1 JOIN table2 ON table1.column = table2.column SET table1.column_to_update = value WHERE condition;

intermediate

Why use UPDATE with JOIN instead of separate SELECT and UPDATE queries?

Because it updates data in one step by combining tables, making it faster and simpler.

beginner

Can you update multiple columns using UPDATE with JOIN?

Yes, you can update multiple columns by separating them with commas in the SET clause.

advanced

What happens if the JOIN condition matches multiple rows in the other table during an UPDATE with JOIN?

The update applies to all matching rows, which can cause multiple updates to the same row if not careful.

Which keyword is used to combine tables in an UPDATE statement in MySQL?

A JOIN

B GROUP BY

C ORDER BY

D WHERE

JOIN is used to combine rows from two or more tables based on a related column.

In an UPDATE with JOIN, where do you specify the condition to match rows between tables?

A In the ON clause

B In the GROUP BY clause

C In the SET clause

D In the WHERE clause

The ON clause defines how rows from the tables are matched in the JOIN.

Which of the following is a correct way to update a column using JOIN?

A UPDATE t1 SET name = 'new' WHERE id IN t2;

B UPDATE t1 SET t1.name = t2.name FROM t2 WHERE t1.id = t2.id;

C UPDATE t1 JOIN t2 ON t1.id = t2.id SET t1.name = t2.name;

D UPDATE t1 SET name = t2.name WHERE t1.id = t2.id;

Option C uses the correct MySQL syntax for UPDATE with JOIN.

Can you update columns in both tables simultaneously using UPDATE with JOIN in MySQL?

A Yes, you can update both tables at once

B No, only one table can be updated at a time

C Only if you use a transaction

D Only if the tables have the same structure

No, MySQL UPDATE with JOIN allows updating only one table at a time.

What is the effect of omitting the WHERE clause in an UPDATE with JOIN?

A The query will fail with an error

B No rows will be updated

C Only the first matching row will be updated

D All rows matching the JOIN condition will be updated

Without WHERE, all rows that match the JOIN condition are updated.

Explain how to use UPDATE with JOIN to change data in one table based on another table.

Think about how two tables connect and how you tell SQL which rows to update.

Describe a real-life example where UPDATE with JOIN would be useful.

Imagine you have a list of products and a list of new prices in another table.

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