Believing JOINs and subqueries have identical performance always
2. Which of the following is the correct syntax to write a JOIN between tables employees and departments on the column department_id?
easy
A. SELECT * FROM employees JOIN departments USING employees.department_id;
B. SELECT * FROM employees JOIN departments WHERE employees.department_id = departments.department_id;
C. SELECT * FROM employees, departments ON employees.department_id = departments.department_id;
D. SELECT * FROM employees JOIN departments ON employees.department_id = departments.department_id;
Solution
Step 1: Identify correct JOIN syntax
The correct JOIN syntax uses the ON keyword to specify the matching columns between tables.
Step 2: Check each option
SELECT * FROM employees JOIN departments ON employees.department_id = departments.department_id; uses ON correctly. SELECT * FROM employees JOIN departments WHERE employees.department_id = departments.department_id; incorrectly uses WHERE instead of ON. SELECT * FROM employees, departments ON employees.department_id = departments.department_id; uses comma join but incorrectly adds ON. SELECT * FROM employees JOIN departments USING employees.department_id; misuses USING with full column name.
Final Answer:
SELECT * FROM employees JOIN departments ON employees.department_id = departments.department_id; -> Option D
Quick Check:
JOIN requires ON for matching columns [OK]
Hint: Use ON to specify join condition, not WHERE [OK]
Common Mistakes:
Using WHERE instead of ON for JOIN condition
Using comma join with ON clause
Incorrect USING syntax with full column names
3. Given the tables orders(order_id, customer_id, amount) and customers(customer_id, name), what will the following query return?
SELECT name FROM customers WHERE customer_id IN (SELECT customer_id FROM orders WHERE amount > 100);
medium
A. Names of customers who have orders with amount greater than 100.
B. All customer names regardless of orders.
C. Names of customers who have no orders.
D. Syntax error due to subquery.
Solution
Step 1: Analyze the subquery
The subquery selects customer_id from orders where amount is greater than 100.
Step 2: Analyze the main query
The main query selects names from customers where customer_id is in the list returned by the subquery.
Final Answer:
Names of customers who have orders with amount greater than 100. -> Option A
Quick Check:
Subquery filters customers by order amount [OK]
Hint: IN with subquery filters matching rows [OK]
Common Mistakes:
Assuming all customers are returned
Thinking it returns customers without orders
Believing subquery causes syntax error
4. Identify the error in the following query that tries to list all employees with their department names:
SELECT e.name, d.department_name FROM employees e JOIN departments d WHERE e.department_id = d.department_id;
medium
A. Using WHERE instead of HAVING for join condition.
B. Missing ON keyword before the join condition.
C. Incorrect table aliases used.
D. No error; query is correct.
Solution
Step 1: Check JOIN syntax
JOIN requires the ON keyword to specify the join condition between tables.
Step 2: Identify the mistake
The query uses WHERE instead of ON to specify the join condition, which is incorrect syntax.
Final Answer:
Missing ON keyword before the join condition. -> Option B
Quick Check:
JOIN needs ON, not WHERE, for conditions [OK]
Hint: JOIN conditions must use ON, not WHERE [OK]
Common Mistakes:
Using WHERE instead of ON for JOIN
Confusing HAVING with JOIN condition
Assuming aliases cause error
5. You want to find all customers who have placed orders but only want to use a JOIN instead of a subquery. Which query correctly replaces this subquery?
SELECT name FROM customers WHERE customer_id IN (SELECT customer_id FROM orders);
hard
A. SELECT DISTINCT c.name FROM customers c JOIN orders o ON c.customer_id = o.customer_id;
B. SELECT c.name FROM customers c JOIN orders o WHERE c.customer_id = o.customer_id;
C. SELECT c.name FROM customers c, orders o WHERE c.customer_id = o.customer_id;
D. SELECT name FROM customers LEFT JOIN orders ON customer_id = customer_id;
Solution
Step 1: Understand the original subquery
The subquery selects customer_ids from orders, filtering customers who placed orders.
Step 2: Translate to JOIN
Using JOIN on customer_id between customers and orders returns matching customers. DISTINCT is needed to avoid duplicates.
Step 3: Evaluate options
SELECT DISTINCT c.name FROM customers c JOIN orders o ON c.customer_id = o.customer_id; uses JOIN with ON and DISTINCT correctly. SELECT c.name FROM customers c JOIN orders o WHERE c.customer_id = o.customer_id; misses ON keyword. SELECT c.name FROM customers c, orders o WHERE c.customer_id = o.customer_id; uses old-style join but is valid but may return duplicates. SELECT name FROM customers LEFT JOIN orders ON customer_id = customer_id; uses LEFT JOIN but ON condition is ambiguous.
Final Answer:
SELECT DISTINCT c.name FROM customers c JOIN orders o ON c.customer_id = o.customer_id; -> Option A
Quick Check:
JOIN with ON and DISTINCT matches subquery result [OK]
Hint: Use JOIN with ON and DISTINCT to replace IN subquery [OK]
Common Mistakes:
Omitting ON keyword in JOIN
Not using DISTINCT causing duplicates
Using ambiguous ON conditions
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