Challenge! �
Overview
Challenge
Hints
You will need to use the payment table
You will also need to use COUNT and WHERE along with some comparison operator
SOLUTION
SELECT COUNT(amount) FROM payment
WHERE amount > 5;
Challenge
Hints
You will need to use the actor table
You will also need to use LIKE and a wildcard operator, such as % or _
SOLUTION
SELECT COUNT(*) FROM actor
WHERE first_name LIKE 'P%';
Challenge
Hints
You will need to use the address table
You will also need to use COUNT and DISTINCT
SOLUTION
SELECT COUNT(DISTINCT(district))
FROM address;
Challenge
Hints
You will again need to use the address table
This will be very similar to the previous challenge
SOLUTION
SELECT DISTINCT(district) FROM address;
Challenge
Hints
You will need to use the film table.
You may also need to use BETWEEN and a WHERE statement with a comparison operator.
SOLUTION
SELECT COUNT(*) FROM film
WHERE rating = 'R'
AND replacement_cost BETWEEN 5 AND 15;
Challenge
Hints
You will need to use the film table
You will also need to use LIKE with a wildcard operator
SOLUTION
SELECT COUNT(*) FROM film
WHERE title LIKE '%Truman%';
Great job!