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answers/2602.sql

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SELECT name
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FROM customers
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WHERE state = 'RS';

answers/2603.sql

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SELECT name, street
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FROM customers
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WHERE city = 'Porto Alegre'

answers/2604.sql

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SELECT id, name
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FROM products
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WHERE price < 10 OR price > 100;

answers/2605.sql

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SELECT products.name, providers.name
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FROM products
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JOIN providers ON products.id_providers = providers.id
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JOIN categories ON products.id_categories = categories.id
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WHERE categories.id = 6;

answers/2606.sql

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SELECT id, name
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FROM products
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WHERE id_categories IN (
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SELECT id
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FROM categories
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WHERE name LIKE 'super%'
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);

answers/2607.sql

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SELECT DISTINCT city FROM providers ORDER BY city;

answers/2608.sql

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SELECT MAX(price), MIN(price) AS min_price
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FROM products;

answers/2609.sql

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SELECT c.name, SUM(p.amount) AS sum
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FROM products p
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JOIN categories c ON p.id_categories = c.id
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GROUP BY c.name;

answers/2610.sql

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SELECT ROUND(AVG(price), 2) AS avg_price
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FROM products;

answers/2611.sql

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SELECT m.id, m.name
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FROM movies m
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JOIN genres g ON m.id_genres = g.id
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WHERE g.description = 'Action'

answers/2613.sql

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SELECT movies.id, movies.name
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FROM movies
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JOIN prices ON movies.id_prices = prices.id
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WHERE prices.value < 2.00;

answers/2614.sql

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SELECT customers.name, rentals.rentals_date
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FROM rentals
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JOIN customers ON rentals.id_customers = customers.id
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WHERE EXTRACT(MONTH FROM rentals.rentals_date) = 9
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AND EXTRACT(YEAR FROM rentals.rentals_date) = 2016;

answers/2615.sql

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SELECT DISTINCT city
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FROM customers

answers/2616.sql

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SELECT customers.id, customers.name
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FROM customers
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WHERE customers.id NOT IN (SELECT id_customers FROM locations)
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ORDER BY customers.id ASC

answers/2617.sql

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SELECT p.name, pr.name
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FROM products p
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JOIN providers pr ON p.id_providers = pr.id
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WHERE pr.name = 'Ajax SA'

answers/2618.sql

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SELECT products.name, providers.name, categories.name
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FROM products
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JOIN providers ON products.id_providers = providers.id
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JOIN categories ON products.id_categories = categories.id
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WHERE providers.name = 'Sansul SA' AND categories.name = 'Imported';

answers/2619.sql

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SELECT products.name, providers.name, products.price
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FROM products, providers, categories
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WHERE products.id_providers = providers.id
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AND products.id_categories = categories.id
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AND categories.name LIKE 'Super Luxury'
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AND price > 1000;

answers/2620.sql

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SELECT c.name,
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o.id
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FROM customers c
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INNER JOIN orders o ON o.id_customers = c.id
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WHERE extract(YEAR FROM orders_date) = 2016 AND extract(MONTH FROM orders_date) IN (1,2,3,4,5,6)

answers/2621.sql

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SELECT prod.name
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FROM products prod
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INNER JOIN providers prov ON prov.id = prod.id_providers
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WHERE prod.amount BETWEEN 10 AND 20
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AND prov.name LIKE 'P%'

answers/2622.sql

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SELECT cus.name
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FROM legal_person lp
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LEFT JOIN customers cus ON cus.id = lp.id_customers

answers/2623.sql

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SELECT prod.name,
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cat.name
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FROM products prod
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INNER JOIN categories cat ON cat.id = prod.id_categories
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WHERE prod.amount > 100
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AND prod.id_categories IN (1,2,3,6,9)

answers/2624.sql

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SELECT count(DISTINCT city)
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FROM customers

answers/2625.sql

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SELECT regexp_replace(np.cpf, '(\d{3})(\d{3})(\d{3})', '\1.\2.\3-')
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FROM natural_person np
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LEFT JOIN customers cus ON cus.id = np.id_customers

answers/2737.sql

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(SELECT name, customers_number
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FROM lawyers
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ORDER BY customers_number DESC
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LIMIT 1)
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UNION ALL
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(SELECT name, customers_number
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FROM lawyers
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ORDER BY customers_number ASC
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LIMIT 1)
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UNION ALL
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(SELECT 'Average', round(AVG(customers_number), 0)
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FROM lawyers)

answers/2738.sql

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SELECT c.name,
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round (((s.math * 2) + (s.specific * 3) + (s.project_plan*5)) / 10, 2) AS media
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FROM candidate c
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INNER JOIN score s ON s.candidate_id = c.id
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ORDER BY media DESC

answers/2739.sql

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SELECT NAME, Cast(extract (DAY FROM payday) AS Integer)
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FROM loan

answers/2740.sql

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(SELECT 'Podium: ' || team FROM league ORDER BY position LIMIT 3)
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UNION ALL
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(WITH t AS (SELECT * FROM league ORDER BY position DESC LIMIT 2)
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SELECT 'Demoted: '|| team FROM t ORDER BY position ASC);

answers/2741.sql

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SELECT concat ('Approved:',' ', NAME), grade
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FROM students
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WHERE grade > 6
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ORDER BY grade DESC

answers/2742.sql

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SELECT lr.name, round (lr.omega * 1.618, 3) AS n
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FROM dimensions d
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INNER JOIN life_registry lr ON lr.dimensions_id = d.id
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WHERE lr.name LIKE 'Richard%' AND d.name IN ('C875', 'C774')
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ORDER BY lr.omega ASC

answers/2743.sql

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SELECT NAME, CHAR_LENGTH(NAME) AS tam
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FROM people
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ORDER BY tam DESC

answers/2744.sql

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SELECT id, password, md5(password) FROM account;

answers/2745.sql

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SELECT NAME, round (salary * 0.1, 2)
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FROM people
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WHERE salary > 3000

answers/2746.sql

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SELECT REPLACE(name, 'H1', 'X') AS name
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FROM virus;

answers/2988.sql

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SELECT t.name,
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COUNT(m.id) AS matches,
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SUM(CASE
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WHEN m.team_1 = t.id AND m.team_1_goals > m.team_2_goals THEN 1
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WHEN m.team_2 = t.id AND m.team_2_goals > m.team_1_goals THEN 1
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ELSE 0
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END) AS victories,
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SUM(CASE
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WHEN m.team_1 = t.id AND m.team_1_goals < m.team_2_goals THEN 1
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WHEN m.team_2 = t.id AND m.team_2_goals < m.team_1_goals THEN 1
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ELSE 0
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END) AS defeats,
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SUM(CASE
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WHEN m.team_1 = t.id AND m.team_1_goals = m.team_2_goals THEN 1
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WHEN m.team_2 = t.id AND m.team_2_goals = m.team_1_goals THEN 1
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ELSE 0
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END) AS draws,
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SUM(CASE
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WHEN m.team_1 = t.id AND m.team_1_goals > m.team_2_goals THEN 3
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WHEN m.team_2 = t.id AND m.team_2_goals > m.team_1_goals THEN 3
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WHEN m.team_1 = t.id AND m.team_1_goals = m.team_2_goals THEN 1
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WHEN m.team_2 = t.id AND m.team_2_goals = m.team_1_goals THEN 1
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ELSE 0
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END) AS score
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FROM teams t
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LEFT JOIN matches m ON t.id IN (m.team_1, m.team_2)
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GROUP BY t.id
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ORDER BY score DESC;

answers/2990.sql

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SELECT emp.cpf, emp.enome, dep.dnome
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FROM empregados emp
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INNER JOIN departamentos dep ON emp.dnumero = dep.dnumero
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LEFT JOIN trabalha t ON emp.cpf = t.cpf_emp
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WHERE t.pnumero IS NULL
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ORDER BY emp.cpf

answers/2993.sql

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SELECT
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amount AS most_frequent_value
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FROM
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value_table
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GROUP BY
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amount
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ORDER BY
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(COUNT(amount)) DESC
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LIMIT 1

answers/2994.sql

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SELECT
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d.name,
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SUM(ROUND(((150*a.hours)*(1+(w.bonus/100))), 1)) AS salary
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FROM
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attendances a
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INNER JOIN doctors d ON a.id_doctor = d.id
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INNER JOIN work_shifts w ON a.id_work_shift = w.id
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GROUP BY
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d.name
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ORDER BY
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salary DESC

answers/2995.sql

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SELECT
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temperature,
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COUNT(mark) AS number_of_records
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FROM
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records
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GROUP BY
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mark,
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temperature
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ORDER BY
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mark

answers/2996.sql

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SELECT packages.year, u1.name AS sender, u2.name AS receiver
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FROM packages
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JOIN users u1 ON packages.id_user_sender = u1.id
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JOIN users u2 ON packages.id_user_receiver = u2.id
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WHERE (packages.year = 2015 OR packages.color LIKE 'blue')
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AND u2.address <> 'Taiwan' ORDER BY packages.year DESC;

answers/2998.sql

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WITH cumulative_profit AS (
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SELECT
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c.name,
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c.investment,
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o.month,
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SUM(o.profit) OVER (PARTITION BY o.client_id ORDER BY o.month) AS cumulative_profit
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FROM
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clients c
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JOIN operations o ON c.id = o.client_id
10+
),
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payback AS (
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SELECT
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name,
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investment,
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MIN(month) AS payback_month,
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MIN(cumulative_profit) - investment AS payback_value
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FROM
18+
cumulative_profit
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WHERE
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cumulative_profit >= investment
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GROUP BY
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name,
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investment
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)
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SELECT *
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FROM payback
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ORDER BY payback_value DESC;

answers/3001.sql

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SELECT NAME,
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CASE WHEN TYPE = 'A' then 20.0
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WHEN TYPE = 'B' then 70.0
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WHEN TYPE = 'C' then 530.5
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END AS price
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FROM products
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ORDER BY TYPE, id DESC

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