SELECT
    u.id,
    last_order.id,
    last_order.created_at
FROM users AS u
LEFT JOIN LATERAL (
    SELECT id, created_at
    FROM orders
    WHERE user_id = u.id
    ORDER BY created_at DESC
    LIMIT 1
) AS last_order ON true;

CREATE INDEX ON orders (user_id, created_at DESC);

from graphforge import GraphForge

graph = GraphForge("research/")

result = graph.execute("""
    MATCH (a)-[:CITES]->(b)
    RETURN a.title, b.title
""")

print(result.to_pandas())

SELECT NULL = NULL;

NULL

WHERE a = b
   OR (a IS NULL AND b IS NULL)

a IS NOT DISTINCT FROM b

SELECT NULL IS NOT DISTINCT FROM NULL; -- true
SELECT 10   IS NOT DISTINCT FROM 10;   -- true
SELECT 10   IS NOT DISTINCT FROM NULL; -- false

a IS DISTINCT FROM b

SELECT *
FROM old_data o
JOIN new_data n USING (id)
WHERE o.email IS DISTINCT FROM n.email;

o.email <> n.email

SELECT country, city, SUM(revenue)
FROM sales
GROUP BY country, city

UNION ALL

SELECT country, NULL, SUM(revenue)
FROM sales
GROUP BY country

UNION ALL

SELECT NULL, NULL, SUM(revenue)
FROM sales;

SELECT
    country,
    city,
    SUM(revenue) AS revenue
FROM sales
GROUP BY GROUPING SETS (
    (country, city),
    (country),
    ()
);

GROUPING(country)
GROUPING(city)

ROLLUP(...)
CUBE(...)

SELECT COUNT(*)
FROM orders
WHERE user_id = 42;

SELECT EXISTS (
    SELECT 1
    FROM orders
    WHERE user_id = 42
);

CREATE INDEX idx_users_status ON users(status);

SELECT *
FROM users
WHERE status != 'active';

status != 'active'

EXPLAIN ANALYZE
SELECT *
FROM users
WHERE status != 'active';

<>
NOT IN
IS NOT NULL

«Просто гарантируй, что в каждый момент времени пишет только одна машина».