data/knowledge/skills/sql_skills.md

SQL Skills

Basic Queries

SELECT

-- All columns
SELECT * FROM users;

-- Specific columns
SELECT name, email, created_at FROM users;

-- With alias
SELECT name AS user_name FROM users;

WHERE

SELECT * FROM products WHERE price > 100;

SELECT * FROM users 
WHERE age >= 18 AND status = 'active';

SELECT * FROM orders 
WHERE created_at BETWEEN '2024-01-01' AND '2024-12-31';

ORDER BY

SELECT * FROM products 
ORDER BY price ASC;

SELECT * FROM users 
ORDER BY created_at DESC, name ASC;

LIMIT

SELECT * FROM users LIMIT 10;

SELECT * FROM products 
ORDER BY price DESC LIMIT 5;

JOIN Operations

INNER JOIN

SELECT u.name, o.order_id, o.total
FROM users u
INNER JOIN orders o ON u.id = o.user_id;

LEFT JOIN

SELECT u.name, o.order_id
FROM users u
LEFT JOIN orders o ON u.id = o.user_id;

RIGHT JOIN

SELECT u.name, o.order_id
FROM users u
RIGHT JOIN orders o ON u.id = o.user_id;

Multiple JOINs

SELECT u.name, o.order_id, p.product_name
FROM users u
INNER JOIN orders o ON u.id = o.user_id
INNER JOIN order_items oi ON o.id = oi.order_id
INNER JOIN products p ON oi.product_id = p.id;

Aggregations

COUNT, SUM, AVG, MIN, MAX

SELECT COUNT(*) FROM users;
SELECT SUM(amount) FROM orders;
SELECT AVG(price) FROM products;
SELECT MIN(created_at) FROM orders;
SELECT MAX(price) FROM products;

GROUP BY

SELECT category, COUNT(*) as count
FROM products
GROUP BY category;

SELECT user_id, SUM(total) as total_spent
FROM orders
GROUP BY user_id
HAVING SUM(total) > 1000;

Subqueries

-- In WHERE
SELECT * FROM products
WHERE price > (SELECT AVG(price) FROM products);

-- In FROM
SELECT category, avg_price
FROM (
    SELECT category, AVG(price) as avg_price
    FROM products
    GROUP BY category
) AS stats;

INSERT, UPDATE, DELETE

INSERT

INSERT INTO users (name, email, age)
VALUES ('John', 'john@email.com', 25);

INSERT INTO users (name, email)
SELECT name, email FROM temp_users;

UPDATE

UPDATE users 
SET age = 26, status = 'active'
WHERE id = 1;

DELETE

DELETE FROM users WHERE id = 1;

DELETE FROM orders 
WHERE created_at < '2024-01-01';

Common Patterns

Finding Duplicates

SELECT email, COUNT(*) as count
FROM users
GROUP BY email
HAVING COUNT(*) > 1;

Second Highest

SELECT MAX(salary) FROM employees
WHERE salary < (SELECT MAX(salary) FROM employees);

-- Or using OFFSET
SELECT salary FROM employees
ORDER BY salary DESC LIMIT 1 OFFSET 1;

Date Functions

SELECT DATE(created_at) as date, COUNT(*) as orders
FROM orders
GROUP BY DATE(created_at);

SELECT * FROM orders
WHERE EXTRACT(YEAR FROM created_at) = 2024;

Case Statement

SELECT name,
    CASE 
        WHEN age < 18 THEN 'Minor'
        WHEN age < 65 THEN 'Adult'
        ELSE 'Senior'
    END as age_group
FROM users;