SQL Programming Resources on Telegram: “SELECT c.customer_id, c.customer

SELECT c.customer_id, c.customer_name, SUM(o.amount) AS total_spending 
FROM customers c 
INNER JOIN orders o ON c.customer_id = o.customer_id 
GROUP BY c.customer_id, c.customer_name;

💰 11. Include Customers with Zero Spending

SELECT c.customer_id, c.customer_name, COALESCE(SUM(o.amount), 0) AS total_spending 
FROM customers c 
LEFT JOIN orders o ON c.customer_id = o.customer_id 
GROUP BY c.customer_id, c.customer_name;

🔢 12. JOIN + COUNT()

SELECT c.customer_id, c.customer_name, COUNT(o.order_id) AS order_count 
FROM customers c 
LEFT JOIN orders o ON c.customer_id = o.customer_id 
GROUP BY c.customer_id, c.customer_name;

Why COUNT(o.order_id) instead of COUNT(*)?

Because COUNT(*) would count the LEFT JOIN row even when the customer has no matching order.

⚠️ 13. A Very Common JOIN Mistake

SELECT ... WHERE o.amount > 500; -- This removes NULLs and behaves like INNER JOIN

Correct:

LEFT JOIN orders o ON c.customer_id = o.customer_id AND o.amount > 500;

Important concept: With an OUTER JOIN, the location of a filter can change the result.

🔗 14. Joining More Than Two Tables

SELECT c.customer_name, o.order_id, p.product_name, o.amount 
FROM customers c 
JOIN orders o ON c.customer_id = o.customer_id 
JOIN products p ON o.product_id = p.product_id;

🏢 15. Real-World Business Example

SELECT p.category, SUM(o.amount) AS total_revenue 
FROM orders o 
JOIN products p ON o.product_id = p.product_id 
GROUP BY p.category 
ORDER BY total_revenue DESC;

This is a typical Data Analyst query.

📈 16. JOIN + WHERE + GROUP BY + HAVING

Question:

«Find customers who spent more than ₹50,000.»

SELECT c.customer_id, c.customer_name, SUM(o.amount) AS total_spending 
FROM customers c 
JOIN orders o ON c.customer_id = o.customer_id 
GROUP BY c.customer_id, c.customer_name 
HAVING SUM(o.amount) > 50000 
ORDER BY total_spending DESC;

Logical flow: JOIN → GROUP BY → HAVING → ORDER BY

🪞 17. SELF JOIN

A table can also be joined to itself.

SELECT e.employee_name AS employee, m.employee_name AS manager 
FROM employees e 
LEFT JOIN employees m ON e.manager_id = m.employee_id;

🔢 18. CROSS JOIN

"CROSS JOIN" produces every possible combination of rows.

5 products x 4 regions = 20 rows

🚨 19. The Biggest JOIN Problem: Duplicate Rows

One customer has five orders → customer appears five times. This is the natural result of a one-to-many relationship.

If you want unique customers:

SELECT COUNT(DISTINCT c.customer_id)