SQL Programming Resources
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Find top SQL resources from global universities, cool projects, and learning materials for data analytics. Admin: @coderfun Useful links: heylink.me/DataAnalytics Promotions: @love_data
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Red:📈 Análisis del canal de Telegram SQL Programming Resources
El canal SQL Programming Resources (@sqlanalyst) en el segmento lingüístico de Inglés es un actor destacado. Actualmente la comunidad reúne a 76 651 suscriptores, ocupando la posición 1 636 en la categoría Tecnologías y Aplicaciones y el puesto 3 966 en la región India.
📊 Métricas de audiencia y dinámica
Desde su creación el невідомо, el proyecto ha mostrado un crecimiento acelerado, reuniendo a 76 651 suscriptores.
Según los últimos datos del 15 septiembre, 2026, el canal mantiene una actividad estable. En los últimos 30 días la variación de miembros fue de 17, y en las últimas 24 horas de -6, conservando un alto alcance.
- Estado de verificación: No verificado
- Tasa de interacción (ER): El promedio de interacción de la audiencia es 1.50%. Durante las primeras 24 horas tras publicar, el contenido suele obtener 0.81% de reacciones respecto al total de suscriptores.
- Alcance de las publicaciones: Cada publicación recibe en promedio 1 148 visualizaciones. En el primer día suele acumular 621 visualizaciones.
- Reacciones e interacción: La audiencia responde de forma activa: el promedio de reacciones por publicación es 3.
- Intereses temáticos: El contenido se centra en temas clave como row, sql, customer_id, logic, desc.
📝 Descripción y política de contenido
El autor describe el recurso como un espacio para expresar opiniones subjetivas:
“Find top SQL resources from global universities, cool projects, and learning materials for data analytics.
Admin: @coderfun
Useful links: heylink.me/DataAnalytics
Promotions: @love_data”
Gracias a la alta frecuencia de actualizaciones (últimos datos recibidos el 16 septiembre, 2026), el canal mantiene la vigencia y un amplio alcance. La analítica demuestra que la audiencia interactúa activamente con el contenido, lo que lo convierte en un punto de referencia dentro de la categoría Tecnologías y Aplicaciones.
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Publicaciones del Canal
⚠️ 23. Common Subquery Mistakes
Mistake 1 — Returning multiple rows with
=
Use IN instead of = when multiple rows are expected.
Mistake 2 — Forgetting NULL behavior with NOT IN
Mistake 3 — Ignoring duplicates
Mistake 4 — Making the query unnecessarily complicated
🎤 SQL Interview Questions
Q1. What is a subquery?
A query nested inside another query.
Q2. What is a scalar subquery?
A subquery that returns a single value.
Q3. When should you use IN?
When the subquery returns a set of values.
Q4. What does EXISTS do?
It checks whether at least one matching row exists.
Q5. What is a correlated subquery?
A subquery that references a column from the outer query.
Q6. What is a derived table?
A subquery in the FROM clause treated as a temporary result set.
Q7. Can a subquery be used in SELECT?
Yes.
Q8. What is the difference between IN and EXISTS?
IN compares against a set, while EXISTS checks for existence.
Q9. Why is NOT IN dangerous with NULL?
Three-valued logic can cause unexpected results.
Q10. Can every subquery be replaced with a JOIN?
Many can, but the best approach depends on the logic.
📝 Practice Questions
Practice 1 — Employees earning more than average:
SELECT employee_name, salary
FROM employees
WHERE salary > (
SELECT AVG(salary)
FROM employees
);SELECT customer_id, customer_name
FROM customers
WHERE customer_id IN (
SELECT customer_id
FROM orders
);SELECT customer_id, customer_name
FROM customers
WHERE customer_id IN (
SELECT customer_id
FROM orders
GROUP BY customer_id
HAVING COUNT(*) > 5
);SELECT product_name, price
FROM products
WHERE price > (
SELECT AVG(price)
FROM products
);SELECT c.customer_id, c.customer_name
FROM customers c
WHERE NOT EXISTS (
SELECT 1
FROM orders o
WHERE o.customer_id = c.customer_id
);SELECT customer_id, total_spending
FROM (
SELECT
customer_id,
SUM(amount) AS total_spending
FROM orders
GROUP BY customer_id
) AS customer_totals
WHERE total_spending > (
SELECT AVG(total_spending)
FROM (
SELECT
customer_id,
SUM(amount) AS total_spending
FROM orders
GROUP BY customer_id
) AS totals
);| 2 | Means: Return customers for whom no matching order exists.
🧠 12. EXISTS vs IN
IN → Compare against a set of returned values
EXISTS → Check whether a matching row exists
🔄 13. Correlated Subquery
Depends on the current row of the outer query.
SELECT
e.employee_name,
e.salary,
e.department_id
FROM employees e
WHERE e.salary > (
SELECT AVG(e2.salary)
FROM employees e2
WHERE e2.department_id = e.department_id
);
«Is this employee's salary higher than the average salary of their own department?»
🏢 14. Above-Department-Average Salary
This is a correlated subquery. Bob is compared against the Sales average, while David is compared against the IT average.
📦 15. Subquery in FROM
It can appear in the FROM clause.
SELECT
customer_id,
total_spending
FROM (
SELECT
customer_id,
SUM(amount) AS total_spending
FROM orders
GROUP BY customer_id
) AS customer_sales;
This is often called a derived table.
📈 16. Finding High-Value Customers
SELECT
customer_id,
total_spending
FROM (
SELECT
customer_id,
SUM(amount) AS total_spending
FROM orders
GROUP BY customer_id
) AS customer_sales
WHERE total_spending > 50000;
🧩 17. Subquery in SELECT
SELECT
c.customer_name,
(
SELECT COUNT(*)
FROM orders o
WHERE o.customer_id = c.customer_id
) AS order_count
FROM customers c;
⚠️ 18. But Be Careful with Correlated Subqueries
Instead of a correlated subquery in SELECT, you could use:
SELECT
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;
The better choice depends on the database optimizer, table size, indexes, and other factors.
🧱 19. Nested Subqueries
SELECT *
FROM products
WHERE price > (
SELECT AVG(price)
FROM products
WHERE category_id IN (
SELECT category_id
FROM categories
WHERE category_name = 'Electronics'
)
);
Deeply nested queries can become difficult to read. Use CTEs instead.
🧠 20. Subqueries vs JOINs
Subquery:
SELECT *
FROM customers
WHERE customer_id IN (
SELECT customer_id
FROM orders
);
JOIN:
SELECT DISTINCT
c.*
FROM customers c
JOIN orders o
ON c.customer_id = o.customer_id;
Choose based on readability, business logic, and performance.
🏆 21. Subqueries for Business Analysis
Find products above average:
SELECT product_name, price
FROM products
WHERE price > (
SELECT AVG(price)
FROM products
);
Find customers with at least one order:
SELECT customer_id, customer_name
FROM customers c
WHERE EXISTS (
SELECT 1
FROM orders o
WHERE o.customer_id = c.customer_id
);
Find customers with more than 10 orders:
SELECT customer_id, customer_name
FROM customers
WHERE customer_id IN (
SELECT customer_id
FROM orders
GROUP BY customer_id
HAVING COUNT(*) > 10
);
🧮 22. Subquery for KPI Comparison
SELECT order_id, customer_id, amount
FROM orders
WHERE amount > (
SELECT AVG(amount)
FROM orders
); | 449 |
| 3 | 🚀 SQL Roadmap 2026 — Part 12
🧩 SQL Subqueries — Using One Query Inside Another
So far, we've learned how to retrieve, filter, group, transform, and combine data.
But sometimes a business question requires one query to use the result of another query.
For example:
«Find employees whose salary is higher than the average salary.»
First, we need to calculate:
SELECT AVG(salary)
FROM employees;
Then compare every employee against that result.
A subquery allows us to do both inside one SQL statement.
🧠 1. What Is a Subquery?
A subquery is a SQL query written inside another SQL query.
SELECT *
FROM employees
WHERE salary > (
SELECT AVG(salary)
FROM employees
);
The inner query:
SELECT AVG(salary) FROM employees
calculates the average salary. The outer query then uses that result.
Think of it as:
Outer Query → Needs an answer → Subquery calculates the answer → Outer Query uses it
🔹 2. Basic Subquery Structure
SELECT column_name
FROM table_name
WHERE column_name operator (
SELECT ...
);
The inner query is enclosed in ( ... ).
📊 3. Subquery Returning One Value
A scalar subquery returns a single value.
SELECT
employee_name,
salary
FROM employees
WHERE salary > (
SELECT AVG(salary)
FROM employees
);
The inner query returns something like 65000. The outer query then finds employees earning more than 65000.
💰 4. Employees Earning Above Average
Classic interview question.
SELECT
employee_name,
salary
FROM employees
WHERE salary > (
SELECT AVG(salary)
FROM employees
);
Logic: Calculate average salary → Compare every employee → Keep salary > average
🏆 5. Products More Expensive Than Average
SELECT
product_name,
price
FROM products
WHERE price > (
SELECT AVG(price)
FROM products
);
🔢 6. Subquery with COUNT()
Customers who have placed more than 5 orders:
SELECT
customer_id,
customer_name
FROM customers
WHERE customer_id IN (
SELECT customer_id
FROM orders
GROUP BY customer_id
HAVING COUNT(*) > 5
);
🔗 7. Subquery with IN
Useful when a subquery returns multiple values.
SELECT *
FROM customers
WHERE customer_id IN (
SELECT customer_id
FROM orders
);
Returns customers who have at least one order.
⚠️ 8. Single Value vs Multiple Values
One value → Use =, >, <, >=, <=
WHERE salary > (
SELECT AVG(salary)
FROM employees
)
Multiple values → Use IN
WHERE customer_id IN (
SELECT customer_id
FROM orders
)
🚫 9. NOT IN
SELECT
customer_id,
customer_name
FROM customers
WHERE customer_id NOT IN (
SELECT customer_id
FROM orders
);
⚠️ Important NULL Warning: NOT IN can produce unexpected results if the subquery contains NULL. For anti-matching, NOT EXISTS is often safer.
🔎 10. EXISTS
SELECT
c.customer_id,
c.customer_name
FROM customers c
WHERE EXISTS (
SELECT 1
FROM orders o
WHERE o.customer_id = c.customer_id
);
Means: Return the customer if at least one matching order exists.
🚫 11. NOT EXISTS
SELECT
c.customer_id,
c.customer_name
FROM customers c
WHERE NOT EXISTS (
SELECT 1
FROM orders o
WHERE o.customer_id = c.customer_id
); | 420 |
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1. Danny’s Diner:
Restaurant analytics to understand the customer orders pattern.
Link: https://8weeksqlchallenge.com/case-study-1/
2. Pizza Runner
Pizza shop analytics to optimize the efficiency of the operation
Link: https://8weeksqlchallenge.com/case-study-2/
3. Foodie Fie
Subscription-based food content platform
Link: https://lnkd.in/gzB39qAT
4. Data Bank: That’s money
Analytics based on customer activities with the digital bank
Link: https://lnkd.in/gH8pKPyv
5. Data Mart: Fresh is Best
Analytics on Online supermarket
Link: https://lnkd.in/gC5bkcDf
6. Clique Bait: Attention capturing
Analytics on the seafood industry
Link: https://lnkd.in/ggP4JiYG
7. Balanced Tree: Clothing Company
Analytics on the sales performance of clothing store
Link: https://8weeksqlchallenge.com/case-study-7
8. Fresh segments: Extract maximum value
Analytics on online advertising
Link: https://8weeksqlchallenge.com/case-study-8 | 2 309 |
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Join for more: https://t.me/sqlanalyst
1. Danny’s Diner:
Restaurant analytics to understand the customer orders pattern.
Link: https://8weeksqlchallenge.com/case-study-1/
2. Pizza Runner
Pizza shop analytics to optimize the efficiency of the operation
Link: https://8weeksqlchallenge.com/case-study-2/
3. Foodie Fie
Subscription-based food content platform
Link: https://lnkd.in/gzB39qAT
4. Data Bank: That’s money
Analytics based on customer activities with the digital bank
Link: https://lnkd.in/gH8pKPyv
5. Data Mart: Fresh is Best
Analytics on Online supermarket
Link: https://lnkd.in/gC5bkcDf
6. Clique Bait: Attention capturing
Analytics on the seafood industry
Link: https://lnkd.in/ggP4JiYG
7. Balanced Tree: Clothing Company
Analytics on the sales performance of clothing store
Link: https://8weeksqlchallenge.com/case-study-7
8. Fresh segments: Extract maximum value
Analytics on online advertising
Link: https://8weeksqlchallenge.com/case-study-8 | 1 |
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⏰ Time: 7:00 PM | 1 915 |
| 11 | A customer has 3 orders. After joining customers with orders, how many rows can that customer produce? | 1 625 |
| 12 | Why can a JOIN cause duplicate rows? | 1 470 |
| 13 | What does a LEFT JOIN return? | 1 393 |
| 14 | Which JOIN returns only matching records from both tables? | 1 316 |
| 15 | Which JOIN returns only matching records from both tables? | 1 |
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| 17 | ⚠️ 20. Double Counting in Multiple JOINs
If both "orders" and "payments" have multiple rows per customer, joining them directly can create a many-to-many multiplication.
Example: 2 orders × 3 payments = 6 joined rows. SUM() will overcount.
Understand the grain of each table before joining.
🧠 21. JOINs and Table Grain
Before writing a JOIN, identify:
Table 1 - One row = one customer,
Table 2 - One row = one order → One-to-Many relationship.
Understanding table grain helps prevent: duplicate counts, inflated revenue, incorrect averages, incorrect KPIs.
🎤 SQL Interview Questions
Q1. What is a JOIN?
Combines rows from multiple tables using a related condition.
Q2. What is the difference between INNER JOIN and LEFT JOIN?
INNER returns only matching, LEFT returns all from left + matching from right.
Q3. How do you find customers who never placed an order?
LEFT JOIN + WHERE o.customer_id IS NULL
Q4. What is a SELF JOIN?
Joins a table to itself, for hierarchical relationships.
Q5. What is a CROSS JOIN?
Creates every possible combination.
Q6. Why can JOINs create duplicate rows?
Because of one-to-many or many-to-many relationships.
Q7. Why should you understand table grain?
Because grain determines how rows multiply and whether aggregations become inaccurate.
Q8. What happens when there is no match in a LEFT JOIN?
Columns from right become NULL.
Q9. How do you count unique customers after a JOIN?
COUNT(DISTINCT customer_id)
Q10. Can a query contain multiple JOINs?
Yes.
📝 Practice Questions
Practice 1: Return customer names and their orders.
SELECT c.customer_name, o.order_id
FROM customers c
JOIN orders o ON c.customer_id = o.customer_id;
Practice 2: Find customers who have never ordered.
SELECT c.customer_id, c.customer_name
FROM customers c
LEFT JOIN orders o ON c.customer_id = o.customer_id
WHERE o.customer_id IS NULL;
Practice 3: Calculate total spending per customer.
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;
Practice 4: Return all customers and their order counts, including zero orders.
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;
Practice 5: Find number of unique customers who placed orders.
SELECT COUNT(DISTINCT c.customer_id) AS unique_customers
FROM customers c
JOIN orders o ON c.customer_id = o.customer_id;
🧪 Mini SQL Challenge
Write a query that returns: Customer name, Product name, Category, Amount - Only orders > ₹1,000.
Solution:
SELECT c.customer_name, p.product_name, p.category, 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
WHERE o.amount > 1000
ORDER BY o.amount DESC;
📌 JOINs are the bridge between database tables. But writing a JOIN is only half the skill. A strong Data Analyst also understands: What each table represents → How tables are related → How rows will multiply → How that affects the KPI.
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| 18 | 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) | 725 |
| 19 | 🚀 SQL Roadmap 2026 — Part 10
SQL JOINs — Combining Data from Multiple Tables
In real-world databases, information is rarely stored in one table.
For example: customers, orders, products, payments, employees, departments
A customer may exist in one table while their orders exist in another. JOINs allow us to combine related data from multiple tables. This is one of the most important SQL concepts for a Data Analyst.
🧠 1. Why Do We Need JOINs?
Suppose we have two tables:
customers
customer_id | customer_name
101 | Alice
102 | Bob
103 | Charlie
orders
order_id | customer_id | amount
1 | 101 | 500
2 | 101 | 800
3 | 102 | 300
The customer name is stored in "customers". The order amount is stored in "orders".
To answer:
«How much did each customer spend?»
We need to combine the tables. That's where JOIN comes in.
🔗 2. Basic JOIN Structure
SELECT
c.customer_name,
o.order_id,
o.amount
FROM customers c
JOIN orders o
ON c.customer_id = o.customer_id;
Here: customers → c, orders → o. These are called table aliases.
The condition: ON c.customer_id = o.customer_id tells SQL how the tables are related.
🔑 3. The JOIN Key
A JOIN usually connects tables through a related column.
customers.customer_id ↓ orders.customer_id
Often: one table contains a primary key, another table contains the corresponding foreign key.
Example: customers.customer_id → Primary Key, orders.customer_id → Foreign Key.
🧩 4. INNER JOIN
"INNER JOIN" returns only rows that have a match in both tables.
SELECT c.customer_name, o.order_id, o.amount
FROM customers c
INNER JOIN orders o ON c.customer_id = o.customer_id;
Result:
Alice | 1 | 500
Alice | 2 | 800
Bob | 3 | 300
Charlie is missing because Charlie has no matching order.
Customers ∩ Orders - Only matching records.
👈 5. LEFT JOIN
"LEFT JOIN" returns: All rows from the left table + matching rows from the right table.
SELECT c.customer_name, o.order_id, o.amount
FROM customers c
LEFT JOIN orders o ON c.customer_id = o.customer_id;
Result includes:
Charlie | NULL | NULL
🎯 6. Finding Customers Who Never Ordered
This is a very common interview and analytics problem.
SELECT c.customer_id, c.customer_name
FROM customers c
LEFT JOIN orders o ON c.customer_id = o.customer_id
WHERE o.customer_id IS NULL;
This technique is often called an anti-join pattern.
👉 7. RIGHT JOIN
"RIGHT JOIN" returns: All rows from the right table + matching rows from the left table.
In practice, many analysts prefer rewriting a RIGHT JOIN as a LEFT JOIN by switching table order because it is often easier to read.
🔄 8. FULL OUTER JOIN
"FULL OUTER JOIN" returns: All rows from both tables, whether they match or not.
Conceptually:
LEFT JOIN + RIGHT JOIN
It can reveal: matching records, customers without orders, orders without matching customers.
⚠️ Not every database supports "FULL OUTER JOIN" directly.
🆚 9. INNER JOIN vs LEFT JOIN
• INNER JOIN = Returns only customers with matching orders.
• LEFT JOIN = Returns all customers, including those without orders.
Simple rule:
INNER JOIN = matching records,
LEFT JOIN = keep everything from the left table.
📊 10. JOIN + Aggregation
Question:
«How much has each customer spent?» | 711 |
| 20 |  🚀 𝗧𝗔𝗧𝗔 𝗚𝗿𝗼𝘂𝗽 𝗙𝗥𝗘𝗘 𝗩𝗶𝗿𝘁𝘂𝗮𝗹 𝗜𝗻𝘁𝗲𝗿𝗻𝘀𝗵𝗶𝗽 𝗣𝗿𝗼𝗴𝗿𝗮𝗺𝘀 😍
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