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SQL Programming Resources

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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📈 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 76 536 suscriptores, ocupando la posición 1 677 en la categoría Tecnologías y Aplicaciones y el puesto 4 007 en la región India.

📊 Métricas de audiencia y dinámica

Desde su creación el невідомо, el proyecto ha mostrado un crecimiento acelerado, reuniendo a 76 536 suscriptores.

Según los últimos datos del 25 octubre, 2026, el canal mantiene una actividad estable. En los últimos 30 días la variación de miembros fue de -171, y en las últimas 24 horas de -10, conservando un alto alcance.

  • Estado de verificación: No verificado
  • Tasa de interacción (ER): El promedio de interacción de la audiencia es 2.69%. Durante las primeras 24 horas tras publicar, el contenido suele obtener 1.02% de reacciones respecto al total de suscriptores.
  • Alcance de las publicaciones: Cada publicación recibe en promedio 2 062 visualizaciones. En el primer día suele acumular 779 visualizaciones.
  • Reacciones e interacción: La audiencia responde de forma activa: el promedio de reacciones por publicación es 5.
  • 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 26 octubre, 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.

76 536
Suscriptores
-1024 horas
-1057 días
-17130 días
Archivo de publicaciones
What happens when you use an INNER JOIN?
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Which window function gives the previous row's value?
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Which clause is evaluated to filter rows before GROUP BY?
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What does DISTINCT do?
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What does COUNT(*) count?
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What is the key difference between UNION and UNION ALL?
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Which clause is evaluated to filter rows before GROUP BY?
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What does DISTINCT do?
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What will this query return? SELECT COUNT(salary) FROM Employee; If the table has 10 rows and 3 employees have NULL salary:
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What does COUNT(*) count?
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SQL Interview Series — Part 13 📌 Question 13: Find Employees Earning More Than the Company Average Suppose you have an Employee table:
employee_id | employee_name | department | salary
------------|---------------|------------|-------
1           | Amit          | IT         | 50000
2           | Rahul         | IT         | 80000
3           | Priya         | HR         | 60000
4           | Neha          | HR         | 90000
5           | Raj           | Finance    | 70000
❓ Find all employees whose salary is greater than the average salary of the entire company. 💡 Approach First, calculate the average salary across all employees. Then compare every employee's salary with that average. Company average: (50000 + 80000 + 60000 + 90000 + 70000) / 5 = 70000 Employees earning more than 70000: Rahul → 80000 Neha → 90000 📌 SQL Solution
SELECT
    employee_id,
    employee_name,
    department,
    salary
FROM Employee
WHERE salary > (
    SELECT AVG(salary)
    FROM Employee
);
🔎 How does it work? The subquery:
SELECT AVG(salary)
FROM Employee;
calculates the company-wide average: 70000 The outer query then checks:
WHERE salary > 70000
and returns only employees earning above that amount. 📌 Expected Result:
employee_name | department | salary
--------------|------------|-------
Rahul         | IT         | 80000
Neha          | HR         | 90000
Double Tap ❤️ For More ----- 1.33 ₽ · /balance_help

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SQL Interview Series — Part 12 📌 Question 12: Find the Department with the Highest Average Salary Suppose you have an Employee table:
employee_id | employee_name | department | salary
------------|---------------|------------|-------
1           | Amit          | IT         | 60000
2           | Rahul         | IT         | 80000
3           | Priya         | HR         | 70000
4           | Neha          | HR         | 90000
5           | Raj           | Finance    | 85000
6           | Karan         | Finance    | 95000
❓ Find the department that has the highest average salary. 💡 Approach First, calculate the average salary for every department. Then, find the department with the highest average. Department averages: IT → (60000 + 80000) / 2 = 70000 HR → (70000 + 90000) / 2 = 80000 Finance → (85000 + 95000) / 2 = 90000 Therefore: Finance → 90000 📌 SQL Solution
SELECT
    department,
    AVG(salary) AS avg_salary
FROM Employee
GROUP BY department
ORDER BY avg_salary DESC
LIMIT 1;
Expected result:
department | avg_salary
-----------|-----------
Finance    | 90000
🔎 How does it work? 1️⃣ "GROUP BY department" Creates one group for each department. 2️⃣ "AVG(salary)" Calculates the average salary within each department. 3️⃣ "ORDER BY avg_salary DESC" Places the department with the highest average first. 4️⃣ "LIMIT 1" Returns only the top department. ⚠️ Interview Trap What if two departments have the same highest average salary? For example: IT → 80000 HR → 80000 Finance → 75000 "LIMIT 1" would return only one department. If the interviewer asks: 👉 "Find all departments having the highest average salary" then use a ranking approach:
SELECT department, avg_salary
FROM (
    SELECT
        department,
        AVG(salary) AS avg_salary,
        DENSE_RANK() OVER (
            ORDER BY AVG(salary) DESC
        ) AS rnk
    FROM Employee
    GROUP BY department
) d
WHERE rnk = 1;
Now both IT and HR would be returned. 🎯 Double Tap ❤️ For More ----- 1.3 ₽ · /balance_help

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SQL Interview Series — Part 11 📌 Question 11: Calculate a Running Total of Sales Suppose you have a Sales table:
sale_id | sale_date  | amount
--------|------------|-------
1       | 2026-01-01 | 1000
2       | 2026-01-02 | 1500
3       | 2026-01-03 | 800
4       | 2026-01-04 | 1200
❓ Calculate the cumulative (running) sales amount for each day. 💡 Approach A running total means: Current day's sales + all previous sales For example: 1000 1000 + 1500 = 2500 2500 + 800 = 3300 3300 + 1200 = 4500 The easiest approach is to use the "SUM()" window function. 📌 SQL Solution
SELECT
    sale_date,
    amount,
    SUM(amount) OVER (
        ORDER BY sale_date
        ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW
    ) AS running_total
FROM Sales
ORDER BY sale_date;
🔎 How does it work? 1️⃣ "ORDER BY sale_date" Determines the order in which the running total is calculated. 2️⃣ "SUM(amount) OVER(...)" Calculates the sum without collapsing the rows like a normal "GROUP BY". 3️⃣ "UNBOUNDED PRECEDING" Means: Start from the first row. 4️⃣ "CURRENT ROW" Means: Include the current row in the total. So the calculation becomes:
Date        Amount    Running Total
----------  --------  -------------
Jan 01       1000        1000
Jan 02       1500        2500
Jan 03        800        3300
Jan 04       1200        4500
⚠️ Interview Trap Don't use: GROUP BY sale_date for a running total. "GROUP BY" collapses rows into groups. A window function allows us to calculate an aggregate while keeping individual rows. Also, if multiple rows can have the same "sale_date", be careful about the ordering. A unique column such as "sale_id" may need to be included:
SUM(amount) OVER (
    ORDER BY sale_date, sale_id
    ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW
)
🎯 Double Tap ❤️ For More ----- 1.31 ₽ · /balance_help