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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

Network:Data Analytics India4 009 Technologies & Applications1 684...

📈 Analytical overview of Telegram channel SQL Programming Resources

Channel SQL Programming Resources (@sqlanalyst) in the English language segment is an active participant. Currently, the community unites 76 536 subscribers, ranking 1 684 in the Technologies & Applications category and 4 009 in the India region.

📊 Audience metrics and dynamics

Since its creation on невідомо, the project has demonstrated rapid growth, gathering an audience of 76 536 subscribers.

According to the latest data from 26 October, 2026, the channel demonstrates stable activity. Although there has been a change in the number of participants by -166 over the last 30 days and by -11 over the last 24 hours, overall reach remains high.

Verification status: Not verified
Engagement rate (ER): The average audience engagement rate is 2.70%. Within the first 24 hours after publication, content typically collects 1.02% reactions from the total number of subscribers.
Post reach: On average, each post receives 2 063 views. Within the first day, a publication typically gains 778 views.
Reactions and interaction: The audience actively supports content: the average number of reactions per post is 5.
Thematic interests: Content is focused on key topics such as row, sql, customer_id, logic, desc.

📝 Description and content policy

The author describes the resource as a platform for expressing subjective opinions:
“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”

Thanks to the high frequency of updates (latest data received on 27 October, 2026), the channel maintains relevance and a high level of publication reach. Analytics show that the audience actively interacts with content, making it an important point of influence in the Technologies & Applications category.

76 536

Subscribers

-1124 hours

-1017 days

-16630 days

2 063

Post views

~ 77824 hours

~ 1 03248 hours

2.70%

Engagement rate

~ 1

Posts per day

Ads index

beta

Posts Archive

76 537

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76 537

What happens when you use an INNER JOIN?

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76 537

Which window function gives the previous row's value?

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76 537

Which clause is evaluated to filter rows before GROUP BY?

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76 537

What does DISTINCT do?

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76 537

What does COUNT(*) count?

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76 537

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76 537

What is the key difference between UNION and UNION ALL?

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76 537

Which clause is evaluated to filter rows before GROUP BY?

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76 537

What does DISTINCT do?

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76 537

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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76 537

What does COUNT(*) count?

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76 537

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76 537

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76 537

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

76 537

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76 537

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

76 537

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76 537

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