SQL Programming Resources
Open in Telegram
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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Network:📈 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
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Channel Posts
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| 4 | Which clause is evaluated to filter rows before GROUP BY? | 1 147 |
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| 6 | What does COUNT(*) count? | 994 |
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| 8 | What is the key difference between UNION and UNION ALL? | 1 278 |
| 9 | Which clause is evaluated to filter rows before GROUP BY? | 1 143 |
| 10 | What does DISTINCT do? | 1 148 |
| 11 | What will this query return?
SELECT COUNT(salary)
FROM Employee; If the table has 10 rows and 3 employees have NULL salary: | 1 145 |
| 12 | What does COUNT(*) count? | 1 134 |
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| 15 | 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
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1.33 ₽ · /balance_help | 1 778 |
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| 18 | 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.
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-----
1.3 ₽ · /balance_help | 1 894 |
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