SQL Programming Resources
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
نمایش بیشتر📈 تحلیل کانال تلگرام SQL Programming Resources
کانال SQL Programming Resources (@sqlanalyst) در بخش زبانی انگلیسی بازیگری فعال است. در حال حاضر جامعه شامل 76 536 مشترک است و جایگاه 1 684 را در دسته فناوری و برنامهها و رتبه 4 009 را در منطقه الهند دارد.
📊 شاخصهای مخاطب و پویایی
از زمان ایجاد در невідомо، پروژه رشد سریعی داشته و 76 536 مشترک جذب کرده است.
بر اساس آخرین دادهها در تاریخ 26 اکتبر, 2026، کانال فعالیت پایداری دارد. در ۳۰ روز گذشته تغییر اعضا برابر -166 و در ۲۴ ساعت گذشته برابر -11 بوده و همچنان دسترسی گستردهای حفظ شده است.
- وضعیت تأیید: تأیید نشده
- نرخ تعامل (ER): میانگین تعامل مخاطب 2.70% است و در ۲۴ ساعت نخست پس از انتشار، محتوا معمولاً 1.02% واکنش نسبت به کل مشترکان کسب میکند.
- دسترسی پستها: هر پست به طور میانگین 2 063 بازدید دریافت میکند. در اولین روز معمولاً 778 بازدید جمعآوری میشود.
- واکنشها و تعامل: مخاطبان بهطور فعال حمایت میکنند؛ میانگین واکنش به هر پست 5 است.
- علایق موضوعی: محتوا بر موضوعات کلیدی مانند row, sql, customer_id, logic, desc تمرکز دارد.
📝 توضیح و سیاست محتوایی
نویسنده این فضا را محل بیان دیدگاههای شخصی توصیف میکند:
“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”
به لطف بهروزرسانیهای پرتکرار (آخرین داده در تاریخ 27 اکتبر, 2026)، کانال همواره بهروز و دارای دسترسی بالاست. تحلیلها نشان میدهد مخاطبان بهطور فعال با محتوا تعامل دارند و آن را به نقطه اثرگذاری مهم در دسته فناوری و برنامهها تبدیل کردهاند.
در حال بارگذاری دادهها
| تاریخ | رشد مشترکین | اشارات | کانالها | |
| 27 اکتبر | +1 | |||
| 26 اکتبر | 0 | |||
| 25 اکتبر | +9 | |||
| 24 اکتبر | 0 | |||
| 23 اکتبر | +2 | |||
| 22 اکتبر | +1 | |||
| 21 اکتبر | 0 | |||
| 20 اکتبر | 0 | |||
| 19 اکتبر | +4 | |||
| 18 اکتبر | 0 | |||
| 17 اکتبر | +2 | |||
| 16 اکتبر | +9 | |||
| 15 اکتبر | 0 | |||
| 14 اکتبر | +3 | |||
| 13 اکتبر | +14 | |||
| 12 اکتبر | 0 | |||
| 11 اکتبر | 0 | |||
| 10 اکتبر | +5 | |||
| 09 اکتبر | 0 | |||
| 08 اکتبر | 0 | |||
| 07 اکتبر | +8 | |||
| 06 اکتبر | +3 | |||
| 05 اکتبر | +4 | |||
| 04 اکتبر | +13 | |||
| 03 اکتبر | +12 | |||
| 02 اکتبر | +5 | |||
| 01 اکتبر | +15 |
| 2 | What happens when you use an INNER JOIN? | 1 303 |
| 3 | Which window function gives the previous row's value? | 1 206 |
| 4 | Which clause is evaluated to filter rows before GROUP BY? | 1 147 |
| 5 | What does DISTINCT do? | 991 |
| 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)
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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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