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SQL Programming Resources

SQL Programming Resources

Kanalga Telegram’da o‘tish

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

Ko'proq ko'rsatish

📈 Telegram kanali SQL Programming Resources analitikasi

SQL Programming Resources (@sqlanalyst) Ingliz til segmentidagi kanali faol ishtirokchi. Hozirda hamjamiyat 76 524 obunachidan iborat bo'lib, Texnologiyalar & Aralashmalar toifasida 1 684-o'rinni va Hindiston mintaqasida 4 009-o'rinni egallagan.

📊 Auditoriya ko‘rsatkichlari va dinamika

невідомо sanasidan buyon loyiha tez o‘sib, 76 524 obunachiga ega bo‘ldi.

26 Oktabr, 2026 dagi oxirgi ma’lumotlarga ko‘ra kanal barqaror faollikka ega. Oxirgi 30 kunda obunachilar soni -166 ga, so‘nggi 24 soatda esa -11 ga o‘zgardi va umumiy qamrov yuqori darajada qolmoqda.

  • Tasdiqlash holati: Tasdiqlanmagan
  • Jalb etish (ER): Auditoriya o‘rtacha 2.70% darajada jalb etiladi. Nashrdan keyingi dastlabki 24 soatda kontent odatda umumiy obunachilar sonining 1.02% ini tashkil etuvchi reaksiyalarni to‘playdi.
  • Post qamrovi: Har bir post o‘rtacha 2 063 marta ko‘riladi; birinchi sutkada odatda 778 ta ko‘rish yig‘iladi.
  • Reaksiyalar va o‘zaro ta’sir: Auditoriya faol: har bir postga o‘rtacha 5 ta reaksiya keladi.
  • Tematik yo‘nalishlar: Kontent row, sql, customer_id, logic, desc kabi asosiy mavzularga jamlangan.

📝 Tavsif va kontent siyosati

Muallif resursni shaxsiy fikrni ifoda etish maydoni sifatida ta’riflaydi:
“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”

Yuqori yangilanish chastotasi (oxirgi ma’lumot 27 Oktabr, 2026 da olingan) sababli kanal doimo dolzarb va katta qamrovli bo‘lib qoladi. Analitika auditoriya kontent bilan faol hamkorlik qilishini, uni Texnologiyalar & Aralashmalar toifasidagi muhim ta’sir nuqtasiga aylantirishini ko‘rsatadi.

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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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