Data Analytics
Perfect channel to learn Data Analytics Learn SQL, Python, Alteryx, Tableau, Power BI and many more For Promotions: @coderfun @love_data
Ko'proq ko'rsatish๐ Telegram kanali Data Analytics analitikasi
Data Analytics (@sqlspecialist) Ingliz til segmentidagi kanali faol ishtirokchi. Hozirda hamjamiyat 109 681 obunachidan iborat bo'lib, Texnologiyalar & Aralashmalar toifasida 1 122-o'rinni va Hindiston mintaqasida 2 340-o'rinni egallagan.
๐ Auditoriya koโrsatkichlari va dinamika
ะฝะตะฒัะดะพะผะพ sanasidan buyon loyiha tez oโsib, 109 681 obunachiga ega boโldi.
24 Iyun, 2026 dagi oxirgi maโlumotlarga koโra kanal barqaror faollikka ega. Oxirgi 30 kunda obunachilar soni 584 ga, soโnggi 24 soatda esa 71 ga oโzgardi va umumiy qamrov yuqori darajada qolmoqda.
- Tasdiqlash holati: Tasdiqlanmagan
- Jalb etish (ER): Auditoriya oโrtacha 2.76% darajada jalb etiladi. Nashrdan keyingi dastlabki 24 soatda kontent odatda umumiy obunachilar sonining 0.68% ini tashkil etuvchi reaksiyalarni toโplaydi.
- Post qamrovi: Har bir post oโrtacha 3 024 marta koโriladi; birinchi sutkada odatda 743 ta koโrish yigโiladi.
- Reaksiyalar va oโzaro taโsir: Auditoriya faol: har bir postga oโrtacha 8 ta reaksiya keladi.
- Tematik yoโnalishlar: Kontent row, sql, analytic, analyst, visualization kabi asosiy mavzularga jamlangan.
๐ Tavsif va kontent siyosati
Muallif resursni shaxsiy fikrni ifoda etish maydoni sifatida taโriflaydi:
โPerfect channel to learn Data Analytics
Learn SQL, Python, Alteryx, Tableau, Power BI and many more
For Promotions: @coderfun @love_dataโ
Yuqori yangilanish chastotasi (oxirgi maโlumot 25 Iyun, 2026 da olingan) sababli kanal doimo dolzarb va katta qamrovli boโlib qoladi. Analitika auditoriya kontent bilan faol hamkorlik qilishini, uni Texnologiyalar & Aralashmalar toifasidagi muhim taโsir nuqtasiga aylantirishini koโrsatadi.
SELECT e1.employee_id AS Employee, e1.name AS Employee_Name, e2.employee_id AS Manager, e2.name AS Manager_Name FROM employees e1 JOIN employees e2 ON e1.manager_id = e2.employee_id;
Explanation:
e1 represents employees.
e2 represents managers.
The join condition e1.manager_id = e2.employee_id matches employees to their managers.
Top 20 SQL Interview Questions
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Hope it helps :)SELECT e1.employee_id AS Employee, e1.name AS Employee_Name, e2.employee_id AS Manager, e2.name AS Manager_Name FROM employees e1 JOIN employees e2 ON e1.manager_id = e2.employee_id;
Explanation:
e1 represents employees.
e2 represents managers.
The join condition e1.manager_id = e2.employee_id matches employees to their managers.
Top 20 SQL Interview Questions
Like this post if you want me to continue this SQL Interview Seriesโฅ๏ธ
Share with credits: https://t.me/sqlspecialist
Hope it helps :)SELECT employee_id, salary, SUM(salary) OVER (ORDER BY employee_id) AS running_total FROM employees;
Explanation:
SUM(salary) OVER (ORDER BY employee_id) calculates a cumulative sum.
The ORDER BY employee_id ensures the total is calculated sequentially.
Running Total Partitioned by a Category
To calculate the running total within groups (e.g., per department): ๐
SELECT department_id, employee_id, salary, SUM(salary) OVER (PARTITION BY department_id ORDER BY employee_id) AS running_total FROM employees;
Top 20 SQL Interview Questions
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Hope it helps :)SELECT employee_id, department_id FROM employees UNION SELECT employee_id, department_id FROM managers;
2๏ธโฃ UNION ALL (Keeps Duplicates)
Combines result sets without removing duplicates.
Faster than UNION because it doesnโt perform duplicate elimination.
SELECT employee_id, department_id FROM employees UNION ALL SELECT employee_id, department_id FROM managers;
Key Differences:
UNION removes duplicates, which may cause performance overhead.
UNION ALL keeps all records, making it more efficient.
Top 20 SQL Interview Questions
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Hope it helps :)SELECT * FROM employees WHERE salary IS NULL;
This retrieves all employees where the salary is missing.
Find Missing Values in Multiple Columns
SELECT * FROM employees WHERE salary IS NULL OR department_id IS NULL;
This checks for NULL values in both the salary and department_id columns.
Count Missing Values in Each Column
SELECT COUNT(*) AS total_rows, COUNT(salary) AS non_null_salaries, COUNT(department_id) AS non_null_departments FROM employees;
Since COUNT(column_name) ignores NULL values, subtracting it from COUNT(*) gives the number of missing values.
Top 20 SQL Interview Questions
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Hope it helps :)
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