Data Analytics
Perfect channel to learn Data Analytics Learn SQL, Python, Alteryx, Tableau, Power BI and many more For Promotions: @coderfun @love_data
Show more๐ Analytical overview of Telegram channel Data Analytics
Channel Data Analytics (@sqlspecialist) in the English language segment is an active participant. Currently, the community unites 109 681 subscribers, ranking 1 122 in the Technologies & Applications category and 2 340 in the India region.
๐ Audience metrics and dynamics
Since its creation on ะฝะตะฒัะดะพะผะพ, the project has demonstrated rapid growth, gathering an audience of 109 681 subscribers.
According to the latest data from 24 June, 2026, the channel demonstrates stable activity. Although there has been a change in the number of participants by 584 over the last 30 days and by 71 over the last 24 hours, overall reach remains high.
- Verification status: Not verified
- Engagement rate (ER): The average audience engagement rate is 2.76%. Within the first 24 hours after publication, content typically collects 0.68% reactions from the total number of subscribers.
- Post reach: On average, each post receives 3 024 views. Within the first day, a publication typically gains 743 views.
- Reactions and interaction: The audience actively supports content: the average number of reactions per post is 8.
- Thematic interests: Content is focused on key topics such as row, sql, analytic, analyst, visualization.
๐ Description and content policy
The author describes the resource as a platform for expressing subjective opinions:
โPerfect channel to learn Data Analytics
Learn SQL, Python, Alteryx, Tableau, Power BI and many more
For Promotions: @coderfun @love_dataโ
Thanks to the high frequency of updates (latest data received on 25 June, 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.
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
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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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