Data Careers Resources & Job Updates | iamrupnath
👉 Connect LinkedIn : https://www.linkedin.com/in/rupnath-shaw Google Search => Techcompreviews IG: @iamrupnath Perfect channel for Data Careers, Job Updates Learn Excel, SQL, Python, Tableau, Power BI, AI tools, AI tips & tricks and many more
显示更多📈 Telegram 频道 Data Careers Resources & Job Updates | iamrupnath 的分析概览
频道 Data Careers Resources & Job Updates | iamrupnath (@codewithrup) 英语 语言赛道中的 是活跃参与者。目前社区聚集了 21 400 名订阅者,在 技术与应用 类别中位列第 6 134,并在 印度 地区排名第 19 532 位。
📊 受众指标与增长动态
自 невідомо 创建以来,项目保持高速增长,吸引了 21 400 名订阅者。
根据 28 七月, 2026 的最新数据,频道保持稳定运转。过去 30 天订阅人数变化为 -413,过去 24 小时变化为 -15,整体触达仍然可观。
- 认证状态: 未认证
- 互动率 (ER): 平均受众互动率为 4.36%。内容发布后 24 小时内通常能获得 1.29% 的反应,占订阅者总量。
- 帖子覆盖: 每篇帖子平均可获得 933 次浏览,首日通常累积 277 次浏览。
- 互动与反馈: 受众积极参与,单帖平均反应数为 1。
- 主题关注点: 内容集中在 apply, qualification, bachelor, degree, engineer 等核心主题上。
📝 描述与内容策略
作者将该频道定位为表达主观观点的平台:
“👉 Connect LinkedIn :
https://www.linkedin.com/in/rupnath-shaw
Google Search => Techcompreviews
IG: @iamrupnath
Perfect channel for Data Careers, Job Updates
Learn Excel, SQL, Python, Tableau, Power BI, AI tools, AI tips & tricks and many more”
凭借高频更新(最新数据采集于 29 七月, 2026),频道始终保持新鲜度与高覆盖。分析显示受众积极互动,使其成为 技术与应用 类别中的关键影响点。
SELECT *, missing indexes, functions in WHERE, unnecessary JOINs, subqueries.
https://www.linkedin.com/feed/update/urn:li:share:7488197667341385728/SELECT
employee_id,
total_projects
FROM (
SELECT
employee_id,
COUNT(DISTINCT project_id) AS total_projects,
DENSE_RANK() OVER (
ORDER BY COUNT(DISTINCT project_id) DESC
) AS rnk
FROM employee_projects
GROUP BY employee_id
) ranked
WHERE rnk = 1;
💡 Explanation:
This query counts the number of unique projects each employee has worked on and identifies those with the highest count.
• COUNT(DISTINCT project_id) counts unique projects for each employee
• GROUP BY employee_id creates one record per employee
• DENSE_RANK() ranks employees based on the number of projects
• The outer query returns all employees tied for the highest number of projects
This question tests your understanding of:
✅ COUNT(DISTINCT)
✅ GROUP BY
✅ Window Functions DENSE_RANK
✅ Ranking Aggregated Results
🎯 Expected Output Example
Employee ID | Total Projects
101 | 12
205 | 12
Both employees have worked on the highest number of distinct projects.
🚀 Alternative Without Window Functions
SELECT
employee_id,
COUNT(DISTINCT project_id) AS total_projects
FROM employee_projects
GROUP BY employee_id
HAVING COUNT(DISTINCT project_id) = (
SELECT MAX(project_count)
FROM (
SELECT
COUNT(DISTINCT project_id) AS project_count
FROM employee_projects
GROUP BY employee_id
) t
);
This solution uses nested subqueries and MAX() instead of window functions.
🚀 Tip for SQL Job Seekers:
Many interview questions involve ranking aggregated results, such as:
Highest number of projects, Most orders, Maximum sales, Highest attendance, Most logins
Practice combining GROUP BY with window functions like DENSE_RANK() to solve these efficiently.
❤️ React with ❤️ for more interview challenges!