اکنون در دسترس! پژوهش تلگرام ۲۰۲۵ — مهمترین بینشهای سال 
Data Analytics
رفتن به کانال در Telegram
Perfect channel to learn Data Analytics Learn SQL, Python, Alteryx, Tableau, Power BI and many more For Promotions: @coderfun @love_data
نمایش بیشتر📈 تحلیل کانال تلگرام Data Analytics
کانال Data Analytics (@sqlspecialist) در بخش زبانی انگلیسی بازیگری فعال است. در حال حاضر جامعه شامل 109 631 مشترک است و جایگاه 1 124 را در دسته فناوری و برنامهها و رتبه 2 395 را در منطقه الهند دارد.
📊 شاخصهای مخاطب و پویایی
از زمان ایجاد در невідомо، پروژه رشد سریعی داشته و 109 631 مشترک جذب کرده است.
بر اساس آخرین دادهها در تاریخ 17 ژوئن, 2026، کانال فعالیت پایداری دارد. در ۳۰ روز گذشته تغییر اعضا برابر 689 و در ۲۴ ساعت گذشته برابر -19 بوده و همچنان دسترسی گستردهای حفظ شده است.
- وضعیت تأیید: تأیید نشده
- نرخ تعامل (ER): میانگین تعامل مخاطب 3.31% است و در ۲۴ ساعت نخست پس از انتشار، محتوا معمولاً 1.51% واکنش نسبت به کل مشترکان کسب میکند.
- دسترسی پستها: هر پست به طور میانگین 3 624 بازدید دریافت میکند. در اولین روز معمولاً 1 658 بازدید جمعآوری میشود.
- واکنشها و تعامل: مخاطبان بهطور فعال حمایت میکنند؛ میانگین واکنش به هر پست 7 است.
- علایق موضوعی: محتوا بر موضوعات کلیدی مانند row, sql, analytic, analyst, visualization تمرکز دارد.
📝 توضیح و سیاست محتوایی
نویسنده این فضا را محل بیان دیدگاههای شخصی توصیف میکند:
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Learn SQL, Python, Alteryx, Tableau, Power BI and many more
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به لطف بهروزرسانیهای پرتکرار (آخرین داده در تاریخ 18 ژوئن, 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.
109 631
مشترکین
-1924 ساعت
+2267 روز
+68930 روز
آرشیو پست ها
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🔥 Now, let’s move to the next topic of SQL Roadmap
✅ Subqueries (Nested Queries)
🧠 1. What is a Subquery?
A subquery is a query inside another query
👉 Think like this:
“First get some data → then use that result in another query”
📊 Basic Example
👉 Find employees earning above average salary
SELECT FROM employees
WHERE salary > (
SELECT AVG(salary) FROM employees
);
👉 Inner query → gives average salary
👉 Outer query → filters employees
⚡ 2. Types of Subqueries
🔹 Single Row Subquery
Returns only one value
SELECT FROM employees
WHERE salary > (SELECT AVG(salary) FROM employees);
🔹 Multiple Row Subquery
Returns multiple values
SELECT FROM employees
WHERE dept_id IN (
SELECT dept_id FROM departments
);
🎯 3. Subquery with IN
SELECT name
FROM employees
WHERE dept_id IN (
SELECT dept_id FROM departments WHERE dept_name = 'IT'
);
✔ Finds employees in IT department
⚡ 4. Subquery with EXISTS
SELECT name
FROM employees e
WHERE EXISTS (
SELECT 1 FROM departments d
WHERE e.dept_id = d.dept_id
);
✔ Checks if matching record exists
🚨 5. Important Difference
IN ->
Compares values & Slower with large data
EXISTS -> Checks existence & Faster with large data
🎯 6. Practice Tasks
1. Find employees with salary > average salary
2. Find employees in IT department using subquery
3. Get departments that have employees
4. Find employees with max salary
5. Get employees not in HR department
🔥 Here are the solutions for Subqueries practice tasks
✅ 1. Find employees with salary > average salary
SELECT FROM employees
WHERE salary > (
SELECT AVG(salary) FROM employees
);
✅ 2. Find employees in IT department using subquery
SELECT FROM employees
WHERE dept_id IN (
SELECT dept_id FROM departments
WHERE dept_name = 'IT'
);
✅ 3. Get departments that have employees
SELECT FROM departments
WHERE dept_id IN (
SELECT dept_id FROM employees
);
👉 Alternative (using EXISTS):
SELECT FROM departments d
WHERE EXISTS (
SELECT 1 FROM employees e
WHERE d.dept_id = e.dept_id
);
✅ 4. Find employees with max salary
SELECT FROM employees
WHERE salary = (
SELECT MAX(salary) FROM employees
);
✅ 5. Get employees not in HR department
SELECT FROM employees
WHERE dept_id NOT IN (
SELECT dept_id FROM departments
WHERE dept_name = 'HR'
);
⚡ Mini Challenge 🔥
👉 Find employees earning second highest salary using subquery
⚡ Mini Challenge Solution 🔥
SELECT FROM employees
WHERE salary = (
SELECT MAX(salary)
FROM employees
WHERE salary < (
SELECT MAX(salary) FROM employees
)
);
🔥 Pro Tip:
If question says:
👉 “above average”, “max”, “second highest”
→ Think Subquery instantly 💯
Double Tap ❤️ For More
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What will this query return?
SELECT e.name, d.dept_name FROM employees e LEFT JOIN departments d ON e.dept_id = d.dept_id WHERE d.dept_name = 'IT';
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What will this query return?
SELECT COUNT(*) FROM employees e LEFT JOIN departments d ON e.dept_id = d.dept_id;
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What is the difference?
-- Query 1 SELECT COUNT(*) FROM employees e INNER JOIN departments d ON e.dept_id = d.dept_id; -- Query 2 SELECT COUNT(*) FROM employees e LEFT JOIN departments d ON e.dept_id = d.dept_id;
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🔥Now, let’s move to the most important SQL topic — JOINS 💯🔥
🧠 1. What is a JOIN?
JOIN is used to combine data from multiple tables
👉 If data is stored in different tables → JOIN helps you connect them
📊 Example Tables
👨💼 employees
emp_id name dept_id
1 Amit 101
2 Neha 102
3 Ravi 101
🏢 departments
dept_id dept_name
101 IT
102 HR
🔗 2. INNER JOIN (Most Used 🔥)
👉 Returns only matching records
SELECT e.name, d.dept_name
FROM employees e
INNER JOIN departments d
ON e.dept_id = d.dept_id;
✔ Only employees with valid department
⬅️ 3. LEFT JOIN
👉 Returns all records from left table + matched from right
SELECT e.name, d.dept_name
FROM employees e
LEFT JOIN departments d
ON e.dept_id = d.dept_id;
✔ Even if department is missing → employee will still show
➡️ 4. RIGHT JOIN
👉 Returns all records from right table + matched from left
SELECT e.name, d.dept_name
FROM employees e
RIGHT JOIN departments d
ON e.dept_id = d.dept_id;
🔄 5. FULL JOIN
👉 Returns all records from both tables
SELECT e.name, d.dept_name
FROM employees e
FULL JOIN departments d
ON e.dept_id = d.dept_id;
(Note: Not supported in MySQL directly — use UNION instead)
⚡ 6. Quick Summary
• INNER -> Only matching rows
• LEFT -> All left + matched right
• RIGHT -> All right + matched left
• FULL -> Everything
🎯 7. Practice Tasks
1. Get employee name with department name
2. Show all employees even if no department
3. Show all departments even if no employees
4. Find employees without department
5. Count employees per department (using JOIN)
🔥 Practice Task Solutions 👇
✅ 1. Get employee name with department name (INNER JOIN)
SELECT e.name, d.dept_name
FROM employees e
INNER JOIN departments d
ON e.dept_id = d.dept_id;
✅ 2. Show all employees even if no department (LEFT JOIN)
SELECT e.name, d.dept_name
FROM employees e
LEFT JOIN departments d
ON e.dept_id = d.dept_id;
✅ 3. Show all departments even if no employees (RIGHT JOIN)
SELECT e.name, d.dept_name
FROM employees e
RIGHT JOIN departments d
ON e.dept_id = d.dept_id;
👉 (Alternative using LEFT JOIN — interview friendly)
SELECT e.name, d.dept_name
FROM departments d
LEFT JOIN employees e
ON e.dept_id = d.dept_id;
✅ 4. Find employees without department
SELECT e.name
FROM employees e
LEFT JOIN departments d
ON e.dept_id = d.dept_id
WHERE d.dept_id IS NULL;
✅ 5. Count employees per department (using JOIN)
SELECT d.dept_name, COUNT(e.emp_id) AS total_emp
FROM departments d
LEFT JOIN employees e
ON d.dept_id = e.dept_id
GROUP BY d.dept_name;
⚡ Mini Challenge 🔥
👉 Find departments with no employees
⚡ Mini Challenge Solution 🔥
👉 Find departments with no employees
SELECT d.dept_name
FROM departments d
LEFT JOIN employees e
ON d.dept_id = e.dept_id
WHERE e.emp_id IS NULL;
Double Tap ❤️ For More
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What does this query return?
SELECT department, AVG(salary) FROM employees GROUP BY department HAVING AVG(salary) > 50000;
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What will this query return?
SELECT department, COUNT(*) FROM employees GROUP BY department HAVING COUNT(*) > 2;
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✅ Business Intelligence (BI) Acronyms You Should Know 📊💡
BI → Business Intelligence
ETL → Extract, Transform, Load
ELT → Extract, Load, Transform
DWH → Data Warehouse
OLAP → Online Analytical Processing
OLTP → Online Transaction Processing
KPI → Key Performance Indicator
SLA → Service Level Agreement
SCD → Slowly Changing Dimension
CDC → Change Data Capture
MDM → Master Data Management
EAV → Entity Attribute Value
FACT → Fact Table
DIM → Dimension Table
STAR → Star Schema
SNOWFLAKE → Snowflake Schema
MTD → Month To Date
QTD → Quarter To Date
YTD → Year To Date
MoM → Month over Month
YoY → Year over Year
ROI → Return on Investment
TAT → Turn Around Time
💡Don’t just expand acronyms — explain where they’re used (ETL in pipelines, KPIs in dashboards, OLAP in analysis).
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Now, let’s move to the next topic of SQL Roadmap:
✅ HAVING Clause
🧠 1. What is HAVING?
HAVING is used to filter grouped data
👉 Think like this:
- WHERE → filters rows (before grouping)
- HAVING → filters groups (after grouping)
⚡ 2. WHERE vs HAVING
Works on
- WHERE: Rows
- HAVING: Groups
Used with
- WHERE: SELECT
- HAVING: GROUP BY
Can use aggregates?
- WHERE: ❌ No
- HAVING: ✅ Yes
💡 3. Basic Syntax
SELECT column, AGG_FUNCTION(column)
FROM table_name
GROUP BY column
HAVING condition;
🎯 4. Examples (Very Important)
👉 Example 1: Departments with more than 2 employees
SELECT department, COUNT() AS total_emp
FROM employees
GROUP BY department
HAVING COUNT() > 2;
👉 Example 2: Departments with average salary > 50k
SELECT department, AVG(salary) AS avg_salary
FROM employees
GROUP BY department
HAVING AVG(salary) > 50000;
⚡ 5. WHERE + HAVING Together
SELECT department, COUNT() AS total_emp
FROM employees
WHERE salary > 30000
GROUP BY department
HAVING COUNT() > 2;
👉 Step-by-step:
1. WHERE → filters employees with salary > 30k
2. GROUP BY → groups by department
3. HAVING → keeps only groups with > 2 employees
🎯 6. Practice Tasks
1. Find departments having more than 3 employees
2. Find departments with average salary > 60k
3. Count employees per department where salary > 40k
4. Show departments with total salary > 1,00,000
5. Find departments with minimum salary > 30k
⚡ Mini Challenge 🔥
👉 Find departments having more than 2 employees AND avg salary > 50k
Double Tap ❤️ For Practice Task Solution
