DataBolt

All Lessons

Introduction to SQL

SQL Lesson 1: Your Sample Database

SQL Lesson 2: SELECT — Reading data

SQL Lesson 3: WHERE — Filtering rows

SQL Lesson 4: AND, OR, NOT

SQL Lesson 5: BETWEEN, IN, LIKE

SQL Lesson 6: NULL — The Mystery Value

SQL Lesson 7: ORDER BY

SQL Lesson 8: LIMIT & OFFSET

SQL Lesson 9: Aggregate Functions

SQL Lesson 10: GROUP BY

SQL Lesson 11: HAVING

SQL Lesson 12: INNER JOINs

SQL Lesson 13: LEFT JOINs

SQL Lesson 14: RIGHT JOINs

SQL Lesson 15: SELF JOINs

SQL Lesson 16: UNION JOINs

SQL Lesson 17: Joining Multiple Tables

SQL Lesson 18: Subqueries

SQL Lesson 19: CTEs (WITH)

SQL Lesson 20: CASE Statements

SQL Lesson 21: Window Functions

SQL Lesson 22: String Functions

SQL Lesson 23: Date & Time Functions

SQL Lesson 24: INSERT, UPDATE, DELETE

SQL Lesson 25: CREATE TABLE & DDL

SQL Lesson 26: Indexes & Performance

SQL Lesson 27: Transactions & ACID

SQL Lesson 28: SQL Execution Order

SQL Lesson 29: 50 Practice Problems

CHAPTER 13

INNER JOINs

Aggregating by category — the analytics workhorse.

IN THIS CHAPTER

Why JOINs exist and what problem they solve
INNER JOIN — only rows with a match in both tables
LEFT JOIN — all left rows, matched right rows (NULLs where no match)
RIGHT JOIN — the mirror of LEFT JOIN
SELF JOIN — a table joined to itself
Finding rows with no match using LEFT JOIN + IS NULL

🌟 Think of it this way: Two spreadsheets: one with customer names+IDs, another with order amounts+customer IDs. A JOIN merges them so you can see 'Ananya spent ₹1,45,148'. Without JOINs, you would have to manually cross-reference the sheets.

13.1 - INNER JOIN

Returns rows only where there is a match in BOTH tables. Lisa (no orders) is excluded.

SQL

SELECT c.name, o.order_id, o.total, o.status
FROM customers c
INNER JOIN orders o ON c.customer_id = o.customer_id
ORDER BY c.name, o.order_id;

13.2 - INNER JOIN (Customers with High Orders)

Find customers and their order details only for orders greater than 3000. This shows how INNER JOIN works with filtering conditions.

SQL

SELECT c.name, o.order_id, o.total
FROM customers c
INNER JOIN orders o
ON c.customer_id = o.customer_id
WHERE o.total > 3000;

✅ Pro Tip: This is commonly used in business to identify high-value customers.

Included Tables in Exercise

Table: customers

customer_id	name	city	country
1	Ananya	Hyderabad	India
2	Rohan	Bangalore	India
3	Sam	Mumbai	India
4	Lisa	London	UK
5	Ravi	Delhi	India

Table: orders

order_id	customer_id	order_date	total	status
1	1	2024-01-15	79999	completed
2	2	2024-01-20	3498	completed
3	1	2024-02-10	65000	completed
4	3	2024-02-14	499	pending
5	2	2024-03-01	248	completed
6	5	2024-03-15	2999	shipped
7	1	2024-04-05	149	completed

Exercise 👇

Exercise:

Tasks

1.👉Get customer names with their order IDs

2.Show order totals with customer names

3.Orders above 5000

4.Only completed orders

5.Sort by order amount

6.Count orders per customer

7.Total spending per customer

8.Customers with more than 1 order

9.Latest order date per customer

10.Average order value per customer

11.Join 3 tables (customers + orders + order_items)

12.Product names with customer orders

13.Total quantity ordered per product

14.Customers who bought Electronics

15.Highest order per customer

Stuck? Read this task's