INNER JOIN syntax in SQL - Step-by-Step Execution

Concept Flow - INNER JOIN syntax

Start with Table A

↓

Start with Table B

↓

Compare rows from A and B

↓

Match rows where join condition is true

↓

Combine matched rows into one result row

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Output combined rows as result set

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End

INNER JOIN takes two tables, compares rows based on a condition, and outputs only rows where the condition matches.

Execution Sample

SQL

SELECT A.id, A.name, B.order_id
FROM Customers A
INNER JOIN Orders B ON A.id = B.customer_id;

This query joins Customers and Orders tables, showing customers with their orders only if they have matching IDs.

Execution Table

Step	Action	Table A Row	Table B Row	Join Condition (A.id = B.customer_id)	Result Row Produced
1	Compare first row of A with first row of B	id=1, name='Alice'	order_id=101, customer_id=1	True	id=1, name='Alice', order_id=101
2	Compare first row of A with second row of B	id=1, name='Alice'	order_id=102, customer_id=2	False	No row
3	Compare second row of A with first row of B	id=2, name='Bob'	order_id=101, customer_id=1	False	No row
4	Compare second row of A with second row of B	id=2, name='Bob'	order_id=102, customer_id=2	True	id=2, name='Bob', order_id=102
5	Compare third row of A with first row of B	id=3, name='Carol'	order_id=101, customer_id=1	False	No row
6	Compare third row of A with second row of B	id=3, name='Carol'	order_id=102, customer_id=2	False	No row
7	No more rows to compare				End of join

💡 All rows compared; only rows with matching customer IDs included in result.

Variable Tracker

Variable	Start	After Step 1	After Step 2	After Step 4	Final
Current Table A Row	None	id=1, name='Alice'	id=1, name='Alice'	id=2, name='Bob'	id=3, name='Carol'
Current Table B Row	None	order_id=101, customer_id=1	order_id=102, customer_id=2	order_id=102, customer_id=2	order_id=102, customer_id=2
Join Condition Result	None	True	False	True	False
Result Rows Produced	Empty	[id=1, Alice, 101]	[id=1, Alice, 101]	[id=1, Alice, 101], [id=2, Bob, 102]	[id=1, Alice, 101], [id=2, Bob, 102]

Key Moments - 2 Insights

Why are some rows from Table A not in the result?

Rows from Table A only appear if they match a row in Table B on the join condition. See execution_table rows 2, 3, 5, 6 where condition is False, so no result row is produced.

Does INNER JOIN include rows with no matching partner in the other table?

No. INNER JOIN only outputs rows where the join condition is True. Rows without matches are skipped, as shown in execution_table where some comparisons produce no result.

Visual Quiz - 3 Questions

Look at the execution_table, what is the result row produced at Step 4?

A id=1, name='Alice', order_id=102

B id=2, name='Bob', order_id=102

C id=3, name='Carol', order_id=101

D No row

💡 Hint

Check the 'Result Row Produced' column at Step 4 in the execution_table.

At which step does the join condition first become False?

A Step 2

B Step 1

C Step 3

D Step 5

💡 Hint

Look at the 'Join Condition' column in execution_table rows 1-3.

If a new row in Table B had customer_id=3, how would the result change?

A No change, because INNER JOIN only matches existing rows

B All rows from Table A would appear in the result

C A new result row would appear matching Table A's id=3

D The query would return an error

💡 Hint

Refer to variable_tracker showing how matching customer_id produces result rows.

Concept Snapshot

INNER JOIN syntax:
SELECT columns
FROM TableA
INNER JOIN TableB ON TableA.key = TableB.key;

It returns rows where keys match in both tables.
Rows without matches are excluded.
Used to combine related data from two tables.

Full Transcript

INNER JOIN combines two tables by matching rows where a specified condition is true. It compares each row from the first table with each row from the second table. Only rows where the join condition matches are included in the output. Rows without a matching partner in the other table are not shown. This is useful to find related data that exists in both tables. The example query joins Customers and Orders on customer ID, showing only customers who have orders. The execution table shows step-by-step how each row pair is compared and which produce results. Variables track the current rows and condition results. Key moments clarify why some rows are excluded and how matching works. The quiz tests understanding of join results and condition evaluation. The snapshot summarizes the syntax and behavior of INNER JOIN.

Practice

1. What does an INNER JOIN do in SQL?

easy

A. It returns all rows from the first table only.

B. It returns rows that have matching values in both tables.

C. It returns all rows from the second table only.

D. It returns all rows from both tables, matching or not.

Solution

1. Step 1: Understand the purpose of INNER JOIN

   INNER JOIN combines rows from two tables where the join condition matches in both tables.

2. Step 2: Compare options with INNER JOIN behavior

   Only the description "It returns rows that have matching values in both tables." correctly states that it returns rows with matching values in both tables.

3. Final Answer:

   It returns rows that have matching values in both tables. -> Option B

4. Quick Check:

   INNER JOIN = matching rows only [OK]

Hint: INNER JOIN returns only matching rows from both tables [OK]

Common Mistakes:

• Thinking INNER JOIN returns all rows from one table
• Confusing INNER JOIN with LEFT or RIGHT JOIN
• Assuming it returns unmatched rows

2. Which of the following is the correct syntax for an INNER JOIN between tables Employees and Departments on the column DeptID using the ON clause?

easy

A. SELECT * FROM Employees INNER JOIN Departments ON Employees.DeptID == Departments.DeptID;

B. SELECT * FROM Employees JOIN Departments WHERE Employees.DeptID = Departments.DeptID;

C. SELECT * FROM Employees INNER JOIN Departments USING DeptID;

D. SELECT * FROM Employees INNER JOIN Departments ON Employees.DeptID = Departments.DeptID;

Solution

1. Step 1: Review INNER JOIN syntax

   The correct syntax uses INNER JOIN with ON and a single equals sign (=) for comparison.

2. Step 2: Check each option

   SELECT * FROM Employees INNER JOIN Departments ON Employees.DeptID = Departments.DeptID; uses correct INNER JOIN syntax with ON and =. SELECT * FROM Employees JOIN Departments WHERE Employees.DeptID = Departments.DeptID; uses WHERE instead of ON. SELECT * FROM Employees INNER JOIN Departments USING DeptID; uses USING instead of ON. SELECT * FROM Employees INNER JOIN Departments ON Employees.DeptID == Departments.DeptID; uses double equals (==), which is not valid in SQL.

3. Final Answer:

   SELECT * FROM Employees INNER JOIN Departments ON Employees.DeptID = Departments.DeptID; -> Option D

4. Quick Check:

   INNER JOIN syntax = ON with single = [OK]

Hint: Use ON with single = for INNER JOIN conditions [OK]

Common Mistakes:

• Using WHERE instead of ON for join condition
• Using double equals (==) instead of single equals (=)
• Using USING instead of ON

3. Given the tables:
Employees(id, name, dept_id)
Departments(dept_id, dept_name)
What is the result of this query?

SELECT name, dept_name FROM Employees INNER JOIN Departments ON Employees.dept_id = Departments.dept_id;

Assuming:
Employees: (1, 'Alice', 10), (2, 'Bob', 20), (3, 'Carol', 30)
Departments: (10, 'HR'), (20, 'Sales')

medium

A. [('Alice', 'HR'), ('Bob', 'Sales')]

B. [('Alice', 'HR'), ('Bob', 'Sales'), ('Carol', '30')]

C. [('Alice', 'HR'), ('Bob', 'Sales'), ('Carol', NULL)]

D. [('Alice', 'HR'), ('Bob', 'Sales'), ('Carol', 'Finance')]

Solution

1. Step 1: Identify matching rows by dept_id

   Employees with dept_id 10 and 20 match Departments with same dept_id. Carol's dept_id 30 has no match.

2. Step 2: Understand INNER JOIN output

   INNER JOIN returns only rows with matching dept_id in both tables, so Carol is excluded.

3. Final Answer:

   [('Alice', 'HR'), ('Bob', 'Sales')] -> Option A

4. Quick Check:

   INNER JOIN excludes unmatched rows [OK]

Hint: INNER JOIN excludes rows without matching keys [OK]

Common Mistakes:

• Including unmatched rows in result
• Assuming NULL values appear for unmatched rows
• Confusing INNER JOIN with LEFT JOIN behavior

4. Identify the error in this SQL query:

SELECT e.name, d.dept_name FROM Employees e INNER JOIN Departments d ON e.dept_id == d.dept_id;

medium

A. The join condition uses '==' instead of '='.

B. Using alias names for tables is not allowed.

C. Missing WHERE clause for filtering.

D. INNER JOIN requires USING instead of ON.

Solution

1. Step 1: Check join condition syntax

   SQL uses a single equals sign (=) for comparison, not double equals (==).

2. Step 2: Verify other parts of the query

   Aliases e and d are valid. WHERE clause is optional. INNER JOIN can use ON.

3. Final Answer:

   The join condition uses '==' instead of '='. -> Option A

4. Quick Check:

   Use = for join conditions, not == [OK]

Hint: Use single = in ON clause, not double == [OK]

Common Mistakes:

• Using == instead of = in join condition
• Thinking aliases are disallowed
• Confusing ON with WHERE clause necessity

5. You have two tables:
Orders(order_id, customer_id, amount)
Customers(customer_id, customer_name)
You want to find all customers who have placed orders and the total amount they spent.
Which query correctly uses INNER JOIN and aggregation to get this result?

hard

A. SELECT c.customer_name, SUM(o.amount) AS total_spent FROM Customers c LEFT JOIN Orders o ON c.customer_id = o.customer_id GROUP BY c.customer_name;

B. SELECT c.customer_name, o.amount FROM Customers c INNER JOIN Orders o ON c.customer_id = o.customer_id;

C. SELECT c.customer_name, SUM(o.amount) AS total_spent FROM Customers c INNER JOIN Orders o ON c.customer_id = o.customer_id GROUP BY c.customer_name;

D. SELECT customer_name, amount FROM Customers INNER JOIN Orders ON customer_id = customer_id;

Solution

1. Step 1: Understand the requirement

   We want customers who placed orders and the total amount spent, so INNER JOIN with aggregation is needed.

2. Step 2: Analyze each option

   SELECT c.customer_name, SUM(o.amount) AS total_spent FROM Customers c INNER JOIN Orders o ON c.customer_id = o.customer_id GROUP BY c.customer_name; correctly uses INNER JOIN on customer_id and sums amounts grouped by customer_name. SELECT c.customer_name, SUM(o.amount) AS total_spent FROM Customers c LEFT JOIN Orders o ON c.customer_id = o.customer_id GROUP BY c.customer_name; uses LEFT JOIN, which includes customers without orders. SELECT c.customer_name, o.amount FROM Customers c INNER JOIN Orders o ON c.customer_id = o.customer_id; does not aggregate amounts. SELECT customer_name, amount FROM Customers INNER JOIN Orders ON customer_id = customer_id; has ambiguous join condition and no aggregation.

3. Final Answer:

   SELECT c.customer_name, SUM(o.amount) AS total_spent FROM Customers c INNER JOIN Orders o ON c.customer_id = o.customer_id GROUP BY c.customer_name; -> Option C

4. Quick Check:

   INNER JOIN + GROUP BY + SUM = total spent per customer [OK]

Hint: Use INNER JOIN with GROUP BY and SUM for totals [OK]

Common Mistakes:

• Using LEFT JOIN instead of INNER JOIN when only matching rows needed
• Forgetting GROUP BY with aggregation
• Incorrect join condition syntax