NTH_VALUE function in SQL - Mini Project: Build & Apply

Using the NTH_VALUE Function in SQL

📖 Scenario: You work in a sales department and want to analyze sales data to find specific sales amounts for each salesperson.

🎯 Goal: Build a SQL query that uses the NTH_VALUE function to find the 2nd highest sale amount for each salesperson.

📋 What You'll Learn

Create a table called sales with columns salesperson (text) and amount (integer).

Insert the exact sales data provided.

Write a query that uses NTH_VALUE(amount, 2) to find the 2nd highest sale per salesperson.

Partition the data by salesperson and order by amount descending.

💡 Why This Matters

🌍 Real World

Sales teams often need to analyze specific ranked sales amounts per salesperson to understand performance beyond just the top sale.

💼 Career

Knowing how to use window functions like NTH_VALUE is valuable for data analysts and database developers working with complex data queries.

1

Create the sales table and insert data

Create a table called sales with columns salesperson (text) and amount (integer). Then insert these rows exactly: ('Alice', 500), ('Alice', 700), ('Alice', 600), ('Bob', 300), ('Bob', 400), ('Bob', 200).

SQL

-- Create the sales table and insert the data
-- Your code here

Hint

Use CREATE TABLE to define the table and INSERT INTO with multiple rows to add the data.

2

Set up the window specification

Write a SQL query selecting salesperson and amount from sales. Define a window specification called sales_window that partitions by salesperson and orders by amount descending.

SQL

CREATE TABLE sales (salesperson TEXT, amount INTEGER);
INSERT INTO sales (salesperson, amount) VALUES
('Alice', 500),
('Alice', 700),
('Alice', 600),
('Bob', 300),
('Bob', 400),
('Bob', 200);

-- Write SELECT with window specification named sales_window
-- Your code here

Hint

Use the WINDOW clause to define sales_window with partition and order.

3

Use NTH_VALUE to find the 2nd highest sale

Modify the query to select salesperson and use NTH_VALUE(amount, 2) OVER sales_window as second_highest_sale to find the 2nd highest sale amount per salesperson.

SQL

CREATE TABLE sales (salesperson TEXT, amount INTEGER);
INSERT INTO sales (salesperson, amount) VALUES
('Alice', 500),
('Alice', 700),
('Alice', 600),
('Bob', 300),
('Bob', 400),
('Bob', 200);

-- Select salesperson and NTH_VALUE(amount, 2) OVER sales_window as second_highest_sale
-- Your code here

Hint

Use NTH_VALUE(amount, 2) OVER sales_window to get the 2nd highest sale per salesperson.

4

Complete the query with DISTINCT to show one row per salesperson

Wrap the previous query as a subquery and select distinct salesperson and second_highest_sale to show only one row per salesperson with their 2nd highest sale amount.

SQL

CREATE TABLE sales (salesperson TEXT, amount INTEGER);
INSERT INTO sales (salesperson, amount) VALUES
('Alice', 500),
('Alice', 700),
('Alice', 600),
('Bob', 300),
('Bob', 400),
('Bob', 200);

-- Select distinct salesperson and second_highest_sale from subquery
-- Your code here

Hint

Use a subquery and SELECT DISTINCT to get one row per salesperson.

Practice

1. What does the SQL function NTH_VALUE(column, n) do when used with an OVER() clause?

easy

A. Returns the maximum value in the column.

B. Returns the total count of rows in the table.

C. Returns the value at the nth position in the ordered set of rows.

D. Returns the first value in the column without ordering.

Solution

1. Step 1: Understand the function purpose

   The NTH_VALUE function is designed to return a value from a specific position (n) in an ordered set of rows.

2. Step 2: Recognize the role of OVER()

   The OVER() clause defines how rows are ordered and grouped, which is essential for NTH_VALUE to pick the correct nth value.

3. Final Answer:

   Returns the value at the nth position in the ordered set of rows. -> Option C

4. Quick Check:

   NTH_VALUE returns nth value in order [OK]

Hint: NTH_VALUE picks the nth row's value after ordering [OK]

Common Mistakes:

• Confusing NTH_VALUE with aggregate functions like COUNT or MAX
• Ignoring the ordering inside OVER() clause
• Assuming it returns the nth row without ordering

2. Which of the following is the correct syntax to get the 3rd highest salary using NTH_VALUE in SQL?

easy

A. SELECT NTH_VALUE(salary, 3) OVER (PARTITION BY salary) FROM employees;

B. SELECT NTH_VALUE(salary, 3) FROM employees ORDER BY salary DESC;

C. SELECT NTH_VALUE(3, salary) OVER (ORDER BY salary DESC) FROM employees;

D. SELECT NTH_VALUE(salary, 3) OVER (ORDER BY salary DESC) FROM employees;

Solution

1. Step 1: Check the function parameters

   The correct order is NTH_VALUE(column, n), so salary first, then 3.

2. Step 2: Verify the OVER clause

   Ordering by salary descending (ORDER BY salary DESC) ensures the highest salaries come first, so the 3rd highest is the 3rd row.

3. Final Answer:

   SELECT NTH_VALUE(salary, 3) OVER (ORDER BY salary DESC) FROM employees; -> Option D

4. Quick Check:

   Syntax: NTH_VALUE(column, n) OVER (ORDER BY col DESC) [OK]

Hint: NTH_VALUE(column, n) needs OVER with ORDER BY for ranking [OK]

Common Mistakes:

• Swapping parameters inside NTH_VALUE
• Missing OVER() clause or ORDER BY inside it
• Using PARTITION BY incorrectly without ORDER BY

3. Given the table Scores with columns student and score, what will this query return?

SELECT student, score, NTH_VALUE(score, 2) OVER (ORDER BY score DESC) AS second_highest FROM Scores;

medium

A. Each row shows the student's score and the 2nd highest score overall repeated.

B. Each row shows the student's score and their own 2nd highest score.

C. The query returns only the row with the 2nd highest score.

D. The query returns an error because NTH_VALUE needs PARTITION BY.

Solution

1. Step 1: Understand the window function without PARTITION

   Since there is no PARTITION BY, the entire table is treated as one group ordered by score descending.

2. Step 2: Effect of NTH_VALUE(score, 2) OVER (ORDER BY score DESC)

   This returns the 2nd highest score value from the whole table, repeated for every row.

3. Final Answer:

   Each row shows the student's score and the 2nd highest score overall repeated. -> Option A

4. Quick Check:

   NTH_VALUE without PARTITION repeats nth value for all rows [OK]

Hint: Without PARTITION, NTH_VALUE repeats same nth value for all rows [OK]

Common Mistakes:

• Thinking NTH_VALUE filters rows to nth only
• Assuming PARTITION BY is mandatory
• Confusing NTH_VALUE with ROW_NUMBER or RANK

4. Identify the error in this SQL query:

SELECT employee_id, NTH_VALUE(salary, 1) OVER (PARTITION BY department_id) FROM employees;

medium

A. Missing ORDER BY clause inside OVER() causes error.

B. NTH_VALUE cannot be used with PARTITION BY.

C. The function parameters are in wrong order.

D. No error; query is correct.

Solution

1. Step 1: Check the OVER() clause requirements

   NTH_VALUE requires an ORDER BY clause inside OVER() to define the row order for nth value selection.

2. Step 2: Identify missing ORDER BY

   The query uses PARTITION BY but lacks ORDER BY, which causes a syntax error.

3. Final Answer:

   Missing ORDER BY clause inside OVER() causes error. -> Option A

4. Quick Check:

   NTH_VALUE needs ORDER BY in OVER() [OK]

Hint: Always include ORDER BY inside OVER() with NTH_VALUE [OK]

Common Mistakes:

• Omitting ORDER BY inside OVER()
• Thinking PARTITION BY alone orders rows
• Swapping function parameters

5. You want to find the 2nd earliest order date for each customer in the Orders table with columns customer_id and order_date. Which query correctly uses NTH_VALUE to achieve this?

hard

A. SELECT customer_id, order_date, NTH_VALUE(order_date, 2) FROM Orders GROUP BY customer_id;

B. SELECT customer_id, order_date, NTH_VALUE(order_date, 2) OVER (PARTITION BY customer_id ORDER BY order_date ASC) AS second_earliest FROM Orders;

C. SELECT customer_id, order_date, NTH_VALUE(order_date, 2) OVER (ORDER BY customer_id, order_date ASC) AS second_earliest FROM Orders;

D. SELECT customer_id, NTH_VALUE(order_date, 2) FROM Orders ORDER BY order_date ASC;

Solution

1. Step 1: Use PARTITION BY to group by customer

   To find the 2nd earliest order per customer, partition rows by customer_id.

2. Step 2: Order each partition by order_date ascending

   Ordering by order_date ASC ensures earliest dates come first.

3. Step 3: Apply NTH_VALUE to get 2nd earliest date

   NTH_VALUE with n=2 picks the 2nd row's order_date in each partition.

4. Final Answer:

   SELECT customer_id, order_date, NTH_VALUE(order_date, 2) OVER (PARTITION BY customer_id ORDER BY order_date ASC) AS second_earliest FROM Orders; -> Option B

5. Quick Check:

   Partition by customer, order by date ASC, NTH_VALUE(2) [OK]

Hint: Partition by group, order ascending, then NTH_VALUE(n) [OK]

Common Mistakes:

• Missing PARTITION BY to group customers
• Ordering by wrong columns or descending order
• Using NTH_VALUE without OVER() clause