FIRST_VALUE and LAST_VALUE in PostgreSQL

Introduction

These functions help you find the first or last value in a group of rows, making it easy to pick important data points without extra work.

When you want to find the earliest or latest date in a list of events.

When you need the first or last price of a product in a sales report.

When you want to show the first or last comment made by a user in a chat.

When you want to compare the first and last scores of a player in a game.

When you want to find the first or last entry in a sorted list of records.

Syntax

PostgreSQL

FIRST_VALUE(column_name) OVER (PARTITION BY partition_column ORDER BY order_column [ROWS BETWEEN ...])

LAST_VALUE(column_name) OVER (PARTITION BY partition_column ORDER BY order_column [ROWS BETWEEN ...])

These functions are used with the OVER() clause to define how to group and order rows.

You can use PARTITION BY to group rows and ORDER BY to sort them before picking the first or last value.

Examples

Finds the lowest salary in the employees table.

PostgreSQL

SELECT FIRST_VALUE(salary) OVER (ORDER BY salary) AS lowest_salary FROM employees;

Finds the first hired employee in each department.

PostgreSQL

SELECT department, FIRST_VALUE(employee_name) OVER (PARTITION BY department ORDER BY hire_date) AS first_hired FROM employees;

Finds the last hired employee in each department. The ROWS clause ensures the window covers all rows.

PostgreSQL

SELECT department, LAST_VALUE(employee_name) OVER (PARTITION BY department ORDER BY hire_date ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING) AS last_hired FROM employees;

Sample Program

This query shows each employee with their department and hire date, plus the first and last hired employee in their department.

PostgreSQL

CREATE TABLE employees (
  id SERIAL PRIMARY KEY,
  employee_name TEXT,
  department TEXT,
  hire_date DATE,
  salary INT
);

INSERT INTO employees (employee_name, department, hire_date, salary) VALUES
('Alice', 'Sales', '2020-01-15', 50000),
('Bob', 'Sales', '2021-03-10', 52000),
('Charlie', 'HR', '2019-07-01', 48000),
('Diana', 'HR', '2022-05-20', 47000),
('Eve', 'IT', '2021-11-11', 60000);

SELECT department, employee_name, hire_date,
  FIRST_VALUE(employee_name) OVER (PARTITION BY department ORDER BY hire_date) AS first_hired,
  LAST_VALUE(employee_name) OVER (PARTITION BY department ORDER BY hire_date ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING) AS last_hired
FROM employees
ORDER BY department, hire_date;

Important Notes

Without the ROWS BETWEEN clause, LAST_VALUE may return unexpected results because it looks at the current row and following rows by default.

Use ORDER BY inside OVER() to define which row is considered first or last.

FIRST_VALUE and LAST_VALUE work well with PARTITION BY to handle groups separately.

Summary

FIRST_VALUE returns the first value in a sorted group of rows.

LAST_VALUE returns the last value in a sorted group of rows, but needs careful window framing.

Both use the OVER() clause to define grouping and order.

Practice

1. What does the FIRST_VALUE function do in PostgreSQL?

easy

A. Returns the first value in a sorted group of rows.

B. Returns the last value in a sorted group of rows.

C. Returns the average value of a column.

D. Returns the total count of rows.

Solution

1. Step 1: Understand the purpose of FIRST_VALUE

   FIRST_VALUE returns the first value in a sorted window or group of rows.

2. Step 2: Compare with other options

   Options B, C, and D describe different functions or aggregates, not FIRST_VALUE.

3. Final Answer:

   Returns the first value in a sorted group of rows. -> Option A

4. Quick Check:

   FIRST_VALUE = first value in sorted rows [OK]

Hint: FIRST_VALUE always picks the first row's value in order [OK]

Common Mistakes:

• Confusing FIRST_VALUE with LAST_VALUE
• Thinking it returns an aggregate like sum or count
• Ignoring the sorting order in the window

2. Which of the following is the correct syntax to get the last value of a column score ordered by date using LAST_VALUE in PostgreSQL?

easy

A. SELECT LAST_VALUE(score) FROM results ORDER BY date;

B. SELECT LAST_VALUE(score) OVER (ORDER BY date ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING) FROM results;

C. SELECT LAST_VALUE(score) OVER (PARTITION BY date) FROM results;

D. SELECT LAST_VALUE(score) OVER (ORDER BY date) FROM results;

Solution

1. Step 1: Understand LAST_VALUE default frame

   LAST_VALUE by default looks from current row to end, so it may not return the last value overall without framing.

2. Step 2: Use explicit window frame to get true last value

   Adding ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING ensures the entire partition is considered.

3. Final Answer:

   SELECT LAST_VALUE(score) OVER (ORDER BY date ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING) FROM results; -> Option B

4. Quick Check:

   LAST_VALUE needs full frame to get last value [OK]

Hint: Add full window frame to LAST_VALUE for correct last row [OK]

Common Mistakes:

• Omitting window frame causing wrong last value
• Using PARTITION BY incorrectly
• Using LAST_VALUE without OVER() clause

3. Given the table sales with columns region, month, and revenue, what will this query return?

SELECT region, month, revenue, FIRST_VALUE(revenue) OVER (PARTITION BY region ORDER BY month) AS first_rev FROM sales ORDER BY region, month;

medium

A. The total revenue summed for each region.

B. The revenue of the last month for each region repeated on every row of that region.

C. The revenue of the first month for each region repeated on every row of that region.

D. The revenue of the current month only.

Solution

1. Step 1: Analyze FIRST_VALUE with PARTITION and ORDER BY

   FIRST_VALUE(revenue) OVER (PARTITION BY region ORDER BY month) returns the first revenue value per region ordered by month.

2. Step 2: Understand output per row

   For each row in a region, the first month's revenue is repeated as first_rev.

3. Final Answer:

   The revenue of the first month for each region repeated on every row of that region. -> Option C

4. Quick Check:

   FIRST_VALUE repeats first row's value per partition [OK]

Hint: FIRST_VALUE repeats first sorted value per group [OK]

Common Mistakes:

• Thinking it returns last value instead
• Assuming it sums or aggregates
• Ignoring PARTITION BY effect

4. You wrote this query but it returns unexpected results:

SELECT employee_id, salary, LAST_VALUE(salary) OVER (ORDER BY employee_id) AS last_sal FROM employees;

What is the likely cause?

medium

A. LAST_VALUE cannot be used without PARTITION BY clause.

B. ORDER BY employee_id is invalid in window functions.

C. salary column must be aggregated with SUM before LAST_VALUE.

D. LAST_VALUE needs a window frame to include all rows; default frame causes wrong output.

Solution

1. Step 1: Recall LAST_VALUE default frame behavior

   By default, LAST_VALUE looks from current row to end, so it may not return the last row's value overall.

2. Step 2: Identify missing window frame

   Without specifying ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING, the last value is not the last row's value.

3. Final Answer:

   LAST_VALUE needs a window frame to include all rows; default frame causes wrong output. -> Option D

4. Quick Check:

   LAST_VALUE requires full frame for correct last row [OK]

Hint: Add full frame to LAST_VALUE to fix wrong results [OK]

Common Mistakes:

• Assuming ORDER BY alone fixes LAST_VALUE
• Thinking PARTITION BY is mandatory
• Trying to aggregate salary before LAST_VALUE

5. You want to find the first and last sale amount for each customer ordered by sale date. Which query correctly returns both values per customer in PostgreSQL?

hard

A. SELECT customer_id, sale_date, sale_amount, FIRST_VALUE(sale_amount) OVER (PARTITION BY customer_id ORDER BY sale_date), LAST_VALUE(sale_amount) OVER (PARTITION BY customer_id ORDER BY sale_date ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING) FROM sales;

B. SELECT customer_id, sale_date, sale_amount, FIRST_VALUE(sale_amount), LAST_VALUE(sale_amount) FROM sales GROUP BY customer_id;

C. SELECT customer_id, sale_date, sale_amount, FIRST_VALUE(sale_amount) OVER (ORDER BY sale_date), LAST_VALUE(sale_amount) OVER (ORDER BY sale_date) FROM sales;

D. SELECT customer_id, sale_date, sale_amount, MIN(sale_amount) OVER (PARTITION BY customer_id), MAX(sale_amount) OVER (PARTITION BY customer_id) FROM sales;

Solution

1. Step 1: Use FIRST_VALUE with PARTITION and ORDER BY

   FIRST_VALUE(sale_amount) OVER (PARTITION BY customer_id ORDER BY sale_date) gets first sale per customer.

2. Step 2: Use LAST_VALUE with full window frame

   LAST_VALUE needs ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING to get last sale per customer.

3. Step 3: Check other options

   SELECT customer_id, sale_date, sale_amount, FIRST_VALUE(sale_amount), LAST_VALUE(sale_amount) FROM sales GROUP BY customer_id; lacks OVER() and ORDER BY, C lacks PARTITION BY, D uses MIN/MAX which is different.

4. Final Answer:

   SELECT customer_id, sale_date, sale_amount, FIRST_VALUE(sale_amount) OVER (PARTITION BY customer_id ORDER BY sale_date), LAST_VALUE(sale_amount) OVER (PARTITION BY customer_id ORDER BY sale_date ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING) FROM sales; -> Option A

5. Quick Check:

   Combine FIRST_VALUE and LAST_VALUE with proper framing [OK]

Hint: Use full frame for LAST_VALUE with PARTITION and ORDER BY [OK]

Common Mistakes:

• Omitting window frame for LAST_VALUE
• Using aggregate functions instead of window functions
• Not partitioning by customer_id