FIRST_VALUE and LAST_VALUE in PostgreSQL - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to select the first salary value in each department.

PostgreSQL

SELECT department_id, [1](salary) OVER (PARTITION BY department_id ORDER BY salary) AS first_salary FROM employees;

Drag options to blanks, or click blank then click option'

AFIRST_VALUE

BLAST_VALUE

CMAX

DMIN

Attempts:

3 left

💡 Hint

Common Mistakes

Using LAST_VALUE instead of FIRST_VALUE.

Using aggregate functions like MAX or MIN without window clause.

The FIRST_VALUE function returns the first value in the ordered partition, which here is the lowest salary per department.

2fill in blank

medium

Complete the code to select the last salary value in each department.

PostgreSQL

SELECT department_id, [1](salary) OVER (PARTITION BY department_id ORDER BY salary ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING) AS last_salary FROM employees;

Drag options to blanks, or click blank then click option'

ALAST_VALUE

BFIRST_VALUE

CMAX

DMIN

Attempts:

3 left

💡 Hint

Common Mistakes

Omitting the frame clause, which causes LAST_VALUE to return unexpected results.

Using FIRST_VALUE instead of LAST_VALUE.

The LAST_VALUE function returns the last value in the ordered partition. The frame clause ensures it looks at the entire partition.

3fill in blank

hard

Fix the error in the code to correctly get the last salary per department.

PostgreSQL

SELECT department_id, LAST_VALUE(salary) OVER (PARTITION BY department_id ORDER BY salary [1]) AS last_salary FROM employees;

Drag options to blanks, or click blank then click option'

AROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW

BRANGE BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING

CRANGE BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW

DROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING

Attempts:

3 left

💡 Hint

Common Mistakes

Using default frame which ends at current row.

Using RANGE frame which behaves differently.

To get the correct last value over the entire partition, the frame must be set to ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING. Otherwise, LAST_VALUE returns the current row's value.

4fill in blank

hard

Fill both blanks to select the first and last hire dates per department.

PostgreSQL

SELECT department_id, [1](hire_date) OVER (PARTITION BY department_id ORDER BY hire_date) AS first_hire, [2](hire_date) OVER (PARTITION BY department_id ORDER BY hire_date ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING) AS last_hire FROM employees;

Drag options to blanks, or click blank then click option'

AFIRST_VALUE

BMAX

CLAST_VALUE

DMIN

Attempts:

3 left

💡 Hint

Common Mistakes

Using aggregate functions like MAX or MIN instead of window functions.

Omitting the frame clause for LAST_VALUE.

FIRST_VALUE gets the earliest hire date per department, and LAST_VALUE gets the latest hire date with the correct frame clause.

5fill in blank

hard

Fill all three blanks to create a query that shows employee name, first salary, last salary, and the difference between them per department.

PostgreSQL

SELECT employee_name, department_id, [1](salary) OVER (PARTITION BY department_id ORDER BY salary) AS first_salary, [2](salary) OVER (PARTITION BY department_id ORDER BY salary ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING) AS last_salary, [3] - [2] AS salary_diff FROM employees;

Drag options to blanks, or click blank then click option'

AFIRST_VALUE

Blast_salary

Csalary

DLAST_VALUE

Attempts:

3 left

💡 Hint

Common Mistakes

Mixing up the order of FIRST_VALUE and LAST_VALUE.

Using column names incorrectly in the difference calculation.

The query uses FIRST_VALUE and LAST_VALUE to get the first and last salaries per department. The difference is calculated by subtracting the last salary from the current salary.

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