FIRST_VALUE and LAST_VALUE in SQL - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to select the first salary in each department using FIRST_VALUE.

SQL

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

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

A3000

B2000

C10000

D5000

Attempts:

3 left

💡 Hint

Common Mistakes

Using a salary threshold that excludes all rows.

Not using the PARTITION BY clause.

The query filters employees with salary greater than 3000 and uses FIRST_VALUE to get the highest salary per department.

2fill in blank

medium

Complete the code to select the last hire date in each department using LAST_VALUE.

SQL

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

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

ACURRENT ROW

B1 PRECEDING

CUNBOUNDED FOLLOWING

D1 FOLLOWING

Attempts:

3 left

💡 Hint

Common Mistakes

Omitting the window frame clause causes unexpected results.

Using CURRENT ROW causes LAST_VALUE to return the current row's value.

LAST_VALUE requires the window frame to extend to UNBOUNDED FOLLOWING to return the last value in the partition correctly.

3fill in blank

hard

Fix the error in the query to correctly get the first salary per department.

SQL

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

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

Ahire_date ASC

Bsalary DESC

Cdepartment_id DESC

Dsalary ASC

Attempts:

3 left

💡 Hint

Common Mistakes

Ordering descending returns the highest salary, not the first.

Ordering by department_id does not affect salary order.

Ordering by salary ASC returns the smallest salary as the first value, which is the usual intent for FIRST_VALUE.

4fill in blank

hard

Fill both blanks to get the last hire date per department correctly.

SQL

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

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

AUNBOUNDED PRECEDING

BCURRENT ROW

CUNBOUNDED FOLLOWING

D1 PRECEDING

Attempts:

3 left

💡 Hint

Common Mistakes

Using CURRENT ROW as the end frame causes incorrect results.

Omitting the frame clause leads to unexpected last values.

The window frame must start from UNBOUNDED PRECEDING and end at UNBOUNDED FOLLOWING for LAST_VALUE to work as expected.

5fill in blank

hard

Fill all three blanks to select the first and last salary per department and filter salaries above 4000.

SQL

SELECT department_id, FIRST_VALUE(salary) OVER (PARTITION BY department_id ORDER BY salary [1]) AS first_salary, LAST_VALUE(salary) OVER (PARTITION BY department_id ORDER BY salary ROWS BETWEEN [2] AND [3]) AS last_salary FROM employees WHERE salary > 4000;

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

AASC

BUNBOUNDED PRECEDING

CUNBOUNDED FOLLOWING

DDESC

Attempts:

3 left

💡 Hint

Common Mistakes

Using DESC ordering for FIRST_VALUE returns the highest salary first.

Omitting the window frame for LAST_VALUE causes wrong results.

Ordering salary ascending gets the lowest salary first. The window frame from UNBOUNDED PRECEDING to UNBOUNDED FOLLOWING ensures LAST_VALUE returns the correct last salary.

Practice

1. What does the SQL function FIRST_VALUE(column_name) OVER (ORDER BY column_name) return?

easy

A. The average value of the column

B. The first value in the ordered set of rows for that column

C. The last value in the ordered set of rows for that column

D. The total count of rows in the table

Solution

1. Step 1: Understand FIRST_VALUE function

   It returns the first value in the ordered set of rows defined by the OVER clause.

2. Step 2: Analyze the ORDER BY clause

   The ORDER BY inside OVER() defines the order to pick the first value from.

3. Final Answer:

   The first value in the ordered set of rows for that column -> Option B

4. Quick Check:

   FIRST_VALUE returns first row value [OK]

Hint: FIRST_VALUE picks the first row after ordering [OK]

Common Mistakes:

• Confusing FIRST_VALUE with LAST_VALUE
• Thinking it returns an aggregate like average
• Ignoring the ORDER BY inside OVER()

2. Which of the following is the correct syntax to get the last value of salary ordered by hire_date in SQL?

easy

A. SELECT LAST_VALUE(salary) OVER (ORDER BY hire_date) FROM employees;

B. SELECT LAST_VALUE(salary) FROM employees ORDER BY hire_date;

C. SELECT LAST_VALUE(salary) OVER (PARTITION BY hire_date) FROM employees;

D. SELECT LAST_VALUE(salary) OVER (ORDER BY hire_date ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING) FROM employees;

Solution

1. Step 1: Understand LAST_VALUE default frame

   LAST_VALUE by default looks only at rows up to current row, so it may not return the true last value.

2. Step 2: Use frame clause to include all rows

   Adding ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING ensures LAST_VALUE looks at the entire partition.

3. Final Answer:

   SELECT LAST_VALUE(salary) OVER (ORDER BY hire_date ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING) FROM employees; -> Option D

4. Quick Check:

   LAST_VALUE needs full frame to get last row [OK]

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

Common Mistakes:

• Omitting frame clause with LAST_VALUE
• Using PARTITION BY incorrectly
• Trying to use LAST_VALUE without OVER()

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

SELECT region, amount, FIRST_VALUE(amount) OVER (PARTITION BY region ORDER BY amount DESC) AS max_amount FROM sales;

medium

A. The first amount in the whole table regardless of region

B. The lowest amount per region repeated on each row of that region

C. The highest amount per region repeated on each row of that region

D. Syntax error due to missing frame clause

Solution

1. Step 1: Analyze PARTITION BY and ORDER BY

   The window partitions rows by region and orders amounts descending within each region.

2. Step 2: Understand FIRST_VALUE with ordering

   FIRST_VALUE picks the first row's amount in each partition ordered by amount DESC, which is the max amount.

3. Final Answer:

   The highest amount per region repeated on each row of that region -> Option C

4. Quick Check:

   FIRST_VALUE with DESC order gives max per partition [OK]

Hint: Use ORDER BY DESC with FIRST_VALUE for max per group [OK]

Common Mistakes:

• Assuming FIRST_VALUE always returns lowest value
• Confusing PARTITION BY with filtering
• Expecting syntax error without frame clause here

4. Identify the error in this query:

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

medium

A. LAST_VALUE needs a frame clause to get correct last value

B. Missing PARTITION BY clause causes error

C. ORDER BY cannot be used inside OVER()

D. salary column must be aggregated with GROUP BY

Solution

1. Step 1: Check LAST_VALUE default behavior

   By default, LAST_VALUE looks only up to current row, so it may not return the true last value.

2. Step 2: Add frame clause to fix

   Adding ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING makes LAST_VALUE consider all rows.

3. Final Answer:

   LAST_VALUE needs a frame clause to get correct last value -> Option A

4. Quick Check:

   LAST_VALUE without frame returns wrong last row [OK]

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

Common Mistakes:

• Thinking PARTITION BY is mandatory
• Believing ORDER BY is invalid inside OVER()
• Confusing LAST_VALUE with aggregate functions

5. You want to find the first and last sale amounts per customer ordered by sale date from the sales table. Which query correctly returns both values on each row?

hard

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

B. SELECT customer_id, sale_date, FIRST_VALUE(amount) OVER (ORDER BY sale_date) AS first_sale, LAST_VALUE(amount) OVER (ORDER BY sale_date) AS last_sale FROM sales;

C. SELECT customer_id, sale_date, FIRST_VALUE(amount) AS first_sale, LAST_VALUE(amount) AS last_sale FROM sales GROUP BY customer_id;

D. SELECT customer_id, sale_date, MIN(amount) OVER (PARTITION BY customer_id) AS first_sale, MAX(amount) OVER (PARTITION BY customer_id) AS last_sale FROM sales;

Solution

1. Step 1: Use PARTITION BY customer_id

   This groups rows by customer to get first and last sales per customer.

2. Step 2: Use ORDER BY sale_date for chronological order

   Ordering by sale_date ensures first and last are by date.

3. Step 3: Add frame clause for LAST_VALUE

   LAST_VALUE needs ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING to get true last value.

4. Final Answer:

   Query in option A correctly returns first and last sale amounts per customer -> Option A

5. Quick Check:

   FIRST_VALUE and LAST_VALUE with proper frame and partition [OK]

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

Common Mistakes:

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