Bird
Raised Fist0
SQLquery~5 mins

FIRST_VALUE and LAST_VALUE in SQL - Cheat Sheet & Quick Revision

Choose your learning style10 modes available
Recall & Review
beginner
What does the SQL function FIRST_VALUE() do?
It returns the first value in an ordered set of rows within a window or partition.
Click to reveal answer
beginner
What is the purpose of LAST_VALUE() in SQL?
It returns the last value in an ordered set of rows within a window or partition.
Click to reveal answer
intermediate
How do FIRST_VALUE() and LAST_VALUE() differ from aggregate functions like MIN() and MAX()?
FIRST_VALUE() and LAST_VALUE() return values based on row order within a window, while MIN() and MAX() return the smallest or largest value regardless of order.
Click to reveal answer
beginner
In SQL, what clause is required to use FIRST_VALUE() and LAST_VALUE() properly?
The OVER() clause with an ORDER BY to define the window or partition and the order of rows.
Click to reveal answer
advanced
Why might LAST_VALUE() return unexpected results without proper framing?
Because by default the window frame ends at the current row, LAST_VALUE() may return the current row's value instead of the last in the partition unless frame boundaries are set.
Click to reveal answer
What does FIRST_VALUE(column) OVER (ORDER BY date) return?
AThe earliest value of column based on date order
BThe smallest value of column overall
CThe last value of column in the table
DThe average value of column
Which clause is mandatory to use with LAST_VALUE()?
AHAVING
BGROUP BY
CORDER BY inside OVER()
DWHERE
What happens if you use LAST_VALUE() without specifying ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING?
AIt returns the first value
BIt returns NULL
CIt returns the last value in the entire partition
DIt returns the value of the current row
Which function would you use to get the first value in each group of rows?
AFIRST_VALUE()
BMAX()
CMIN()
DCOUNT()
Can FIRST_VALUE() and LAST_VALUE() be used without window functions?
AYes, they work like aggregate functions
BNo, they require OVER() clause
CYes, but only in WHERE clause
DNo, they are not SQL functions
Explain how FIRST_VALUE() and LAST_VALUE() work in SQL and when you would use them.
Think about how you find the first or last item in a sorted list.
You got /4 concepts.
    Describe the importance of window framing when using LAST_VALUE() and how to fix common issues.
    Consider how the window frame limits which rows LAST_VALUE() sees.
    You got /4 concepts.

      Practice

      (1/5)
      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
      Recommended

      Ready to prove it?

      Pass the test and this pattern moves unknown strong while earning +1 Grip toward your certificate.

      10 questions~5 mineasy hard