LAG function for previous row access in SQL - Cheat Sheet & Quick Revision

Recall & Review

beginner

What does the LAG function do in SQL?

The LAG function lets you look at the value from a previous row in the same result set without using a self-join. It helps compare current row data with previous row data easily.

beginner

Write the basic syntax of the LAG function.

LAG(column_name, offset, default_value) OVER (PARTITION BY partition_column ORDER BY order_column) Here, offset is how many rows back to look (default 1), and default_value is what to return if no previous row exists.

intermediate

Why do we use the OVER clause with LAG?

The OVER clause defines how to group and order rows for the LAG function. It tells SQL which rows to compare and in what order, so it knows which previous row to look at.

beginner

What happens if there is no previous row when using LAG?

If there is no previous row, LAG returns NULL by default or the specified default value if provided. This prevents errors and lets you handle edge cases.

beginner

Give a real-life example where LAG is useful.

LAG is useful to compare sales of a product this month with last month. For example, you can find the difference in sales between current and previous month to see growth or decline.

What does the LAG function return by default if there is no previous row?

A NULL

B 0

C An error

D The current row value

By default, LAG returns NULL if there is no previous row to look back to.

Which clause is required to specify the order of rows for the LAG function?

A GROUP BY

B ORDER BY inside OVER()

C WHERE

D HAVING

The ORDER BY clause inside the OVER() clause defines the order of rows for LAG to look back.

What is the default offset value for LAG if not specified?

A 0

B 2

C 1

D NULL

The default offset for LAG is 1, meaning it looks at the immediately previous row.

Which of these is a valid use of LAG?

A To get the previous row's value

B To filter rows

C To sum all rows

D To get the next row's value

LAG is used to access the previous row's value in the result set.

Can LAG be used without the OVER() clause?

A Only with GROUP BY

B Yes, always

C Only in MySQL

D No, OVER() is required

LAG is a window function and requires the OVER() clause to define the window.

Explain how the LAG function works and why it is useful.

Think about comparing current data with past data in a list.

Describe the syntax of the LAG function and the role of each part.

Break down the function into its parameters and the window definition.

Practice

1. What does the SQL LAG function do?

easy

A. Returns the value from the next row in the result set

B. Calculates the sum of all previous rows

C. Returns the value from the previous row in the result set

D. Filters rows based on a condition

Solution

1. Step 1: Understand the purpose of LAG

   The LAG function is used to access data from a previous row relative to the current row in a query result.

2. Step 2: Compare options with definition

   Returns the value from the previous row in the result set correctly states that LAG returns the previous row's value. Other options describe different functions or behaviors.

3. Final Answer:

   Returns the value from the previous row in the result set -> Option C

4. Quick Check:

   LAG = previous row value [OK]

Hint: LAG looks backward one row by default [OK]

Common Mistakes:

• Confusing LAG with LEAD (next row)
• Thinking LAG sums values
• Assuming LAG filters rows

2. Which of the following is the correct syntax to use LAG to get the previous salary ordered by employee ID?

easy

A. LAG(salary) OVER (ORDER BY employee_id)

B. LAG(salary) BY employee_id

C. LAG(salary) ORDER employee_id

D. LAG(salary) PARTITION BY employee_id

Solution

1. Step 1: Recall LAG syntax

   LAG requires an OVER clause with ORDER BY to define the row order for previous row access.

2. Step 2: Match syntax with options

   LAG(salary) OVER (ORDER BY employee_id) correctly uses OVER (ORDER BY employee_id). The other options misuse syntax or keywords.

3. Final Answer:

   LAG(salary) OVER (ORDER BY employee_id) -> Option A

4. Quick Check:

   LAG syntax = LAG() OVER (ORDER BY ...) [OK]

Hint: Always use OVER with ORDER BY for LAG [OK]

Common Mistakes:

• Omitting OVER clause
• Using PARTITION BY without ORDER BY
• Incorrect keyword order

3. Given the table Employees with columns id and salary, what will this query return?

SELECT id, salary, LAG(salary) OVER (ORDER BY id) AS prev_salary FROM Employees;

Assuming data:
id | salary
1 | 5000
2 | 6000
3 | 5500

medium

A. [{id:1, salary:5000, prev_salary:NULL}, {id:2, salary:6000, prev_salary:5000}, {id:3, salary:5500, prev_salary:6000}]

B. [{id:1, salary:5000, prev_salary:5000}, {id:2, salary:6000, prev_salary:6000}, {id:3, salary:5500, prev_salary:5500}]

C. [{id:1, salary:5000, prev_salary:6000}, {id:2, salary:6000, prev_salary:5500}, {id:3, salary:5500, prev_salary:NULL}]

D. [{id:1, salary:5000, prev_salary:NULL}, {id:2, salary:6000, prev_salary:NULL}, {id:3, salary:5500, prev_salary:NULL}]

Solution

1. Step 1: Understand LAG with ORDER BY id

   LAG(salary) returns the salary from the previous row ordered by id. For id=1, no previous row exists, so NULL is returned.

2. Step 2: Apply LAG to each row

   Row 1: prev_salary = NULL
   Row 2: prev_salary = salary of id=1 = 5000
   Row 3: prev_salary = salary of id=2 = 6000

3. Final Answer:

   [{id:1, salary:5000, prev_salary:NULL}, {id:2, salary:6000, prev_salary:5000}, {id:3, salary:5500, prev_salary:6000}] -> Option A

4. Quick Check:

   LAG returns previous row's value or NULL if none [OK]

Hint: First row prev value is always NULL [OK]

Common Mistakes:

• Assuming first row has previous value
• Mixing up current and previous rows
• Ignoring ORDER BY effect

4. Identify the error in this query:

SELECT id, salary, LAG(salary) ORDER BY id AS prev_salary FROM Employees;

medium

A. AS alias must come before LAG

B. ORDER BY cannot be used inside LAG

C. LAG cannot be used with salary column

D. Missing OVER() clause for LAG function

Solution

1. Step 1: Check LAG function syntax

   LAG requires an OVER() clause to specify ordering. The query uses ORDER BY without OVER(), which is invalid.

2. Step 2: Confirm correct syntax

   The correct syntax is LAG(salary) OVER (ORDER BY id). The query misses OVER(), causing a syntax error.

3. Final Answer:

   Missing OVER() clause for LAG function -> Option D

4. Quick Check:

   LAG needs OVER() with ORDER BY [OK]

Hint: LAG always needs OVER() clause [OK]

Common Mistakes:

• Using ORDER BY without OVER()
• Placing alias before function
• Misunderstanding LAG usage

5. You want to find the difference between each employee's salary and the previous employee's salary ordered by id. Which query correctly calculates this difference?

hard

A. SELECT id, salary, LAG(salary) - salary OVER (ORDER BY id) AS diff FROM Employees;

B. SELECT id, salary, salary - LAG(salary) OVER (ORDER BY id) AS diff FROM Employees;

C. SELECT id, salary, salary - LAG(salary) FROM Employees ORDER BY id;

D. SELECT id, salary, salary - LAG(salary) PARTITION BY id AS diff FROM Employees;

Solution

1. Step 1: Understand the goal

   We want to subtract the previous salary from the current salary, ordered by id.

2. Step 2: Check each option's correctness

   SELECT id, salary, salary - LAG(salary) OVER (ORDER BY id) AS diff FROM Employees; correctly uses LAG(salary) OVER (ORDER BY id) and subtracts it from salary.
   SELECT id, salary, LAG(salary) - salary OVER (ORDER BY id) AS diff FROM Employees; reverses subtraction order.
   SELECT id, salary, salary - LAG(salary) FROM Employees ORDER BY id; misses OVER clause.
   SELECT id, salary, salary - LAG(salary) PARTITION BY id AS diff FROM Employees; misuses PARTITION BY without OVER.

3. Final Answer:

   SELECT id, salary, salary - LAG(salary) OVER (ORDER BY id) AS diff FROM Employees; -> Option B

4. Quick Check:

   Difference = current - previous salary using LAG [OK]

Hint: Use LAG with OVER and subtract current minus previous [OK]

Common Mistakes:

• Forgetting OVER() clause
• Subtracting in wrong order
• Misusing PARTITION BY without OVER