Why LAG function for previous row access in SQL? - Purpose & Use Cases

The Big Idea

What if you could instantly see yesterday's data next to today's without flipping pages or writing notes?

The Scenario

Imagine you have a list of daily sales numbers on paper, and you want to compare each day's sales to the previous day. You try to do this by flipping back and forth between pages or writing down extra notes to remember the previous day's number.

The Problem

This manual way is slow and confusing. You might lose track of which day you are on, make mistakes copying numbers, or waste time flipping pages. It's hard to keep everything organized and accurate.

The Solution

The LAG function in SQL lets you look at the previous row's value easily, right next to the current row. It automatically remembers the previous data for you, so you can compare or calculate differences without any extra effort.

Before vs After

✗ Before

SELECT date, sales FROM daily_sales;
-- Then manually compare each row with the previous one outside SQL

✓ After

SELECT date, sales, LAG(sales) OVER (ORDER BY date) AS previous_day_sales FROM daily_sales;

What It Enables

It makes comparing current and past data simple and fast, unlocking powerful trend analysis and insights.

Real Life Example

A store manager can quickly see how today's sales compare to yesterday's, helping decide if a promotion worked or if changes are needed.

Key Takeaways

Manually tracking previous rows is slow and error-prone.

LAG function automatically accesses previous row data in SQL.

This enables easy comparisons and better data insights.

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