LAG function for previous row access in SQL - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to select the previous salary using LAG function.

SQL

SELECT employee_id, salary, LAG(salary) OVER (ORDER BY employee_id) AS previous_salary FROM employees WHERE department_id = [1];

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

A10

B'10'

Cdepartment_id

Dsalary

Attempts:

3 left

💡 Hint

Common Mistakes

Using quotes around numeric values in WHERE clause.

Using column names instead of values in WHERE clause.

The department_id is a number, so use 10 without quotes to filter correctly.

2fill in blank

medium

Complete the code to get the previous order date for each customer.

SQL

SELECT customer_id, order_date, LAG(order_date) OVER (ORDER BY [1]) AS previous_order FROM orders;

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

Acustomer_id

Border_date

Corder_id

Dorder_status

Attempts:

3 left

💡 Hint

Common Mistakes

Ordering by customer_id instead of order_date.

Ordering by unrelated columns.

Ordering by order_date ensures the previous order date is the correct previous row.

3fill in blank

hard

Fix the error in the LAG function usage to get previous sales amount per region.

SQL

SELECT region, sales_date, LAG(sales_amount) OVER (PARTITION BY [1] ORDER BY sales_date) AS prev_sales FROM sales_data;

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

Aemployee_id

Bsales_date

Csales_amount

Dregion

Attempts:

3 left

💡 Hint

Common Mistakes

Partitioning by sales_date which is ordering column.

Partitioning by sales_amount which is a value column.

Partitioning by region groups rows by region to get previous sales within each region.

4fill in blank

hard

Fill both blanks to calculate the difference between current and previous salary per department.

SQL

SELECT employee_id, salary, LAG(salary) OVER (PARTITION BY [1] ORDER BY [2]) AS prev_salary, salary - LAG(salary) OVER (PARTITION BY [1] ORDER BY [2]) AS salary_diff FROM employees;

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

Adepartment_id

Bsalary

Cemployee_id

Dhire_date

Attempts:

3 left

💡 Hint

Common Mistakes

Ordering by salary which may not reflect sequence.

Partitioning by employee_id which is unique per row.

Partition by department_id to group employees by department. Order by hire_date to get previous salary in hire order.

5fill in blank

hard

Fill all three blanks to get previous and next order dates per customer and calculate gap days.

SQL

SELECT customer_id, order_date, LAG(order_date) OVER (PARTITION BY [1] ORDER BY [2]) AS prev_order, LEAD(order_date) OVER (PARTITION BY [1] ORDER BY [2]) AS next_order, DATEDIFF(day, LAG(order_date) OVER (PARTITION BY [1] ORDER BY [2]), order_date) AS days_since_prev FROM orders;

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

Acustomer_id

Border_date

Corder_id

Dsales_amount

Attempts:

3 left

💡 Hint

Common Mistakes

Partitioning by order_id which is unique per order.

Ordering by sales_amount which is unrelated to order sequence.

Using different columns for LAG and LEAD ordering.

Partition by customer_id to group orders per customer. Order by order_date to get previous and next orders chronologically. Use order_date in DATEDIFF to calculate days difference.

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