OVER clause with ORDER BY in SQL - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ query_result

intermediate

Calculate running total of sales per employee

Given the Sales table with columns employee_id, sale_date, and amount, what is the output of this query?

SELECT employee_id, sale_date, amount, SUM(amount) OVER (PARTITION BY employee_id ORDER BY sale_date) AS running_total FROM Sales ORDER BY employee_id, sale_date;

SQL

SELECT employee_id, sale_date, amount, SUM(amount) OVER (PARTITION BY employee_id ORDER BY sale_date) AS running_total FROM Sales ORDER BY employee_id, sale_date;

A [{"employee_id":1,"sale_date":"2023-01-01","amount":100,"running_total":100},{"employee_id":1,"sale_date":"2023-01-05","amount":150,"running_total":250},{"employee_id":2,"sale_date":"2023-01-02","amount":200,"running_total":200},{"employee_id":2,"sale_date":"2023-01-06","amount":100,"running_total":300}]

B [{"employee_id":1,"sale_date":"2023-01-01","amount":100,"running_total":100},{"employee_id":1,"sale_date":"2023-01-05","amount":150,"running_total":150},{"employee_id":2,"sale_date":"2023-01-02","amount":200,"running_total":200},{"employee_id":2,"sale_date":"2023-01-06","amount":100,"running_total":100}]

C [{"employee_id":1,"sale_date":"2023-01-01","amount":100,"running_total":100},{"employee_id":1,"sale_date":"2023-01-05","amount":150,"running_total":250},{"employee_id":2,"sale_date":"2023-01-02","amount":200,"running_total":100},{"employee_id":2,"sale_date":"2023-01-06","amount":100,"running_total":300}]

D [{"employee_id":1,"sale_date":"2023-01-01","amount":100,"running_total":250},{"employee_id":1,"sale_date":"2023-01-05","amount":150,"running_total":250},{"employee_id":2,"sale_date":"2023-01-02","amount":200,"running_total":300},{"employee_id":2,"sale_date":"2023-01-06","amount":100,"running_total":300}]

💡 Hint

Think about how the SUM() function accumulates values ordered by sale_date for each employee separately.

The SUM() OVER clause with PARTITION BY employee_id and ORDER BY sale_date calculates a running total of sales for each employee ordered by the sale date. The running total adds the current amount to the sum of all previous amounts for that employee.

🧠 Conceptual

intermediate

Effect of ORDER BY in OVER clause without PARTITION BY

What does the ORDER BY clause inside the OVER() do when used without PARTITION BY?
Consider this query:

SELECT id, value, ROW_NUMBER() OVER (ORDER BY value DESC) AS rank FROM Items;

A It assigns a unique rank to each row based on descending value across the entire table.

B It resets the rank for each distinct value in the value column.

C It produces the same rank for all rows because there is no partition.

D It causes a syntax error because PARTITION BY is required with ORDER BY in OVER().

💡 Hint

Think about how window functions rank rows when no partition is specified.

Without PARTITION BY, the window function treats the entire table as one group. ORDER BY inside OVER() defines the order in which the function processes rows. ROW_NUMBER() assigns a unique rank based on that order.

📝 Syntax

advanced

Identify the syntax error in this window function

Which option contains the correct syntax for using the OVER clause with ORDER BY?
Given the query:

SELECT name, salary, RANK() OVER (ORDER BY salary DESC) FROM Employees;

A SELECT name, salary, RANK() OVER (PARTITION salary ORDER BY DESC) FROM Employees;

B SELECT name, salary, RANK() OVER ORDER BY salary DESC FROM Employees;

C SELECT name, salary, RANK() OVER (ORDER BY salary DESC) FROM Employees;

D SELECT name, salary, RANK() OVER (ORDER salary DESC) FROM Employees;

💡 Hint

Check the placement and syntax of the OVER clause and ORDER BY inside it.

The correct syntax requires parentheses after OVER and ORDER BY inside those parentheses. Option C follows this syntax exactly.

❓ optimization

advanced

Optimizing window function with ORDER BY on large dataset

You have a large Transactions table and want to calculate a running total of amount ordered by transaction_date. Which approach is best for performance?

A Use a correlated subquery to sum amounts up to each transaction_date.

B Use SUM(amount) OVER (ORDER BY transaction_date) with an index on transaction_date.

C Use GROUP BY transaction_date and then join back to the original table.

D Calculate running total in application code after fetching all rows unordered.

💡 Hint

Window functions with proper indexing can be very efficient.

Using SUM() OVER (ORDER BY ...) with an index on the ordering column allows the database to efficiently compute running totals without expensive joins or subqueries.

🔧 Debug

expert

Why does this query produce unexpected results?

Consider this query:

SELECT employee_id, sale_date, amount, SUM(amount) OVER (ORDER BY sale_date) AS running_total FROM Sales ORDER BY employee_id, sale_date;

Why might the running_total not reset for each employee?

A Because ORDER BY in OVER() must match ORDER BY in the main query to reset the total.

B Because sale_date values are not unique, causing incorrect running totals.

C Because SUM() cannot be used with ORDER BY in OVER() without GROUP BY.

D Because PARTITION BY employee_id is missing, so the running total is calculated over all rows ordered by sale_date.

💡 Hint

Think about how PARTITION BY affects window function grouping.

Without PARTITION BY, the window function treats all rows as one group, so the running total accumulates over all employees combined, not resetting per employee.

Practice

1. What does the OVER clause with ORDER BY do in SQL?

easy

A. It creates a new table sorted by the specified column.

B. It performs calculations on rows in a specific order without grouping them.

C. It filters rows based on order conditions.

D. It groups rows and then orders the groups.

Solution

1. Step 1: Understand the OVER clause

   The OVER clause allows window functions to work on a set of rows related to the current row without grouping them.

2. Step 2: Role of ORDER BY inside OVER

   ORDER BY inside OVER defines the order in which rows are processed for functions like ROW_NUMBER() or SUM() running totals.

3. Final Answer:

   It performs calculations on rows in a specific order without grouping them. -> Option B

4. Quick Check:

   OVER with ORDER BY = ordered calculations without grouping [OK]

Hint: OVER with ORDER BY orders rows for calculations, no grouping needed [OK]

Common Mistakes:

• Thinking OVER groups rows like GROUP BY
• Confusing ORDER BY inside OVER with filtering
• Assuming it creates a new sorted table

2. Which of the following is the correct syntax to assign row numbers ordered by salary using the OVER clause?

easy

A. SELECT ROW_NUMBER(ORDER BY salary) AS rank FROM employees;

B. SELECT ROW_NUMBER() AS rank OVER ORDER BY salary FROM employees;

C. SELECT ROW_NUMBER() ORDER BY salary FROM employees;

D. SELECT ROW_NUMBER() OVER (ORDER BY salary) AS rank FROM employees;

Solution

1. Step 1: Identify correct ROW_NUMBER() syntax

   ROW_NUMBER() requires OVER() with ORDER BY inside parentheses to define ordering.

2. Step 2: Check each option

   SELECT ROW_NUMBER() OVER (ORDER BY salary) AS rank FROM employees; correctly uses ROW_NUMBER() OVER (ORDER BY salary). Others miss parentheses or misplace clauses.

3. Final Answer:

   SELECT ROW_NUMBER() OVER (ORDER BY salary) AS rank FROM employees; -> Option D

4. Quick Check:

   ROW_NUMBER() needs OVER(ORDER BY ...) [OK]

Hint: ROW_NUMBER() must be followed by OVER(ORDER BY column) [OK]

Common Mistakes:

• Omitting OVER() parentheses
• Placing ORDER BY outside OVER()
• Using ROW_NUMBER() without OVER

3. Given the table sales with columns id, amount, what is the output of this query?

SELECT id, amount, SUM(amount) OVER (ORDER BY id) AS running_total FROM sales ORDER BY id;

medium

A. The query returns an error due to missing GROUP BY.

B. Each row shows id, amount, and total sum of all amounts repeated.

C. Each row shows id, amount, and a running total of amounts ordered by id.

D. Rows are grouped by id with sum of amounts per group.

Solution

1. Step 1: Understand SUM() OVER (ORDER BY id)

   This calculates a running total of amount, adding amounts in order of id for each row.

2. Step 2: Analyze output per row

   Each row shows its id, amount, and the sum of amounts from first id up to current id.

3. Final Answer:

   Each row shows id, amount, and a running total of amounts ordered by id. -> Option C

4. Quick Check:

   SUM() OVER with ORDER BY = running total [OK]

Hint: SUM() OVER(ORDER BY) gives running totals, not grouped sums [OK]

Common Mistakes:

• Thinking it sums all rows for each row
• Expecting GROUP BY to be required
• Confusing running total with grouped sum

4. Identify the error in this query:

SELECT id, RANK() OVER ORDER BY score FROM players;

medium

A. Missing parentheses after OVER and ORDER BY inside them.

B. RANK() cannot be used with OVER clause.

C. ORDER BY should be outside the OVER clause.

D. The query is correct and will run without errors.

Solution

1. Step 1: Check syntax of RANK() with OVER

   RANK() requires OVER() with parentheses enclosing ORDER BY clause.

2. Step 2: Identify missing parentheses

   The query misses parentheses after OVER and does not put ORDER BY inside them, causing syntax error.

3. Final Answer:

   Missing parentheses after OVER and ORDER BY inside them. -> Option A

4. Quick Check:

   OVER() must have parentheses with ORDER BY inside [OK]

Hint: Always use OVER(...) with parentheses enclosing ORDER BY [OK]

Common Mistakes:

• Writing OVER without parentheses
• Placing ORDER BY outside OVER()
• Assuming RANK() works without OVER

5. You want to assign a rank to employees based on their salary, but if two employees have the same salary, they should get the same rank and the next rank should skip accordingly. Which query correctly uses the OVER clause with ORDER BY to achieve this?

hard

A. SELECT employee_id, salary, RANK() OVER (ORDER BY salary DESC) AS salary_rank FROM employees;

B. SELECT employee_id, salary, ROW_NUMBER() OVER (ORDER BY salary DESC) AS salary_rank FROM employees;

C. SELECT employee_id, salary, DENSE_RANK() OVER (ORDER BY salary DESC) AS salary_rank FROM employees;

D. SELECT employee_id, salary, RANK() OVER ORDER BY salary DESC AS salary_rank FROM employees;

Solution

1. Step 1: Understand ranking functions

   RANK() assigns same rank to ties and skips ranks after ties; DENSE_RANK() assigns same rank but no skips; ROW_NUMBER() assigns unique numbers.

2. Step 2: Check syntax and behavior

   SELECT employee_id, salary, RANK() OVER (ORDER BY salary DESC) AS salary_rank FROM employees; uses RANK() with correct OVER(ORDER BY salary DESC) syntax, matching requirement. SELECT employee_id, salary, RANK() OVER ORDER BY salary DESC AS salary_rank FROM employees; has syntax error. ROW_NUMBER() and DENSE_RANK() do not skip ranks after ties.

3. Final Answer:

   SELECT employee_id, salary, RANK() OVER (ORDER BY salary DESC) AS salary_rank FROM employees; -> Option A

4. Quick Check:

   RANK() with OVER(ORDER BY) skips ranks after ties [OK]

Hint: Use RANK() OVER(ORDER BY) for ranks with gaps on ties [OK]

Common Mistakes:

• Using ROW_NUMBER() which never skips ranks
• Using DENSE_RANK() which does not skip ranks
• Missing parentheses in OVER clause